Like
Report
Numerade Educator
Problem 81 Hard Difficulty
Assume that all of the functions are twice differentiable and the second derivatives are never $ 0 $.
(a) If $ f $ and $ g $ are positive, increasing, concave upward functions on $ l $, show that the product function of $ fg $ is concave upward on $ l $.
(b) Show that part (a) remains true if $ f $ and $ g $ are both decreasing.
(c) Suppose $ f $ is increasing and $ g $ is decreasing. Show, by giving three examples, that $ fg $ may be concave upward, concave downward, or linear. Why doesn't the argument in parts (a) and (b) work in this case?
Answer
(a) since $f$ and $g$ are positive, increasing, and $\mathrm{CU}$ on $I$ with $f^{\prime \prime}$ and $g^{\prime \prime}$ never equal to 0 , we have $f>0, f^{\prime} \geq 0, f^{\prime \prime}>0$
$g>0, g^{\prime} \geq 0, g^{\prime \prime}>0$ on $I .$ Then $(f g)^{\prime}=f^{\prime} g+f g^{\prime} \Rightarrow(f g)^{\prime \prime}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime} \geq f^{\prime \prime} g+f g^{\prime \prime}>0$ on $I \Rightarrow$
$f g$ is $\mathrm{CU}$ on $I$
(b) In part (a), if $f$ and $g$ are both decreasing instead of increasing, then $f^{\prime} \leq 0$ and $g^{\prime} \leq 0$ on $I,$ so we still have $2 f^{\prime} g^{\prime} \geq 0$
on $I .$ Thus, $(f g)^{\prime \prime}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime} \geq f^{\prime \prime} g+f g^{\prime \prime}>0$ on $I \Rightarrow f g$ is $\mathrm{CU}$ on $I$ as in part (a).
(c) Suppose $f$ is increasing and $g$ is decreasing [with $f$ and $g$ positive and $\mathrm{CU}$ ]. Then $f^{\prime} \geq 0$ and $g^{\prime} \leq 0$ on $I,$ so $2 f^{\prime} g^{\prime} \leq 0$
on $I$ and the argument in parts (a) and (b) fails.
More Answers
Discussion
You must be signed in to discuss.
Top Calculus 2 / BC Educators
Heather Z.
Oregon State University
Caleb E.
Baylor University
Samuel H.
University of Nottingham
Joseph L.
Boston College
Watch More Solved Questions in Chapter 4
Video Transcript
F and G are positive, increasing in concave up on this interval. I. And we want to show that the product of these. So F of G is also concave or F times G is also concave on this um interval. I so I'm just gonna set y equal to F times G. And then what we're gonna do is find the second derivative. So the first derivative, we're going to have to use the product rule to find which is just the derivative of the first, which is f prime times the second, which is just G. And then plus the first times the derivative of the second. So this is our first derivative. And so now we want to find the second derivative. And how we're going to do that is by using the product rule now twice. So we're going to do the product rule for this first product of f prime times G, which we get the derivative of the first, which would be F double prime times a second, plus the first times the derivative of the second. And then we do the same thing over here, which would be the derivative of the first, which would be F prime times a second, which is G prime plus the first times the derivative of the second. And now, if we simplify this, we just get F double prime times G plus two times F prime times G prime plus F times G double prime. And so we know that F F double prime and G double prime. These are both positive. Since we're told that both of these functions are concave up, which means their second derivative is positive. We know that G and effort also positive. Since we're told that both of these functions are positive on this interval. And then we also know that these two F prime and G prime are positive as well since we're increasing um F and G are both increasing on this interval. So we're going to have a positive value plus positive value plus a positive value, which means that our second derivative y double prime is going to be greater than zero on the interval I. Which means that this function why is equals F times G is concave up on I. And now for part B, we're told that the only difference in part A is now F and G are both decreasing. So if we go ahead and look at the second derivative again, which was why prime was equal to F double prime times G plus two times F prime times G prime plus G double prime times F. If we look at this a derivative again, the only difference is that these two F prime and G prime are going to be negative now since we're decreasing and if we're decreasing, that means our first derivative is negative. So we're going to have these two values F prime and G prime be negative. However, if we multiply a negative times a negative, that means that this two times F prime times G prime is going to be positive. So this first term of double prime times G isn't going to change that still positive. The second term is also positive since we have a negative times negative. And this for the third term also doesn't change and is positive as well. So we have a positive plus a positive value plus another positive value, which is always going to be positive. So again, why double prime is greater than zero, which means that why is concave up on I on the interval I and now, for part C, we're told that F is increasing and G is decreasing. And we want to give some examples where we would be concave up, so concave up, concave down and linear. And we're told that F is increasing and G is decreasing. So if we go ahead and look at our second derivative again, which is equal to, we go up here was equal to F double prime times G plus two times F. Prime times G prime plus G double prime times F. So if we suppose that those that F is increasing in G is decreasing, then we know that their first derivative of F is going to be greater than zero and the first derivative of G is going to be less than zero. So that means that this here is going to be positive and this here is going to be negative. And so the only thing that changes is these two the signs of these two values since our second derivative is still positive for both these and G. And efforts still supposedly positive. So these values are still positive. So now we have a positive value plus a positive times a negative value, billion negative value. So we now have a negative value here and then another positive value here. So the reason that we can't actually say whether F times G is going to be positive, negative or linear is because we have a positive plus a negative plus a positive. So depending on how big this negative value is, we are not, we might have a positive Y double prime or we might have a negative Y double prime, or you might even have a zero as our Y double prime. So if we were to try and make this to be concave up, we would need F double prime times G plus G double prime times F to be greater than The absolute value of two times F prime times G prime. And this would be for concave up. And if we wanted concave down the mood need F double prime times G plus G double prime times F to be less than the absolute value of two times F prime times G prime. And the reason I'm using absolute value is because we want just the positive value of this to be less than or greater than F double prime times G plus G double prime times F. And so if this was true, we would be concave down and if we were linear, then our second derivative would actually be equal to zero. So that would mean that F double prime times G plus G, double prime times F would be equal to the absolute value of two times F prime times G prime. This would be the case where we are linear
Oregon State University
Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
Kayleah T.
Harvey Mudd College
Kristen K.
University of Michigan - Ann Arbor
Michael J.
Idaho State University
