Calculus 1 Lecture 3.1: Increasing/Decreasing and Concavity of Functions

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[@eliezerr.johnson9644]

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)?

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+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!

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Relative Extrema 37:01 Absolute extrema 50:00

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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!❤

[@Patt0920]

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.

[@planetvance]

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.

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[@zofa300]

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

[@elizabethatienootieno3083]

what is X^2=0? Or even X^3=0

[@Ohm_Sharma]

@@elizabethatienootieno3083 In both those cases, x = 0.

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Excellent lecture, totally saved my bacon!

[@khairilmuiz2966]

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?

[@fcbetter]

Yes, I noticed that as well.

[@hotman718]

yes professor made a mistake. It happens. x^-2/3 is undefined so x=1/8 is the only critical point

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Really great, Thanks

[@VitusAvitus]

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.

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[@Ohm_Sharma]

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!

[@carultch]

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.

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At 1:15:27 how do we equate x to the negative power to 0. Isnt it undefined?

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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.

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Is this section the Extrema on an Interval because this is what i have for 3.1- Extrema on an Interval

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The fact that I am watching this in my calculus class😌

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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.

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Sir plz add turned on the subtitles on this video..as on some videos subtitles are given..so its easy for students to understand.

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Europe, Africa, and a lot of other country beside USA, use brackets to describe the domain or the range of the function, which makes a big difference between the point and the domain or the range. :)

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aint no one talking about how he cleanly made that dotted line?

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how do you find out the interval? And if it is all real numbers how do you get an absolute max or minEx.) 2x^2 +3x-2 absolute min.= -(25/8) how?

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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 :)

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Yes you are correct.

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Could you please turn on captions for this video?

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relative to what the graph is doing at that part of the graph

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@1:20:15 I think one of his students messed up the f(x) for -1, its supposed to be -3 not 9

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I was going over it again and again thinking the same.

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