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 ...
_______________________________________
Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
/ collections

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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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[@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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[@bruhnish598]

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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[@judyhopps1324]

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

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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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[@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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no its +1 not -1 on the line

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[@dashkaalceus6217]

At 3:49 when finding the Derivative. Did you factor out the 2 to find the dy/dx?

[@gianny2977]

he factored it just to simplify and make it easier to solve for x and find the critical point. you should always simplify as much as you can (unless its a super complicated problem and it'll just take you back to the original answer)

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at min. 10:31 for (x-1)^2 = 0 then (x=1)^2 so is + not -, correct?

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will the critical values be substituted in real function or the marginal function? . If we have only one critical value so are we going to substitute the two extreme values located to the left and side of the critical value? eg if we have the critical value of 4 , the two extreme values located near 4 are 3 and 5 , are we going to substitute these values?

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