In Exercises 67–72, you will explore some functions and their ...

Question

In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:

b. Solve the equation y=f(x) for x as a function of y, and name the resulting inverse function g.

67. y= √(3x-2), 2/3 ≤ x ≤ 4, x_0=3

Accepted Answer

Hello. In this video, we are given the function f x is equal to the square root of 5 x + 1 and is defined on the domain from -15 to 3. We want to find the inverse function f of inverse of x. Let's go ahead and start this problem by defining the inverse function. Now, if we take a look at the given function, we have F of X is equal to the square root of 5 X + 1. Now, in order to start finding the inverse, the first thing we want to do is we want to swap F of X with the value of Y. So we have Y is equal to the square root of 5 X + 1. Then the second step in finding the inverse is to swap the positions of x and y in the function. So that is going to leave us with X is equal to the square root of 5 Y + 1. Now the last thing we need to do is we need to go ahead and solve for Y. We will start by squaring both sides of the equation to remove the square root symbol, and that will leave us with x 2 is equal to 5 y + 1. We will then subtract 1 to isolate y, leaving us with 5 y is equal to x 2 minus 1. And finally, in order to solve for y, we need to remove the coefficient of 5. So we will divide both the sides of the equation by 5. That will leave us with y is equal to x2 minus 1 divided by 5. So the inverse function for the given function is going to be f inverse of x equal to x2 minus 1 divided by 5. But now what we need to do is we need to go ahead and define the domain of the inverse function. Now in order to define the domain, what we should do is we should find the range of the given function. The reason for this is that the inverse function swaps the domain and ranges of the original function. So if we find the range of the original function, then the range of the original is equivalent to the domain of the inverse. So, let's go ahead and take a look at the domain that the original function is defined on. Now the domain again is going to be from -15 to 3. In order to find the range, we want to find the values of the original function at the end points of the domain. So we will start by plugging in -15 into our function. F-15 is equal to the square root of 5 multiplied by -15 + 1. This will leave us with the square root of -1 + 1, which is equivalent to 0. So we know that the lowest point of F inverse or of F-15 is 0. Now let's check the value of the original function at 3. F of 3 is equivalent to the square root of 5 multiplied by 3 + 1. This will leave us with the square root of 15 + 1, which is equivalent to the square root of 16, and this is going to simplify to be 4. So we know that the range. Of f of x is defined as 0 less than or equal to y less than or equal to 4. And what this tells us is that the domain of f inverse of x is going to be the values of x 0. Less than or equal to x less than or equal to 4, and this is going to be the domain that our inverse function is defined on. So now that we have our inverse function and the domain that it's defined on, the overall solution to this problem is going to be option A. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

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