Use ƒ' and ƒ" to complete parts (a) and (b).

Question

a. Find the intervals on which f is increasing and the intervals on which it is decreasing.

b. Find the intervals on which f is concave up and the intervals on which it is concave down.

ƒ(x) = x⁹/9 + 3x⁵ - 16x

Accepted Answer

Welcome back, everyone. In this problem, consider the function F X E equals X 9 divided by 9 + 13 X 5 minus 1,296 X plus 17. Determine the increasing and decreasing intervals of F and the intervals on which the curve is concave up and concave down. Now, if we're going to determine the increasing and decreasing intervals of F, we'll need the first derivative of F because the first derivative can help us to figure out the function's critical points. So let's differentiate F. Now differentiating F. With respect to X, OK. Then we'll get the first derivative to be equal to. Next to the 8th, OK. Plus 65 4 minus 1296. Now why is this first derivative so important? Well, at the critical points, OK, the points we'll need to help us figure out what's happening on our intervals. At the critical points, the first derivative of X is equal to 0. So this means then that we can set X to the 8th plus 65 X to the 4th. Minus 1,296 to 0 to help us solve for X. No, here, if we let. If we let Y or sorry if we let you. Be equal to x 4th, OK, where you is greater or where you is greater than or equal to 0, which makes sense because it's raised to the 4th, so it's going to be positive. Then our equation becomes U2 + 65 U minus 1,296, and this is a quadratic equation. So we can solve by using the quadratic formula. OK? So, by the quadratic formula, then, And by the quadratic formula. Then you. Of course, before we start, we know that A equals 1, B equals 65, and C equals -1,296. Then that means you is going to be equal to. 65 plus R minus the square root of 65 2 + 4 multiplied by 1,296 and that's all divided by 2 multiplied by 1 or 2. Now if we simplify here, this is gonna be -65 plus R minus the square root of 9,409, which is 97, are divided by 2, which then means that you is gonna be equal to -65 plus 97 divided by 2. OK, which equals 16. Or OK. Where can I put that here? Or you is gonna be equal to -65 minus 97 divided by 2. No, we don't even need to go any further because -65 minus 97 will be a negative number. That's gonna be -162 divided by 2. But remember, you cannot be negative. You, by definition has to be greater than our equal to zero. So this will not be a possible value of you, we can discard this value, which means then that you equals 16. If you equal 16, that means that X 4th will be equal to 16, and if X 4 equals 16, that means X is going to be equal to the 4th root of 16. Which equals plus or minus 2? So this is the X value for our critical point. Know that we have the X value, then we can start to test the sign of our first derivative for each interval. Now, what intervals are we going to have? Well, if we try to put our X values on the X-axis, if we think about it, OK. It goes from negative infinity, -22, and then infinity. So we're thinking about three intervals from negative infinity to -2, from negative 2 to 2, and from 2 to infinity. Now, what can we say about uh these three intervals here? Well, if we think about it, OK, we can test each sign by choosing a value. Let's start with when X is less than -2, OK? For X is less than -2, we know that the first derivative will be, sorry, our function will be increasing if the first derivative is negative and decreasing where the first derivative. Sorry, let me say that again. Our first derivative will be. Increasing if. Let me say that one more time. Our function will be increasing where the first derivative is positive, and it will be decreasing where the, where the first derivative is negative, OK? You get that? Good. So here, For X is less than -2, let's choose a value and see if it makes our first derivative positive or negative. Let's use when X equals -3. In that case, our first derivative. It's gonna be equal to -3 to the 8. Plus 65 multiplied by -3 to the 4th minus 1,296, which equals 10,530, and that is greater than 0. So what does this tell us? This tells us that on this interval, our function is increasing. Next, let's test it for X between -2 and 2, yeah. So when X is greater than -2, but less than 2, let's use the value for X being equal to 0, right? When X equals 0, then our first derivative is gonna be equal to 0 to 8 plus 65 multiplied by 0 to the 4th minus 1,296. We'll be left with -2,196. That's negative, so that means our function is decreasing on this interval. Finally, Or X greater than 2, OK? If we choose a value, let's say when X equals 3, OK. Then, our first derivative is gonna be equal to 38 plus 65 multiplied by 34 minus 1,296, which equals 10,530. Again, it's increasing. Because our first derivative is positive, OK? So what does this tell us? Well, no, this tells us that F is increasing under the intervals, negative infinity to -2 and 2 to infinity, but decreasing and negative to 2. Now how can we determine the concavity of uh FF X of our function? Well, we can use the 2nd derivative to help us, OK? Let me put that in blue here. Now if we find a second derivative. That means we're going to differentiate the first derivative with respect to X, and the second derivative is going to be X 7 plus 260 X cubed. OK? And if we factor out our common term here, we can get it as 4 XQed multiplied by 2 X to the 4th plus 65. Now, if we're gonna figure out if it's concave up, if the sign of the second derivative is positive, the function is concave up, and if it's negative, the function is concave down. Now if we think about our expressions here, we could choose some values, but again, we know. That 2 X 4 plus 65 is going to be greater than 0 because X the 4th is also going to be always greater than 0, OK? So that means the sign. Of the 2nd derivative. Is going to be determined by X cubed, OK, because the sign for X cubed is going to change depending on the value of X. Now, when X is positive, OK, when X is positive, that means for Xcu the loss of the positive, which makes the second derivative positive and thus the function is going to be concave up. However, when X is less than 0, where X cubed is going to be less than 0, the second derivative will also be less than 0, or function will be concave down. Thus, based on what we found, our function is increasing on the impervals negative infinity, -2, and 2 infinity, decreasing on 22, and is concave up on 0 to infinity while concave down from negative infinity to 0. Thanks for watching everyone. I hope this video helped.

Use ƒ' and ƒ" to complete parts (a) and (b).

a. Find the intervals on which f is increasing and the intervals on which it is decreasing.

b. Find the intervals on which f is concave up and the intervals on which it is concave down.

ƒ(x) = x⁹/9 + 3x⁵ - 16x

Key Concepts

First Derivative Test

Second Derivative Test

Critical Points

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