Relative Extrema, Local Maximum and Minimum, First Derivative Test, Critical Points- Calculus

This calculus video tutorial explains how to find the relative extrema of a function such as the local maximum and minimum values using the first derivative test.
Introduction to Limits: • Calculus 1 - Introduct...
Derivatives - Fast Review:
• Calculus 1 - Derivatives
Introduction to Related Rates:
• Introduction to Relate...
_____________________________
Extreme Value Theorem:
• Extreme Value Theorem
Finding Critical Numbers:
• Finding Critical Numbers
Local Maximum & Minimum:
• Finding Local Maximum ...
Absolute Extrema:
• Finding Absolute Maxim...
Rolle's Theorem:
• Rolle's Theorem
________________________________
Mean Value Theorem:
• Mean Value Theorem
Increasing and Decreasing Functions:
• Increasing and Decreas...
First Derivative Test:
• First Derivative Test
Concavity & Inflection Points:
• Concavity, Inflection ...
Second Derivative Test:
• Second Derivative Test
_________________________________
L'Hopital's Rule:
• L'hopital's rule
Curve Sketching With Derivatives:
• Curve Sketching - Grap...
Newton's Method:
• Newton's Method
Optimization Problems:
• Optimization Problems ...
_______________________________________
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www.video-tuto...
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[@shiro-om3ny]

If you need to find out if it is a max or min, use this: f’’(x)>0 = min or f’’(x)

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[@99ish91]

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[@ts37924]

if we get a f"(x) = 0 is there still a possibility for x be to be a maxima or a minima if f'(x)=0
is the change in slope (i.e the second derevitive) of a * maxima * or a * minima * must always be * negetive dec. * or
* positive inc. * respectively or can it be * zero * at any of them aswell?

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[@judyhopps1324]

there is a mistake at 10:31. (x-1)^2 has even multiplicity, the sign should be + in the number line

[@hawamusah-opata9799]

that's true cuz i got confused over there

[@kk7420]

@@hawamusah-opata9799 wouldn't it be "-"? If the power is even, the sign stays the same(like he mentioned in the video). When x is greater then 2, it's negative. Therefore when x is lesser than 2, it's still negative.

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9:40 the easiest thing to do is plug in 0. that way, you can cross out all the 'x's and focus only on the signs of the constants. in this case, you have 3 negatives, so you'll get a negative answer after plugging in 0. so the sign between -2 and 1 is '-'. now you can use the multiplicity trick.

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1:06 here lies the issue with my highschool teacher ;D,
they explain very well everything graphically with simple functions, and the calculations are easy, then we go to functions with critical points that are not extremas, and not defined functions and all that sht and they are like, ey analogically from what we've discussed, like i can imagine graphs as easy as a person who worked this field for 40 years;
it is way easier to remember when you study from raw methods, you cant force intiution in student's brain, it is something that develops over time

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[@zef1o338]

Take note that another way to find whether the point is a maximum or minimum is taking the second derivative and plugging the x value. If the result is x>0 then it is a minimum and if x

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Can you find the max and min values of a function with an imaginary critical value? I'm trying to solve a homework problrm but the critical value is the square root of -1 and I'm stuck.

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Hi shouldn’t the value be positive at 1, (x-1)^2 12:00 ???

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