Determine the intervals on which the function g(𝓍) = ∫ₓ⁰ t / ...

Question

Determine the intervals on which the function g(𝓍) = ∫ₓ⁰ t / (t² + 1) dt is concave up or concave down.

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says consider F of X, which is equal to the interval from X to 1 of T divided by the quantity of T2 + 9 in quantity DT. For which intervals is F of X concave upward and downward. So we need to determine the intervals over which F of X is concave up and over which intervals it is concave down. So in order to determine concavity, we'll actually need to take the second derivative of our function F of X with respect to X. So, in order to do that, the first thing we want to do is to rewrite our equation F of X. So, here we're given that F of X is equal to the integral from X to 1 of T, divided by the quantity of T squared plus 9 in quantity DT. And so what we want to do is to switch the limits of integration. Recall that if we switch the limits of integration, we pick up a minus sign. So, this means that this is going to be equal to minus the integral from 1 to X. Of T divided by the quantity of T squared plus 9 in quantity DT. So now we need to take a derivative of this function. So we'll be looking for F of X, which is the derivative with respect to X of F of X, and here we're taking the derivative of an integral, and we're taking the derivative with respect to the variable that is actually the upper limit of our limits of integration. So that means that we can use a fundamental theorem of calculus, since our lower limit is a constant. And so that means that F of X is going to be equal to minus X divided by the quantity of X2 + 9. So now we just need to take one more derivative. So we're we're going to be looking for F of X. And this is going to be equal to the derivative with respect to X of FX, and then taking that derivative, we'll have minus, and here we're taking the derivative of a fraction. So we're going to want to use our quotient rule. So we call for the quotient rule that we'll have our denominator, which is going to be the quantity of X2 + 9 in quantity, and then that gets multiplied by the derivative of our numerator. So we'll have the multiplied in this by the derivative of X with respect to X, which is just 1. Then we'll have minus our numerator, which is X, multiplied by the derivative of our denominator. And so here we're gonna multiply this by the derivative with respect to X of the quantity of X2 + 9, which is 2 X. And that whole quantity is going to be divided by our denominator squared, so we'll have the quantity of X2 plus 9 in quantity squared. Now we just need to simplify this expression. So collecting like terms in the numerator, this is going to be equal to minus the quantity of minus X2 + 9 in quantity divided by the quantity of X2 + 9 in quantity squared. Distributing the my sign through the numerator, we'll see that this is going to be equal to. The quantity of X2 minus 9 in quantity divided by the quantity of X2 + 9 in quantity squared. Still we need to use a second derivative to determine the intervals of concavity, and we call that if our second derivative is positive over an interval, that means that our function is in this case F of X, is going to be concave up. So, in our case, we're looking for F of X to be greater than 0. And this is going to occur when our numerator is positive, since our denominator and our expression for FX is always going to be a positive quantity. So the sign of FX is going to be determined entirely by the numerator. So, FX is going to be greater than 0, when X2 minus 9 is greater than 0. And this actually occurred. When X is less than -3. And when X is greater than 3. So, since F is positive on those intervals, that means that our function F of X is going to be concave up. On the interval. Which is the open interval from minus infinity to -3. Union with the open interval from 3 to infinity. So, that is the first part of the answer that we're looking for. Now, for the second part, we need to actually find the intervals over which our function is concave down. So that is going to occur when F of X. is less than 0, and that's going to occur when our numerator of X2 minus 9 is less than 0. And this actually occurs. When -3 is less than X, which is less than 3, so, when X is between -3 and 3. So, that means that our function F of X is going to be concave down. On the open interval. From -3 to 3. So that is the second part of the question answered. So just to recap, we found that our function F of X is going to be concave up on the open interval from minus infinity to -3, union with the open interval from 3 to infinity, and our function F of X is concave down on the open interval from -3 to 3. So that is the answer that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Determine the intervals on which the function g(𝓍) = ∫ₓ⁰ t / (t² + 1) dt is concave up or concave down.

Key Concepts

Concavity

Second Derivative Test

Fundamental Theorem of Calculus

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