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:

a. Plot the function y=f(x) together with its derivative over the given interval. Explain why you know that f is one-to-one over the interval.

72. y= 2-x-x³, -2 ≤ x ≤ 2, x_0 = 3/2

Accepted Answer

Hello. In this video we are asked to consider the following function Y is equal to 4 minus x minus 2 x cubed on the interval from -1 to 1. And we want to determine if this function is 1 to 1 on the interval. Now, in order for a function to be 1 to 1 on an interval, one of two things must be true. The function must be strictly increasing. Or strictly decreasing. On the interval in order for the function to be 1 to 1 on the given interval. So what we need to do is we need to go ahead and determine the behavior of the function. Now, the main thing that tells us the behaviors of a function is the first derivative of the function. So in order to determine whether this function is going to be strictly increasing or decreasing on the interval, we need to go ahead and find the critical points of the function and determine if all the behavior is going to be the same. So we will start by taking the first derivative of the function. Y is going to equal to -1 minus 6 x 2. Now what we want to do is we want to go ahead and check for any critical points of the function. In order to do this, we will set the function equal to 0 and solve for the values of x. We will add 1 to both sides of the equation, leaving us with -6 x2 equal to 1. Dividing by -6 leaves us with x2 equal to -1 divided by 6. But here's where the issue comes in. If we take the square root of both sides of the equation. We are taking a square root of a negative number, meaning we get an imaginary number. Therefore, there are no critical values of the function, and what we need to do now is we just need to determine the behavior from -1 to 1. So we will go ahead and draw a number line for our entire interval. What we are going to do is we're going to take a testing point on the interval, plug it into our derivative, and determine the behavior based off of the output. Now the easiest number to pick for this interval is going to be the value of 0 since 0 is within -1 and 1. Plugging in 0 into y gives us -1 minus 0, which gives us a negative number, and what that tells us is that the function is going to be decreasing on the entire interval from -1 to 1. Furthermore, because it is strictly decreasing from -1 to 1, our definition of 1 to 1 is satisfied. So the solution to this problem is going to be a. This function is 1 to 1. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

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