Curve Sketching - First & Second Derivatives - Graphing Rational Functions & Asymptotes - Calculus

This calculus video tutorial provides a summary of the techniques of curve sketching. It shows you how to graph polynomials, rational functions with horizontal & vertical asymptotes and square roots using the first derivative and the second derivative of the functions.
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 ...
_______________________________________
Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
/ collections

[@TheOrganicChemistryTutor]

Final Exams and Video Playlists: www.video-tutor.net/

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5:19 I think you meant decreasing at an increasing rate (f' = - and f'' = +)

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No he didn't because it isn't decreasing more its decreasing less

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why did we not use chain rule on that second derivative for the 6/(x+3)^2

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5:19 decreasing at INCREASING rate?

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[@tabuchihara]

When f'(x) = - and f''(x) = - : it is decreasing at an increasing rate
When f'(x) = - and f''(x) = + : it is decreasing at a decreasing rate
I got confused in the beginning too thats what he meant though

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[@folu-ke]

At 5:18 in the video, I think you made a mistake about the function "decreasing at a decreasing rate" when the first derivative is negative but the second derivative is positive. Shouldn't it be decreasing at an increasing rate?

[@asinner]

Yeah, I caught that too. Pretty sure he meant to say at an increasing rate.

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@@asinner no, he spoke correctly. You can see that the function becomes more horizonal as it continues, which means it is decreasing slower

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he is correct there, i think he is wrong at f'(x)= - f''(x)= -. the curve curves down at a higher rate than it started with.

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Wouldn’t you use the chain rule for the derivative of 6(x+3)^-2?

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HE MADE A MISTAKE THERE,U ARE ABSOLUTELY CORRECT,A CHAIN RULE SHOULD HAVE BEEN USED

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the derivative of x+3 is 1 so dont worry

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17:30 isn't the vertical asymptote -3??

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[@carmenmoen170]

So even if the second derivative is a plus and the first derivative is minus (3rd quadrant) the graph is decreasing at a decreasing rate instead of decreasing at an increasing rate?

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Yep, that is correct! The graph would still be decreasing, but the value at which it is decreasing by, becomes less and less for each point.

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@15:40 VA should be -3

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[@jayar6695]

Uhm. 21:45I'm confused on how you get the second derivative, 6(x+3)^-2, you directly use multiplacation rule. How about the chainrule? Comment if you get my point. Correct me also if im wrong😅

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Uhm the derivative of the inside which is x+3 is 1 so it does not change the answer

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[@hankthecanecorso5462]

Probably already noted, but you misspoke - f' negative and f" negative, slope is decreasing at an INCREASING rate

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[@user-xq3jq3cf2f]

VA at x=3 or x=-3

[@irobot4761]

x=-3, your VA will exist on the graph somewhere where the curve does not, since you know the denominator can never be 0, you know that x is -3, not 3. I think he made a mistake haha

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why did we equate the denominator and numerator to zero separately aren't we suppose to equate the entire term all together?

[@Therealtrapanthem]

Sheikh Muazzin Azeem not sure if you ever got the answer to your question but this will drive me nuts if you didn’t lol. So a rational function is equal to zero when the numerator is equal to zero. And, a function is undefined when the denominator is equal to zero. 2 very different pieces of information that are useful for graphing.

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on the step where you got the second derivative 10:15 and set it equal to 0 (2(x-1) = 0) why is 0 not also an inflection point along with 1?

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5:00 shouldn't it be decreasing at increasing rate? Actually you've said decreasing at decreasing rate for two examples.

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Iam sorry byt what if the first derivative =0 and the second derivative is negative then the scetch would be decreasing??

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