Caleb Harrison That's great! I'm really glad to hear that you're happy with your instructor! The more good teachers we have, the better educated future generations will be. Enjoy! P. Leonard
Caleb Harrison That's fantastic but if your instructor/professor is so great then why surf youtube for calculus videos? Just saying. These are invaluable resources for non-traditional students. Much luck in your Calculus career. Cheers.
Professor Leonard , I'm not sure if I got this right, but the graph you were working on from 28:00 and on..., isn't the concavity of that graph increasing from neg-infinity to approximately negative one? If the graph is going from left to right, shouldn't the slope be increasing from (-infnty, -1)?
Leonard, I can't tell you how grateful I am that you exist. let me tell you how much easy you have made my life, maths is my all times favorite subject now. I pray you live a long and happy life, so that you can continue making students' life easy all around the world. please keep posting more brilliant videos. I am struggling a bit in trig: I hope you read this and I'll really appreciate if you reply.
+Maha Agro Hi, thanks for the comment!! Videos are always in the works, and Calc 3 is being posted this semester. I have some other projects in the works, and I hope that Trig becomes a little more understandable for you. Good Luck!
Instead of music when practicing other stuff, I just listen/watch these. When I do come and give 100%, I already understand about 30-60% of the video. Keep being the Calculus 🐐
I feel extremely fortunate to have you for pre-algebra, trig, calc 1, calc 2, calc 3, and differential equations. Maybe linear algebra is coming?! Seriously though, thank you for doing SO much! You’re getting me through my statistics and comp sci degree.
Thanks to you I finished precal with a solid 98, a week ago I got a 98 on the first Cal1 exam. I took Algebra 10 years ago, got an A in it, then dropped out, so I was good but still had to relearn everything when I decided to go back to school. Prof. Leonard and Organic Chemistry Tutor for the win! If you're in a similar situation, go for it and use these resources daily. Here now before we go over this in class today ;)
Prof Leonard, you've got a gift for teaching. Thank you so much for helping us all get through math class! By the way, keep telling jokes because you've got a witty sense of humour.
Sir, you are not only raising the education standard of your own country but the world. I am enjoying Calculus here in India. Your youtube channel is probably the best place on the Internet. Thank you for the lectures.
Thank you so much!!!! Sometimes I need to be taught like I'm a 5 yr old...very simple, absorbent, and concise...I actually have hope of passing my online class.
Thank you Mr. Leonard. I credit your style of explaining Calculus for my new found confidence in applying the Product, Quotient, and Chain Rules, Differentiation by Substitution, Implicit Differentiation, and Related Rates; and now Concavity. Part of it for me, I think, is in the way you speak of the common mistakes you see and warn against doing them. Like when you've warned against distributing terms that are not distributable. With this concavity, there are times when the slope is decreasing in the mathematical sense while it is getting steeper in a graphical sense as at point c at 35:00 in this video. Where the graph is getting more steep in the negative direction the slope is decreasing. To the graphical mind this may be an initial stumbling block. Also your "follow the d/dx" has been very helpful to me. Thanks again.
I think an easy way to represent concavity is through velocity and acceleration. You may have a negative velocity but you can still be accelerating/decelerating and vice-versa.
You are the GOAT. The real MVP. So easy to understand when you explain it. Thank you professor, you saved me and a whole bunch of other students from struggling. I am now striving for an A in my calc math class.
I love maths but then there are most times i can't do things myself. We don't have proper classes (offline or online) in India which sucks honestly. But i love your explanation the most. I've been searching for an english free tutor for a long time, then with God's grace i found you sir Leonard. I am in grade 12 currently and this it so helpful. Thank you sir!❤
at 1:15:32 solution of x^(-2/3)=0 will not give x=0 as solution. In fact it will give 1/x^2=0 so x has to go to infinity for 1/x^2 to be 0. In fact at x=0 the slope is not defined. It is still a critical number
[1:20:45]. Example: Finding Absolute Extrema on a Closed Interval ●[1:21:10]. 𝒇(x) = 1 / (x² - x), on (0, 1) ○ can I check [0, 1]? No, I can only check the open interval. ○ [1:23:15]. We must take one-sided limits ■ Use sign analysis to determine the function's behavior near the asymptotes. ■ lim_ (x → 0⁺) 𝒇(x) = lim_ (x → 0⁺) [1 / (x² - x)] = -∞ ■ lim_ (x → 1⁻) 𝒇(x) = lim_(x → 1⁻) [1 / (x² - x)] = -∞ ■ In this case, the function tends to negative infinity at both endpoint points. ◇ No absolute minimum exists because the function tends to negative infinity. ○ [1:27:30]. First derivative. ○ [1:29:35]. To solve rational equations, set the numerator equal to zero, as the equation can only equal zero when the numerator is zero. ◇ Check if the solution is in the interval. ○ We don't need to check the endpoints because we already did the work by evaluating the limits at 0 and 1. ○ Since it is an open interval and the function is continuous, we know the solution will be an absolute maximum. [1:32:45] Summary of the Procedure to Find Absolute Extrema: ● If the interval is closed, evaluate the function at the critical numbers and at the interval's endpoints. ● If the interval is open, evaluate the function at the critical numbers and perform a sign analysis near the endpoints. ● Compare the values: The highest value corresponds to the absolute maximum, and the lowest value corresponds to the absolute minimum.
At 1:17:56, why is the function not undefined at -1? Wouldnt that give a root with a negative in it? My calculator even gives me a domain error when I put it in.
Leonard, Thank You for making these awesome videos. The way you teach is very effective and fun. Also, making these videos free was the best part, some people are at a financial disadvantaged, and it's good to see there are still people who are this smart and still like to help others without expecting something in return($$) . Honestly I have never had teacher that can teach as effectively as you do and never expect to. Hope the best for your future! Keep on doing GREAT things!!!
There is a break at 19:56 where he comes back and it is clear that some material has been skipped. There was no prior discussion about the 2nd derivative test, so this feels a little confusing... like we missed part of a lecture.
this is awesome!!!!! i can't believe calculus could be this interesting. thanks Sir. i'm going to tell my friends about your videos.... and yeah they will loveeeeee it!!!
[0:00]. Introduction: Increasing and Decreasing Intervals ●[0:47]. Definition of increasing and decreasing intervals. ○ Graphical Example: Identifying intervals where the function rises or falls when *reading the graph from left to right along x-axses*. ■ Increasing and decreasing intervals. ●[3:05]. Relationship between the slope and increasing/decreasing intervals. ○ Analysis of the slope at each point within increasing and decreasing intervals. ○ *An increasing interval has a positive slope and a decreasing interval has a negative slope*. ●[5:56]. Formalization of the concept of increasing/decreasing intervals using the derivative. ○ If 𝒇′(x) > 0 in an interval, then 𝒇(x) is increasing in that interval. ○ If 𝒇′(x) < 0 in an interval, then 𝒇(x) is decreasing in that interval. ○ If 𝒇′(x) = 0 at a point or interval, then 𝒇(x) is constant at that point or interval. ○ Sign analysis [9:39]. Concavity and Points of Inflection. ●[9:58]. Definition of concavity: The direction of the curve's curvature. ○ Examples of how concavity affects the shape of an increasing curve. ○ *Concavity describes how the slope of the function changes*. ○ [11:15]. Graph example: Two different directions of a curve's curvature in a increasing function. ■ 1. Increasing at an increasing rate ■ 2. Increasing at a decreasing rate ●[13:35]. Relationship between the second derivative and concavity. ○ The second derivative represents the rate of change of the slope. ○ Types of Concavity ■ Concave Up Description: The graph of the function curves upward, resembling a cup. Second Derivative: Positive (𝒇′′(x) > 0 Slope Behavior: The slope of the function is increasing. Example: 𝒇(x) = x² ■ Concave Down Description: The graph of the function curves downward, resembling a cap. Second Derivative: Negative (𝒇′′(x) < 0 ) Slope Behavior: The slope of the function is decreasing. Example: 𝒇(x) = −x² ○ [18:30]. Points of inflection: Points where the curve changes concavity. ■ Graphical Example of points of inflection and how concavity changes around them. ○ [20:00].The second derivative represents the rate of change of the slope. ■ If f″(x) > 0 in an interval, then 𝒇(x) is concave upward in that interval. ■ If f″(x) < 0 in an interval, then 𝒇(x) is concave downward in that interval. ■ If f″(x) = 0, it is a possible point of inflection, where concavity may change. [23:00]. Identification of Intervals and Points of Inflection on a Graph. ●[25:50]. Example: Identify increasing/decreasing intervals, concave up/down intervals, and points of inflection on a given graph. ○ Discussion on the importance of analyzing the function along the x-axis. ○ Identification of intervals based on the function's behavior. ○ Differentiation between intervals and specific points in mathematical notation. [32:58] Analyzing the Sign of the First and Second Derivatives at Specific Points ●[33:40]. Example: Determine the sign of 𝒇′(x) and f″(x) at specific points on a graph. ○ Explanation of the relationship between the sign of the first derivative and the function's increasing/decreasing behavior. ○ Explanation of the relationship between the sign of the second derivative and the function's concavity. [36:47]. Relative Extrema: Relative Maximums and Minimums. ●[37:09]. Definition of relative extrema: Relative maximums and minimums within an interval. ○ A relative maximum is a high point in an interval where the function changes from increasing to decreasing. ○ A relative minimum is a low point in an interval where the function changes from decreasing to increasing. ●[40:00]. Example: Graphical Example of relative maximums and minimums. ●[41:40]. The slope at relative extrema is zero. ○[42:03]. This means that the tangent line to the curve at that point is horizontal. [42:22]. Critical Numbers and Their Relationship with Relative Extrema. ●[42:37]. Definition of a critical number: A point where the function's slope is zero or undefined. ○[43:32]. Critical numbers are candidates for being relative maximums or minimums. ●[43:51]. Not all critical numbers are relative maximums or minimums. ○[43:58]. Example: 𝒇(x) = x³; Graph example of a critical number that does not correspond to a relative extremum. ●[44:18]. Procedure to find critical numbers. ○ Calculate the function's derivative and find the x-values where the derivative is zero or undefined. ○ Importance of analyzing points where the derivative is undefined. ■ If the derivative is a fraction, analyze both the numerator and the denominator. ■ [46:19]. Also, *examine the denominator* because it could introduce undefined points for the slope, which is significant. These undefined points can cause the function to change from increasing to decreasing or vice versa. [47:03] Example: Finding Critical Numbers ●[47:23]. Exercise: 𝒇(x) = x³ - 3x + 1; Find the critical numbers of a polynomial function. ○ Calculation of the function's derivative. ○ Set the derivative equal to zero and solve for x. ■ Identification of the *potential* critical numbers. △ [49:10]. Critical numbers do not guarantee relative maximums or minimums. ► Critical points can “mislead” us in the following cases: - Inflection points: As in 𝒇(x) = x³. - Nonexistent derivatives: As in 𝒇(x) = |x|. - Repetitive oscillations: As in 𝒇(x) = sin(x). - Piece wise functions: Where the derivative changes abruptly. [49:44] Absolute Extrema: Absolute Maximums and Minimums ●[50:39]. Definition of absolute extrema: The highest or lowest point of the function within a given interval. ○ The absolute maximum is the *highest* value the function reaches in the interval. ○ The absolute minimum is the *lowest* value the function reaches in the interval. ●[51:44]. Not all functions have absolute maximums or minimums in an *infinite interval* (-∞, ∞). ○[52:35]. Example of function that do not have absolute maximums or minimums in an infinite interval. ○[53:10]. Example of function that have absolute minimum in an infinite interval. ●[54:12]. In a *closed interval*, a continuous function always has an absolute maximum and minimum. ○ Imagine a hallway with two closed doors at the ends. If you walk through the hallway without jumping, at some point you’ll reach the highest point and at another the lowest point. Since you can’t leave the hallway, there will always be a maximum and a minimum ○[55:37]. The importance of the function's continuity in the closed interval. [56:15]. Location of Absolute Extrema. ●[56:36]. Absolute extrema are found at critical numbers or the endpoints of the closed interval. ○ Reasoning why absolute extrema are found at these points: they represent the function's upper and lower limits within the interval. ●[59:00]. From close interval to open interval ●[1:00:55]. In an *open interval*, absolute extrema can only occur at critical numbers. ○ If absolute extrema are not found at critical numbers, they do not exist in the open interval. ●[1:03:00]. Example of how including or excluding endpoint points affects the existence of absolute maximums and minimums. [1:04:50]. Procedure to Find Absolute Extrema. ● ⑴ Find the critical numbers of the function. Evaluate the function at the critical numbers. ● ⑵ Evaluate the function at the interval's endpoints. [1:05:31] Example: Finding Absolute Extrema on a Closed Interval ●[1:05:37]. Exercise:f(x) = 2x³ - 15x² + 36x on [1, 5] ○ Calculation of the function's derivative. ○ Find the critical numbers by setting the derivative equal to zero. ○ Evaluate the function at the critical numbers and the interval's endpoints. ○ Identification of the absolute maximum and absolute minimum. ○ [1:10:00]. f(x) = 2x³ - 15x² + 36x on [1, 5) ○ [1:10:27]. f(x) = 2x³ - 15x² + 36x on (1, 5) [1:10:42] Example: Finding Absolute Extrema on a Closed Interval ●[1:10:49]. y = 6x^(4/3) - 3x^(1/3) on [-1, 1] ○ Calculation of the function's derivative. ○ Find the critical numbers by setting the derivative equal to zero. ○ [1:16:14]. Evaluate the function at the critical numbers amd at endpoints. ○ [1:20:07]. y = 6x^(4/3) - 3x^(1/3) on [-1, 1) ○ [1:20:27]. y = 6x^(4/3) - 3x^(1/3) on (-1, 1]
Calc exam tomorrow, and I'm in the middle of an amphetamine powered study extravaganza that'll last at least 10 hours. BLESS THIS MAN FOR MAKING THESE VIDEOS. These are the damage control for my poor decision making : ' ]
Sir ur way of teaching is fabulous..but as I m learning calculus for the first time..so plz add subtitles on all ur videos for calculus...I will very thankful for this.
Hey Mr. Professor Leonard. Thanks a lot for your lecture, i watch this series really carefully. But i think on the 1:15:40, at x=0 this function is not differentiable. Because x^-2/3--------> 1/x^2/3. Denominator cannot be 0 :)
Respected Professor Leonard, I am very grateful to you for sharing such type of great knowledge. I hope you will must go ahead to keep help of younger. But I am confused on 47:00 to "If you have a denom, set denom=0 as well" What does it mean. I am not a native speaker may be it is due to language difference. If you please could not mind and can explain me I will be very thankful to you.
Does anyone have a link to the type of sign analysis test that's done round 1:25:30? I've done sign analysis for inequalities but I can't see how plugging in two random "test" x-values close to the asymptotes necessitates that the sign of the entire interval is the same as the output of the test values . . .
Wait, at 1:15:32 he says that x^(-2/3) is 0 at x=0 but isn't it actually undefined? Also in general, what about something like an absolute value "curve" that's also continuous but has an undefined slope at x=0? The slope is undefined, not zero, but isn't that also a critical point since we could have a min or a max where the slope is undefined?
It is a point where something interesting happens, because the original function has a vertical slope at this point. It is interesting, but frustrating, behavior of the function.
I'm very confused about the extrema on the endpoints. At 40:14, he says that's not a relative extremum, but in my calc class, we learned that that would be a relative min. Can someone explain this to me, please?
It depends on what application you have in mind. Yes, the end points can be extreme points of the function that can solve an optimization problem. They usually aren't the cases you are most likely interested in finding, but they are local extreme points of the function over the domain. Take the function f(x) = sin(x)^2 + cos(x), over the domain of x=0 to pi/2. Why the domain restriction? Maybe this is applied to a situation where it only makes sense to have angles from 0 to 90 degrees. Suppose you wanted the minimum value of this function. You'd need to enumerate all points with a derivative equal to zero, and the two end points, as candidates for the minimum value. They are as follows: x=0, x=pi/3 and x=pi/2. Both events happen at x=0, a local minimum with a slope equal to zero and the endpoint of the interval. At x=pi/3, we have a local maximum. At x=pi/2, we don't have a stationary point or a local extreme point if you let the function continue, since the derivative isn't anything close to zero. But we do have the minimum possible value of the function over this domain, and if we are interested in the function's minimum value, this would be your answer.
@@AliSeyed I think you're right, instead of factoring I turned the exponents into fractions with surds then added the fractions, now equating to zero, only the numerator would lead to zero, even on the calculator 0^(-2/3) = Math Error. I am not sure how the professor concluded x could equal zero. However, I also took the denominator into consideration to determine where the slope is undefined (equated it to zero). And THAT's where x=0.
@36:45 I get what you're saying and I see and understand why but I feel like without a definite inflexion point then there is ambiguity in when starts concave up or down
Around that part, I thought (-infinity,0) were indicating the coordinate, until he wrote inflection point x =1. Then I know it was increasing from -infinity to 0, but not the function at the point (-infinity,0) was increasing.
wow.... I just came up with a genius idea... the double like button... only the God's can receive such an applause. how freakin awesome is it that an exponential function is constantly climbing? like its a U shape and even when it goes down its really going up bc its just less negative less negative less negative, pause (Nate dogs voice) HOLD UP, positive more positive more positive... poopie caca I say gosh... its not difficult to think of its just very odd its like trying to remember a dream its cloudy and its like wait the truth is right in front of me I just need to clear more of this gases mist smoke in the way
Probably the best calculus instructor in the world
Caleb Harrison
That's great! I'm really glad to hear that you're happy with your instructor! The more good teachers we have, the better educated future generations will be. Enjoy!
P. Leonard
Professor Leonard
I honestly don't think there could be anyone better than this guy P. Len
Caleb Harrison That's fantastic but if your instructor/professor is so great then why surf youtube for calculus videos? Just saying. These are invaluable resources for non-traditional students. Much luck in your Calculus career. Cheers.
Professor Leonard Your a gangster come get a job in colorado.
Professor Leonard , I'm not sure if I got this right, but the graph you were working on from 28:00 and on..., isn't the concavity of that graph increasing from neg-infinity to approximately negative one? If the graph is going from left to right, shouldn't the slope be increasing from (-infnty, -1)?
I'm 43 and taking Calculus for the first time in my life, this guy is helping me out so much
I can’t believe I’m watching this for free. Goodness, I’m so grateful for having access to such tutoring lessons.
Leonard, I can't tell you how grateful I am that you exist. let me tell you how much easy you have made my life, maths is my all times favorite subject now. I pray you live a long and happy life, so that you can continue making students' life easy all around the world. please keep posting more brilliant videos. I am struggling a bit in trig: I hope you read this and I'll really appreciate if you reply.
+Maha Agro Hi, thanks for the comment!! Videos are always in the works, and Calc 3 is being posted this semester. I have some other projects in the works, and I hope that Trig becomes a little more understandable for you. Good Luck!
thanks very much Leonard :)
Instead of music when practicing other stuff, I just listen/watch these. When I do come and give 100%, I already understand about 30-60% of the video. Keep being the Calculus 🐐
I was sick and subsequently missed over a month of class. I was still able to make an 87 on my last exam because of your lectures. Thank you so much!
I feel extremely fortunate to have you for pre-algebra, trig, calc 1, calc 2, calc 3, and differential equations. Maybe linear algebra is coming?!
Seriously though, thank you for doing SO much! You’re getting me through my statistics and comp sci degree.
Thanks to you I finished precal with a solid 98, a week ago I got a 98 on the first Cal1 exam. I took Algebra 10 years ago, got an A in it, then dropped out, so I was good but still had to relearn everything when I decided to go back to school. Prof. Leonard and Organic Chemistry Tutor for the win! If you're in a similar situation, go for it and use these resources daily. Here now before we go over this in class today ;)
Leonardo ,a person with both brain and muscles
All the people
brains and gains
Prof Leonard, you've got a gift for teaching. Thank you so much for helping us all get through math class! By the way, keep telling jokes because you've got a witty sense of humour.
So glad this dude decided to upload all these. I think I am going to go support his patreon bc he is a lifesaver.
did u evr donate
Sir, you are not only raising the education standard of your own country but the world. I am enjoying Calculus here in India. Your youtube channel is probably the best place on the Internet. Thank you for the lectures.
You have great conceptual clarity and thats the way maths should be taught..
Thanks ..Respect (from India)
Thank you so much!!!! Sometimes I need to be taught like I'm a 5 yr old...very simple, absorbent, and concise...I actually have hope of passing my online class.
Professor, you are awesome, your level of enthusiasm and your level of energy for teaching calculus tends to infinity. limit -> infinity. Love you.
Relative Extrema 37:01 Absolute extrema 50:00
Some people where born to impact others Sir, this is yyour calling and you are doing excellently well!
Thank you Mr. Leonard.
I credit your style of explaining Calculus for my new found confidence in applying the Product, Quotient, and Chain Rules, Differentiation by Substitution, Implicit Differentiation, and Related Rates; and now Concavity. Part of it for me, I think, is in the way you speak of the common mistakes you see and warn against doing them. Like when you've warned against distributing terms that are not distributable. With this concavity, there are times when the slope is decreasing in the mathematical sense while it is getting steeper in a graphical sense as at point c at 35:00 in this video. Where the graph is getting more steep in the negative direction the slope is decreasing. To the graphical mind this may be an initial stumbling block. Also your "follow the d/dx" has been very helpful to me. Thanks again.
Your vids never fail to make me concave up 😊
BEST Instructor I have ever seen...
I think an easy way to represent concavity is through velocity and acceleration. You may have a negative velocity but you can still be accelerating/decelerating and vice-versa.
I am having straights A's in Calculus. Thank you sir!
You are the GOAT. The real MVP. So easy to understand when you explain it. Thank you professor, you saved me and a whole bunch of other students from struggling. I am now striving for an A in my calc math class.
This is an AWESOME video for going over these concepts. Easily confused! You made it so easy to follow! Love it!
I love maths but then there are most times i can't do things myself. We don't have proper classes (offline or online) in India which sucks honestly. But i love your explanation the most. I've been searching for an english free tutor for a long time, then with God's grace i found you sir Leonard. I am in grade 12 currently and this it so helpful. Thank you sir!❤
Omg Iam a freshman now and u saved my life thank you so much.
1:19:41 for the x^{-2/3}, isn't x can't be 0 because if it's 0, then it will become 1/0 because of negative exponent?
Yes, I noticed that as well.
yes professor made a mistake. It happens. x^-2/3 is undefined so x=1/8 is the only critical point
You are already doing more work than my actual professor. You may be the only reason I survive this semester.
Best lecturer ever ❤what i failed to learn in 9 weeks learnt it in Just a week long live professor leonard
Omg,such a great teacher... Thanks a lot prof.Respect for you sir #from 🇮🇳
Thank you for all of these, from the bottom of my little black heart.
I just can't emphasis how thankful I am.
at 1:15:32 solution of x^(-2/3)=0 will not give x=0 as solution. In fact it will give 1/x^2=0 so x has to go to infinity for 1/x^2 to be 0. In fact at x=0 the slope is not defined. It is still a critical number
what is X^2=0? Or even X^3=0
@@elizabethatienootieno3083
In both those cases, x = 0.
was looking for this comment thanks!
Mr. Leonard, you're one in a million. I'm getting A in calculus 1 this semester because of you. I have to go for my fees and bring it to you
You are a hero!! The best teacher ever!
I'll be starting chapter 3 of Calc1 this semester, and this thing is a huge help for me! Thank you prof leonard!
I want the autograph of the student who sneezes every video
lol and he never acknowledges the kid either it is so funny
AirIndy5000 I
when did he sneeze
@@ivantasticforever 24:00
And cough too
I love when I catch myself answering his questions when I watch these vids at the library.
i swear to god, you make everything seem so easy when you teach it
[1:20:45]. Example: Finding Absolute Extrema on a Closed Interval
●[1:21:10]. 𝒇(x) = 1 / (x² - x), on (0, 1)
○ can I check [0, 1]? No, I can only check the open interval.
○ [1:23:15]. We must take one-sided limits
■ Use sign analysis to determine the function's behavior near the asymptotes.
■ lim_ (x → 0⁺) 𝒇(x) = lim_ (x → 0⁺) [1 / (x² - x)] = -∞
■ lim_ (x → 1⁻) 𝒇(x) = lim_(x → 1⁻) [1 / (x² - x)] = -∞
■ In this case, the function tends to negative infinity at both endpoint points.
◇ No absolute minimum exists because the function tends to negative infinity.
○ [1:27:30]. First derivative.
○ [1:29:35]. To solve rational equations, set the numerator equal to zero, as the equation can only
equal zero when the numerator is zero.
◇ Check if the solution is in the interval.
○ We don't need to check the endpoints because we already did the work by evaluating the limits at 0 and 1.
○ Since it is an open interval and the function is continuous, we know the solution will be an absolute maximum.
[1:32:45] Summary of the Procedure to Find Absolute Extrema:
● If the interval is closed, evaluate the function at the critical numbers and at the interval's endpoints.
● If the interval is open, evaluate the function at the critical numbers and perform a sign analysis near the endpoints.
● Compare the values: The highest value corresponds to the absolute maximum, and the lowest value corresponds to the absolute minimum.
At 1:17:56, why is the function not undefined at -1? Wouldnt that give a root with a negative in it? My calculator even gives me a domain error when I put it in.
Leonard, Thank You for making these awesome videos. The way you teach is very effective and fun. Also, making these videos free was the best part, some people are at a financial disadvantaged, and it's good to see there are still people who are this smart and still like to help others without expecting something in return($$) . Honestly I have never had teacher that can teach as effectively as you do and never expect to. Hope the best for your future! Keep on doing GREAT things!!!
woaw i can't believe such a thing is here!!!!!!....good work. I now find my studies so easy, keep it up.
I hope linear algebra will be posted soon.
your expertise is not only math, but an outstanding math educator as well. Thanks for sharing your knowledge.
Thank you so much! I finally learned this after 2 months of struggling
hey professor... i loved your videos a lot... thanks for all these stuff.. lots of love from India.
One of the best PRO Professors i have ever seen 👏👏👏👏
one of my favourit teacher
There is a break at 19:56 where he comes back and it is clear that some material has been skipped. There was no prior discussion about the 2nd derivative test, so this feels a little confusing... like we missed part of a lecture.
29:30 with comeback of the year. Good job teach
this is awesome!!!!! i can't believe calculus could be this interesting. thanks Sir. i'm going to tell my friends about your videos.... and yeah they will loveeeeee it!!!
[0:00]. Introduction: Increasing and Decreasing Intervals
●[0:47]. Definition of increasing and decreasing intervals.
○ Graphical Example: Identifying intervals where the function rises or falls when *reading the
graph from left to right along x-axses*.
■ Increasing and decreasing intervals.
●[3:05]. Relationship between the slope and increasing/decreasing intervals.
○ Analysis of the slope at each point within increasing and decreasing intervals.
○ *An increasing interval has a positive slope and a decreasing interval has a negative slope*.
●[5:56]. Formalization of the concept of increasing/decreasing intervals using the derivative.
○ If 𝒇′(x) > 0 in an interval, then 𝒇(x) is increasing in that interval.
○ If 𝒇′(x) < 0 in an interval, then 𝒇(x) is decreasing in that interval.
○ If 𝒇′(x) = 0 at a point or interval, then 𝒇(x) is constant at that point or interval.
○ Sign analysis
[9:39]. Concavity and Points of Inflection.
●[9:58]. Definition of concavity: The direction of the curve's curvature.
○ Examples of how concavity affects the shape of an increasing curve.
○ *Concavity describes how the slope of the function changes*.
○ [11:15]. Graph example: Two different directions of a curve's curvature in a increasing function.
■ 1. Increasing at an increasing rate
■ 2. Increasing at a decreasing rate
●[13:35]. Relationship between the second derivative and concavity.
○ The second derivative represents the rate of change of the slope.
○ Types of Concavity
■ Concave Up
Description: The graph of the function curves upward, resembling a cup.
Second Derivative: Positive (𝒇′′(x) > 0
Slope Behavior: The slope of the function is increasing.
Example: 𝒇(x) = x²
■ Concave Down
Description: The graph of the function curves downward, resembling a cap.
Second Derivative: Negative (𝒇′′(x) < 0 )
Slope Behavior: The slope of the function is decreasing.
Example: 𝒇(x) = −x²
○ [18:30]. Points of inflection: Points where the curve changes concavity.
■ Graphical Example of points of inflection and how concavity changes around them.
○ [20:00].The second derivative represents the rate of change of the slope.
■ If f″(x) > 0 in an interval, then 𝒇(x) is concave upward in that interval.
■ If f″(x) < 0 in an interval, then 𝒇(x) is concave downward in that interval.
■ If f″(x) = 0, it is a possible point of inflection, where concavity may change.
[23:00]. Identification of Intervals and Points of Inflection on a Graph.
●[25:50]. Example: Identify increasing/decreasing intervals, concave up/down intervals,
and points of inflection on a given graph.
○ Discussion on the importance of analyzing the function along the x-axis.
○ Identification of intervals based on the function's behavior.
○ Differentiation between intervals and specific points in mathematical notation.
[32:58] Analyzing the Sign of the First and Second Derivatives at Specific Points
●[33:40]. Example: Determine the sign of 𝒇′(x) and f″(x) at specific points on a graph.
○ Explanation of the relationship between the sign of the first derivative and the
function's increasing/decreasing behavior.
○ Explanation of the relationship between the sign of the second derivative and the function's concavity.
[36:47]. Relative Extrema: Relative Maximums and Minimums.
●[37:09]. Definition of relative extrema: Relative maximums and minimums within an interval.
○ A relative maximum is a high point in an interval where the function changes from increasing to decreasing.
○ A relative minimum is a low point in an interval where the function changes from decreasing to increasing.
●[40:00]. Example: Graphical Example of relative maximums and minimums.
●[41:40]. The slope at relative extrema is zero.
○[42:03]. This means that the tangent line to the curve at that point is horizontal.
[42:22]. Critical Numbers and Their Relationship with Relative Extrema.
●[42:37]. Definition of a critical number: A point where the function's slope is zero or undefined.
○[43:32]. Critical numbers are candidates for being relative maximums or minimums.
●[43:51]. Not all critical numbers are relative maximums or minimums.
○[43:58]. Example: 𝒇(x) = x³; Graph example of a critical number that does not correspond to a relative extremum.
●[44:18]. Procedure to find critical numbers.
○ Calculate the function's derivative and find the x-values where the derivative is zero or undefined.
○ Importance of analyzing points where the derivative is undefined.
■ If the derivative is a fraction, analyze both the numerator and the denominator.
■ [46:19]. Also, *examine the denominator* because it could introduce undefined points for the slope, which is significant.
These undefined points can cause the function to change from increasing to decreasing or vice versa.
[47:03] Example: Finding Critical Numbers
●[47:23]. Exercise: 𝒇(x) = x³ - 3x + 1; Find the critical numbers of a polynomial function.
○ Calculation of the function's derivative.
○ Set the derivative equal to zero and solve for x.
■ Identification of the *potential* critical numbers.
△ [49:10]. Critical numbers do not guarantee relative maximums or minimums.
► Critical points can “mislead” us in the following cases:
- Inflection points: As in 𝒇(x) = x³.
- Nonexistent derivatives: As in 𝒇(x) = |x|.
- Repetitive oscillations: As in 𝒇(x) = sin(x).
- Piece wise functions: Where the derivative changes abruptly.
[49:44] Absolute Extrema: Absolute Maximums and Minimums
●[50:39]. Definition of absolute extrema: The highest or lowest point of the function within a given interval.
○ The absolute maximum is the *highest* value the function reaches in the interval.
○ The absolute minimum is the *lowest* value the function reaches in the interval.
●[51:44]. Not all functions have absolute maximums or minimums in an *infinite interval* (-∞, ∞).
○[52:35]. Example of function that do not have absolute maximums or minimums in an infinite interval.
○[53:10]. Example of function that have absolute minimum in an infinite interval.
●[54:12]. In a *closed interval*, a continuous function always has an absolute maximum and minimum.
○ Imagine a hallway with two closed doors at the ends. If you walk through the hallway without jumping,
at some point you’ll reach the highest point and at another the lowest point. Since you can’t leave the hallway,
there will always be a maximum and a minimum
○[55:37]. The importance of the function's continuity in the closed interval.
[56:15]. Location of Absolute Extrema.
●[56:36]. Absolute extrema are found at critical numbers or the endpoints of the closed interval.
○ Reasoning why absolute extrema are found at these points: they represent the function's upper
and lower limits within the interval.
●[59:00]. From close interval to open interval
●[1:00:55]. In an *open interval*, absolute extrema can only occur at critical numbers.
○ If absolute extrema are not found at critical numbers, they do not exist in the open interval.
●[1:03:00]. Example of how including or excluding endpoint points affects the existence of absolute maximums and minimums.
[1:04:50]. Procedure to Find Absolute Extrema.
● ⑴ Find the critical numbers of the function. Evaluate the function at the critical numbers.
● ⑵ Evaluate the function at the interval's endpoints.
[1:05:31] Example: Finding Absolute Extrema on a Closed Interval
●[1:05:37]. Exercise:f(x) = 2x³ - 15x² + 36x on [1, 5]
○ Calculation of the function's derivative.
○ Find the critical numbers by setting the derivative equal to zero.
○ Evaluate the function at the critical numbers and the interval's endpoints.
○ Identification of the absolute maximum and absolute minimum.
○ [1:10:00]. f(x) = 2x³ - 15x² + 36x on [1, 5)
○ [1:10:27]. f(x) = 2x³ - 15x² + 36x on (1, 5)
[1:10:42] Example: Finding Absolute Extrema on a Closed Interval
●[1:10:49]. y = 6x^(4/3) - 3x^(1/3) on [-1, 1]
○ Calculation of the function's derivative.
○ Find the critical numbers by setting the derivative equal to zero.
○ [1:16:14]. Evaluate the function at the critical numbers amd at endpoints.
○ [1:20:07]. y = 6x^(4/3) - 3x^(1/3) on [-1, 1)
○ [1:20:27]. y = 6x^(4/3) - 3x^(1/3) on (-1, 1]
At 1:15:27 how do we equate x to the negative power to 0. Isnt it undefined?
calculus tastes something else with prof. Leonard ❤
This man is a hero !
Prof Leonard, I just want to say thank you.
Best professor ever
Excellent lecture, totally saved my bacon!
You are the best lecturer in the world
very batter lecture I learn first time in my life, thanks sir ,i see your lecture form Bangladesh.
The greatest prof ever.
Calc exam tomorrow, and I'm in the middle of an amphetamine powered study extravaganza that'll last at least 10 hours. BLESS THIS MAN FOR MAKING THESE VIDEOS. These are the damage control for my poor decision making : ' ]
lol
did u pass
i think u failed
u failed
@@jensmalzer6344 it’s been 3 years dawg are you good
you're the best, professor Leonard
I had to go one chapter without you. Had to put in extra work. (Tho I found some of your stuff in calc 2 lol) but glad to have this back. Yay !
Dude you are SO GOOD at explaining things!!!
Sir ur way of teaching is fabulous..but as I m learning calculus for the first time..so plz add subtitles on all ur videos for calculus...I will very thankful for this.
Hey Mr. Professor Leonard. Thanks a lot for your lecture, i watch this series really carefully. But i think on the 1:15:40, at x=0 this function is not differentiable. Because x^-2/3--------> 1/x^2/3. Denominator cannot be 0 :)
Yes you are correct.
ive got a calc test next week, and while my professor is good, the test will be really hard. thanks for helping me prep for it.
way more helpful than my calculus prof :)
Really great, Thanks
RUclips is my collage
Ali Kasem to bad they don't have English lessons
"to bad" the irony lol
@@andresramirez5306 lmfaooo comedy
@@andresramirez5306 I'm dead 💀
@@javierd2403They do, he just haven't taken it
Respected Professor Leonard, I am very grateful to you for sharing such type of great knowledge. I hope you will must go ahead to keep help of younger. But I am confused on 47:00 to "If you have a denom, set denom=0 as well" What does it mean. I am not a native speaker may be it is due to language difference. If you please could not mind and can explain me I will be very thankful to you.
He means DENOMINATOR of a fraction !!! If you are working on a problem and you have a fraction, you need to set the denominator to zero !!!
I was pretty confused as well. He never gave an example of a function with a denominator, so I'm not entirely sure what he's talking about.
Thank you. Your videos are very helpful
professor did a mistake at 1:27:50 he wrote (x^2-1)^-1 but it should be (x^2-x)^-1
Samsher Singh I was looking for this comment, thank you
That's what you would get after you've taken the derivative. He didn't mess up
@@bobjoe164 but he didn't take derivative yet.... but yea he wrote it correctly in the next step
Which foul creatures dislike the video.
Probably some other math professors
😂😂😂😂😂😂
@@dollieelizabethjames343 dey jelly
@Jose Jimenez LOL!
In the future, I want to be like you Professor Leonard.
Does anyone have a link to the type of sign analysis test that's done round 1:25:30? I've done sign analysis for inequalities but I can't see how plugging in two random "test" x-values close to the asymptotes necessitates that the sign of the entire interval is the same as the output of the test values . . .
Wait, at 1:15:32 he says that x^(-2/3) is 0 at x=0 but isn't it actually undefined? Also in general, what about something like an absolute value "curve" that's also continuous but has an undefined slope at x=0? The slope is undefined, not zero, but isn't that also a critical point since we could have a min or a max where the slope is undefined?
Yes, he made a mistake. Best way to do it is to replace x^(1/3) with variable like 'z' and solve for z, then solve for x.
Care to explain? I'm a noob
Yes undefined, as it 0^(-2/3) is 1/(0^(-2/3)) or 1/0
East or west professor Leonard is the best!!
"You could be straight, but that's not fun." -Professor Leonard 2014
Thank you so muchhhh :< I did not even notice the video was 8 years old already
I might be in love with you Professor. Thank you for making this a hell lot easier!
i don't think the problem he is doing at 1:21:49 has an absolute maximum. the origin of the functions is at 0
Thank you so so much for these videos!! You rock!
x = 0 cannot be a critical point. x^(-2/3) = 0 does not yield any solutions. Other than that, this was a wonderful lecture. Thank you!
It is a point where something interesting happens, because the original function has a vertical slope at this point. It is interesting, but frustrating, behavior of the function.
I'm very confused about the extrema on the endpoints. At 40:14, he says that's not a relative extremum, but in my calc class, we learned that that would be a relative min. Can someone explain this to me, please?
It depends on what application you have in mind. Yes, the end points can be extreme points of the function that can solve an optimization problem. They usually aren't the cases you are most likely interested in finding, but they are local extreme points of the function over the domain.
Take the function f(x) = sin(x)^2 + cos(x), over the domain of x=0 to pi/2. Why the domain restriction? Maybe this is applied to a situation where it only makes sense to have angles from 0 to 90 degrees.
Suppose you wanted the minimum value of this function. You'd need to enumerate all points with a derivative equal to zero, and the two end points, as candidates for the minimum value. They are as follows: x=0, x=pi/3 and x=pi/2. Both events happen at x=0, a local minimum with a slope equal to zero and the endpoint of the interval. At x=pi/3, we have a local maximum.
At x=pi/2, we don't have a stationary point or a local extreme point if you let the function continue, since the derivative isn't anything close to zero. But we do have the minimum possible value of the function over this domain, and if we are interested in the function's minimum value, this would be your answer.
At 1:17:54, when x=-1,shouldn't the y=-3 instead of y=9? Thus, giving you different Absolute Max. and Min.?
it implies, for ex, (-2)squared=4 and -2squared=-4. So try using parenthesis in your calculator
i was wondering the same thing. it's (-1)^4/3 it's not (-1)^2 so that it changes signs to +1. is 4/3 somehow related to squared?
You're as good a teacher as Mitch McDuff
Wow... I am Concaved up 🙂
He saying "THE IDEA" gives me goosebumps
I couldn't understand how this x^(-2/3)=0 has an answer zero at 1:15:25 ? Because x goes to denominator, how come 1/sth be equal to 0?
@Professor Leonard
@@AliSeyed I think you're right, instead of factoring I turned the exponents into fractions with surds then added the fractions, now equating to zero, only the numerator would lead to zero, even on the calculator 0^(-2/3) = Math Error. I am not sure how the professor concluded x could equal zero. However, I also took the denominator into consideration to determine where the slope is undefined (equated it to zero). And THAT's where x=0.
2021 math and we have leonard professor yay!
Thank you so much Professor Leonard, your lectures are a blessing!
Thank you for your videos. They have been extremely helpful.
@36:45 I get what you're saying and I see and understand why but I feel like without a definite inflexion point then there is ambiguity in when starts concave up or down
Around that part, I thought (-infinity,0) were indicating the coordinate, until he wrote inflection point x =1. Then I know it was increasing from -infinity to 0, but not the function at the point (-infinity,0) was increasing.
The fact that I am watching this in my calculus class😌
wow.... I just came up with a genius idea... the double like button... only the God's can receive such an applause. how freakin awesome is it that an exponential function is constantly climbing? like its a U shape and even when it goes down its really going up bc its just less negative less negative less negative, pause (Nate dogs voice) HOLD UP, positive more positive more positive... poopie caca I say gosh... its not difficult to think of its just very odd its like trying to remember a dream its cloudy and its like wait the truth is right in front of me I just need to clear more of this gases mist smoke in the way
2024, and your saving lives.....my hero
Professor Leonard, I am curious to know just how to cube root in my head (LIKE YOU DID) without a calculator? (Serious inquiry)