Absolute & Local Minimum and Maximum Values - Relative Extrema, Critical Numbers / Points Calculus

This calculus video tutorial explains how to find the absolute minimum and maximum values as well as the local max and local min. It explains the extreme value theorem for finding absolute extrema and discusses the first derivative test to identify relative maximum and minimum values using a sign chart on a number line. It also discusses how to find the critical points or critical numbers of a function. This video contains plenty of examples and difficult / hard practice problems.
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-tutor.net/
Full-Length Videos and Worksheets:
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[@TheOrganicChemistryTutor]

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

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that is a historical moment 36:45

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[@Zenbeau]

14:08 you said if the leading coefficient has a odd power like x^3 then it will have no max or minimums. but at 22:21 that function also has a leading coefficient with odd power of ^3 but that one has a max and minimum for some reason?

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[@hopecritical-ops5554]

Quick question: Wouldn't the square root of 9-(3)^2=0, which would create a domain issue?

[@saadalikhan494]

I have a query. at 3:58 we have a graph which does not have a minima as last point is encircled and therefore is not part of the graph but the point before that is included in the graph so why we don't consider it as a minima. Thanks in advance

[@danjumaameh1601]

@36.12 it should be minus 12(0) i.e -0 and not +0 as stated in the question but more importantly perhaps @ 3.27 which should be -12(2) and not +12(2) ...... -24 and not +24 hence final answer @36.45 should be -7 and not +41.

[@jimhalpert9803]

Bruh he corrected it in the video itself.

[@nyomimorris6364]

41:33, I am confused why 0 is part of the critical points. Wouldnt it just be -1 and 1?

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[@samadkamran5303]

So like... does chain rule not apply to this because when you bring the exponent down don't you have to multiply the expression by its derivative-

[@wildlifeadventuregirl]

For the last problem I dont think you would include -3/2^1/2 because it is equal to -2.1 which is outside the restricted domain. so there would be no local min and -2 would be the absolute min. Correct me if Im wrong.

[@SM-qn7sr]

At 1:09:00 he said it’s not part of the interval

[@SM-qn7sr]

on 57:26, shouldn't the graph be restricted on the interval [-1,2]? Why did you make it continuous on all real numbers?

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[@christanellylaboga8279]

how is 5/3 present in 2/3? please explain

[@vctoriaotumdi9991]

Thank you

[@user-bw1jn1yc1i]

Thank you for this video, nice explanations. One correction: f(2) = 3(2)^2 - 12(2) + 5 = -7 actually, not 41. Cheers!

[@user-bw1jn1yc1i]

Whoops, just saw your in-video correction. No worries!

[@magnus4537]

What if we have two minimums, but they are "the same"? (-1, -1) and (1, -1). Are both absolute minimum? Or are they both local minimum?

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[@rami5155]

@1:04:30 you keep using -3/sqrt(2) wich is -2.12 so its out of our intervall. Or did I do something wrong here?

[@Loots1]

no youre right, i think he just made a mistake

[@misutokiriya6547]

It's been like a year but for future uses, he did remove it in the end; thus, not making it the absolute minimum value in the interval [-2,3]

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[@rickybobby5584]

Very nice but on 7:20 you said if f'(c) doesn't exist, we have a cusp. Not necessarily coz we might have vertical tangent just like in f(x)=x^(1/3) at x=0 . So how can we tell if we have a cusp or vertical tangent?

[@SM-qn7sr]

its a cusp because if f'(c) doesnt exist at x=a certain number then its not differentiable at that number and not differentiable means the first derivative isn't continuous on x=that number. Its only a verticle tangent if f(x) is undefined, not f'(c)

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what will we do when we have modulus function

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[@mr.9754]

52:10 If you want to shave (valuable) seconds off (while getting a better idea visually/kinestheticly), don't bother wasting time "testing points" to find out where the derivative changes sign. Simply quickly plot the (*numerator of the) derivative, ie plot f' vs x. Just use end behavior and the zeros to do this. (*Yes, there is something in the denominator in this particular example because we are dealing with a rational function, but in this example the denominator is always positive, so it won't affect finding where f' changes sign. Keep in mind, that you can avoid "testing the points" 100% of the time if dealing with a polynomial function.)

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[@adityawankhede7260]

8:82 Shouldn't the lower local min be a critical number instead?

[@phyoeminnkyaw9551]

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[@RiusII401]

Hi, for the first problem you did, does it matter which number you plug into the first derivative? You did plugged in a # that is smaller than 1.5 into 2(2x-3). If I plug in a # that is smaller that 1.5 into 6x-6, would I still get the same answer as you? Thank you!

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[@emkayvevo4950]

Please, do you have to find the X and Y intercepts to draw the graph with accuracy? If so, then how do I find my X_intercept for 2x^4 - 2x^2 - 1=0? Thank you

[@TheOrganicChemistryTutor]

It doesn't look factorable. You can use the quadratic formula where a = 2, b = -2, and c =-1. The result that you get will equal x^2 and not x so you will have to take the square root of that result to get the final answer. You can also graph it using an online graphing calculator and find where it touches the x axis. Go to google and type in 'online graphing calculator' and the choose the first link that comes up which is desmos.com. Type in the expression "2x^4-2x^2-1" without the =0 part and you should see the graph. Move the mouse to the point where the graph touches the x - axis and you should see the answer. x = +1.169 and -1.169. I hope that helps :)

[@emkayvevo4950]

Thank you.

[@sebastiangiray3614]

question on the laast problem..................................why would you include -3/2^1/2 ???? its already beyond -2???? please answer??organic chem plssss

[@suga_tofu]

Does absolute max/min mean global max/min?

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[@3B-OD]

52:25 if you plug in 1 the answer is gonna be negative no ?

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57:50

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[@MostafaMahmoud-tw7xi]

36:45 he switch the -12 to +12 did anyone noticed ?

[@MaurickSh]

When you plug in a 2 into 2(2x-3), wouldn't you get a negative 2 instead by following order of operations, so instead of 2X2 = 4 -3 = 1, you'd do 2 - 3 = -1 X 2 = -2?

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that isnt how order of operation works....

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