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Show that $ \tan x > x $ for $ 0 < x < \pi /2 $. …

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Problem 82 Medium Difficulty

Suppose $ f $ and $ g $ are both concave upward on $ (-\infty, \infty) $. Under what condition on $ f $ will the composite function $ h(x) = f(g(x)) $ be concave upward?


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Fahad Paryani

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Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

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Derivatives

Differentiation

Volume

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04:35

Volume - Intro

In mathematics, the volume of a solid object is the amount of three-dimensional space enclosed by the boundaries of the object. The volume of a solid of revolution (such as a sphere or cylinder) is calculated by multiplying the area of the base by the height of the solid.

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06:14

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Video Transcript

So we're told the functions F and G are concave up for this interval from negative infinity to infinity or for their whole domains. And we're trying to find what are the conditions on F. That we need for this function H. Of X, which is equal to F of G. Of X. Or the composite function of F. G. And we want to find what are the conditions of F. That we need for H. Of X to be concave up. So we want H. Of X to be concave up, which I'm just gonna know as see you. So we're gonna want its second derivative to be positive. So what we're gonna do is find the second derivative of each of X. So the first derivative, we're gonna have to use the chain rule for which we do the derivative of the outside. And then multiply by the derivative of the inside. So we have F. Prime of G. Of X, times G. Prime of X. And now to find the second derivative of H of X, we're gonna have to use the chain rule and the product rule. So the product rule is the derivative of the first times a second. Plus this plus the first times the derivative of the second. So the derivative of F prime of G of X would be F double prime of G of X, multiplied by the derivative of what's inside here, which would be G prime of X. And then we still have another G prime of X. So this is G prime of X squared. And then we add the first times the derivative of the second, which would just be G double prime of X. So now we have our second derivative of H of X. And we're told that F is concave up on all real values are from negative infinity to infinity. So we know that this is positive for all values of X. We know that this is also going to be positive since it's some value of X squared in anything squared is going to give us a positive value. And so we know that this is also positive, so this is a positive times positive, which is positive. And then this term, we see this is the first derivative of F at G of X. And we don't actually, we weren't given any um parameters on what that would be, so this is an unknown, so this is unknown and then G double prime of X, we know is actually positive since we were told that G of X is also concave up. So we know that this first term is going to be positive in the second term is a positive value times this f prime of G of X. So what we're gonna need is we're gonna need f prime of G F X to be greater than or equal to zero, since if it's equal to zero, this term over here goes to zero. However, this term before it was already positive, so we'll still have a positive second derivative. So we can have F prime of G of X being greater than equal to zero. And F prime of G of X is just F of X at some value or F prime of X at some value. Since we're looking at the whole interval from negative infinity to infinity. So really what we need is F prime of X to be greater than or equal to zero for ex being negative infinity to infinity. So this is the parameter or the condition that we need in order for our function H of X to be concave up.

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Review

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