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) = x√(x +9)

Accepted Answer

Hello, in this video, we are going to by finding the intervals of increasing and decreasing and the intervals of concave up and concave down. The problem asks us to consider the function F of X equal to -2 X multiplied by the square root of X + 12. We want to determine the intervals where the graph of F is increasing and decreasing, and we want to determine the intervals on which the curve is concave up and concave down. So, let's go ahead and approach the first part of this problem. We want to determine the intervals of increasing and decreasing. Now, in order to determine these intervals, we want to go ahead and use the first derivative test. So, we will start by finding the critical points of the given graph, which means that we need to take the derivative of the given function. So again, we can rewrite. The given function as F of X equal to -2 X multiplied by the quantity X + 12 raised to the half power. We will take the derivative by using the product rule. That will be F of X equal to -2. -2 X, then multiplied by the derivative of X + 12 raises the half power. By the chain rule, this is going to be 1/2 multiplied by X + 12, raised to the negative half power, multiplied by the derivative of the innermost function, which will be 1. Then we will add X + 12, raise to the half power, and multiply this by the derivative of -2 X, which is negative 2. Now we need to simplify the derivative. We can start by combining the terms together, and this will give us negative X divided by the square root. Of X + 12, then we will subtract 2 multiplied by the square root of X + 12. We can then combine the terms into a single fraction, and that will give us negative X minus 2 multiplied by the quantity X + 12, divided by the square root of X + 12. Distributing -2 into X + 12, and combining like terms will give us a final derivative of -3X minus 24, divided by the square root of X + 12. Then we can factor out a -3 from the numerator, leaving us with -3 multiplied by the quantity X + 8, divided by the square root of X + 12. This is going to be the derivative F of X. And now what we want to do is we want to find the critical points of this function. Now, there are 2 things to note. First, our derivative is in the form of a rational function. So first we need to ask what are the restrictions of the domain. Now, normally, when we have a square root function, in order to find the domain of that square root, we will take what is inside the square root, which in this case is X + 12, and this must be greater than or equal to 0. However, because this square root is in the denominator of a rational function, we cannot include the value of zero, otherwise, this will be an undefined function. So, in order to find the domain of this function, we want to strictly have X + 12 greater than 0. Now, when we solve for the domain, we will subtract 12 to both sides of the inequality, and this will tell us that the domain must be values of X that are strictly greater than -12. So, now that we have the idea of what the domain looks like, we will go ahead and create a number line and use the first derivative test. For the critical points. Now, what about the critical points? Well, in order to find the critical points, we will find the values of X that cause the numerator of the derivative to be zero. If the numerator is 0, then the entire function is 0. So, that will be -3 multiplied by X + 8, and we'll set it equal to 0, and by solving for X, we will be left with the value X is equal to -8. This is going to be the critical point that we apply on our number line. So, we will represent -12 as an open circle, as we don't want to include -12 in our domain. Now, since we are allowed to have any value that is greater than -12, this domain will be from -12 to positive infinity. Then we will label our critical point, which is -8. Now what we want to do is we want to take testing points to the left of -8 and to the right of -8, and that will tell us that the behaviors of the derivative. So, let's go ahead and pick a testing point to the left of -8 on the interval from -12 to -8. A testing point that we can pick is -9. If we take this testing point and plug it into the derivative, F of -9 will give us the value. will give us a positive value. If we plug in -9 into the function, that will give us the product of a negative number with another negative number, which is positive, and this will then be divided by a positive number. So we end up getting a positive output, telling us that the graph will be increasing from -12 to -8. Then if we take a pet testing point to the right of -8, such as 0, and we plug it into the function F of 0 will produce a negative number. What this means is that the graph will be decreasing from -8 to positive infinity. So what does this tell us about the intervals of increasing and decreasing? Well, this tells us that the intervals of increasing will be from -12 to -8. And the intervals of decreasing. will be from -8. To positive infinity, and this is going to be the solution to the first part of the problem. Now let's go ahead and approach the second part of the problem. The second part of the problem asks us to determine the intervals of concave up and concave down. Now, when we are talking about concavity, what we need to do is we need to calculate the second derivative. So, let's go ahead and calculate the second derivative. In order to calculate the second derivative, the first thing we can do is we can rewrite F of X in a product form. We can let F of X equal. To the quantity, -3, multiplied by X + 8, multiplied by X + 12, raised to the negative half power. By using the product rule, F of X will equal to -3. Multiplied by X + 12 raised to the negative 3 halves power, multiplied by -12. Multiplied by X + 8 + X + 12, and what we can do here is we can go ahead and combine the like terms and reorder the terms. This will allow us to simplify the double derivative to be -3, multiplied by X + 12, raised to the negative 3 haves power, and this will be multiplied by X + 16 divided by 2. And combining terms together will give us -3 halves multiplied by X + 16, divided by X + 12, raised to the half power. Now, what we want to do is we want to use the 2nd derivative test in order to determine the intervals of concave up and concave down. So, just like before, we want to find the critical point of the second derivative. In order to do this, we will figure out the values in which X + 16, the numerator is 0. We take the numerator set it equal to 0, and this gives us X's equal to -16. But there is a small problem here. Recall that the domain of the second derivative is the same as the domain as the first derivative, which is also the same as the domain of the original function. X must be strictly greater than -12, since X equal to -16 is outside of the interval or the domain. That means that we will not consider -16 as a critical point to the second derivative. However, we can still check the concavity of the portions of the domain that we do have. Now, the domain for the second derivative is still -12 to positive infinity. If we take a testing point and plug this into the second derivative, let's say our testing point is 0, then F of 0 will equal to negative. 3 divided by 2 multiplied by 16 divided by the square root of 12, which is going to give us a negative output. What this means is that the interval of concave down is going to be from -12 to positive infinity, and that is going to be the final part to the solution of this problem. So, I hope this video helps you in understanding how to use the intervals of increased decrease, and to find the intervals of concave down, and I will go ahead and see you all in the next video.

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) = x√(x +9)

Key Concepts

First Derivative Test

Second Derivative Test

Critical Points

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