Increasing and decreasing functions. Find the intervals on whi...

Question

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

f(x) = √(9 - x²) + sin⁻¹ (x/3)

Accepted Answer

Hello, in this video, we are going to be determining the intervals of the given function where the function is increasing and decreasing. Now, the problem gives us the function F of X equal to the square root of 25 minus X2, plus sine inverse of X divided by 5. Now, before we continue with this problem, let's just quickly write down the domain of each portion of the given function. So, notice that the first part of the function, the first term, is a square root function. What this means is that the inside argument of the function 25 minus X2d must be greater than or equal to 0. So to solve for the domain for the first term of the function, we will go ahead and solve the inequality. We will take 25 and subtract it to both sides of the inequality, leaving us with negative X2d greater equal to -25. If we divide both sides by -1, we get to X2 less than or equal to 25. And this is going to simplify to give us the value of X between -5 and 5. Now, what about sine inverse? Well, sine inverse has a domain from -1 to 1 on the on the rectangular axis. In order to solve for the domain, we will go ahead and multiply the entirety of the inequality by 5. This leaves us with -5, less than or equal to X, less than or equal to 5. So notice that both of the terms in the given function share the same domain, and since we only want the values of X from -5 to 5, that means that the interval that we are working on for the For the intervals of increase and decrease are from -5 to 5. OK, now that we've identified the domain of this function, what we want to go ahead and do is find the critical points in order to obtain the intervals of increase and decrease. In order to obtain the critical points, we need to calculate the derivative of the function. Now, we are going to rewrite the function as F of X equal to 25 minus X2, raise the power of 1/2, and we will add sine inverse of X divided by 5. Now what we are going to do is we are going to calculate the derivative. The derivative of the first term is going to be 1/2 multiplied by 25 minus X2, raise the power of negative/. Then we will multiply this by the derivative of the innermost function, which is going to be negative to X. By definition, the derivative of sine inverse. is going to be 1 divided by the square root of 1 minus X2. This argument of X is X divided by 5. So we have 1 divided by the square root. Of one Minus X divided by 5 quantity squared. Then by the chain rule, we have to multiply this by the derivative of the innermost function, which is X divided by 5, and that is going to be 15th. Now we will go ahead and simplify the derivative. By combining all of the terms together in the first term, we can simplify the first part of the derivative to be negative X divided by the square root of 25 minus X2. For the second term, we can go ahead and simplify this to be 1 divided by 5 multiplied by the square root of 1 minus X2 divided by 25. If we were to push in 5 into this square root, we can rewrite the square root to be the square root of 25 multiplied by 1 minus X2 divided by 25, and this will further simplify to give us the square root of 25 minus X2. So we can rewrite the denominator of the second term to be the square root of 25 minus X squared. And if we combine both of these terms together, the derivative of the given function will be 1 minus X divided by the square root of 25 minus X squared. No, Now, what we want to do is we want to go ahead and calculate the critical points of this function. Now, since we have a rational function, the way we obtain the critical points is by solving for the values of X that cause the numerator to be zero. In order to do this, we will take the numerator and set this equal to 0. We have 1 minus X equal to 0, which will further simplify to be X's equal to 1. So we only have one critical point for the given function. Now, what we are going to do is we are going to create a number line. Now recall that we calculated earlier that the domain is from -5 to 5. So, we will go ahead and label the boundary points of this number line as -5 and 5. Then we will go ahead and label the critical point, which is 1. Now, what we want to do is we want to take testing points on each of the intervals of the number lines to determine the behaviors of the graph, and plug this into the derivative. A testing point that we can take on the interval from -5 to 1 is 0. If we plug in the value of 0 into the derivative, F of 0 is equal to 1 divided by the square root of 25, and that is going to give us a positive value. Since that value was positive, that tells us that the graph of F of X will be increasing on the interval from -5 to 1. We will go ahead and repeat this process for the interval of 1 to 5. A testing point in this region will be positive too. F of 2 is equal to -1 divided by the square root of 21. Since this value is negative, that tells us that the graph F of X is decreasing on the interval from 1 to 5. So to summarize. The interval of increasing of F of X is going to be the interval from -5 to 2. And and the interval of decreasing. Is going to be the interval from 1. To 5, and I made an error on the interval of increasing, it should be from -5 to 1. And these are going to be the answers to the problem. And with that being said, the overall solution to this problem is going to be D. So, I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

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

f(x) = √(9 - x²) + sin⁻¹ (x/3)

Key Concepts

Derivative

Critical Points

Increasing and Decreasing Intervals

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