Video Transcript
Let 𝑓 of 𝑥 be equal to 𝑥 cubed minus two 𝑥 squared plus 𝑥 minus three. Which of the following numbers is the 𝑥-coordinate of a point where the instantaneous rate of change of 𝑓 is the same as the average rate of change of 𝑓 over the interval negative two is less than 𝑥 which is less than two? Is it a) negative two-thirds, b) negative three over two, c) three over two, or d) two-thirds?
In this equation, we have a function in 𝑥. And we’re trying to find the value of the 𝑥-coordinate at the point where the instantaneous rate of change of a function is the same as the average rate of change of the function over the interval negative two is less than 𝑥 which is less than two. So we’re going to need to calculate two things. We’re going to need to find an expression for the instantaneous rate of change of 𝑓. And we’ll need to find the value of the average rate of change of 𝑓 over our interval.
Now, there’s a number of ways to think about the formula for the average rate of change of our function. When we’re thinking about the average rate of change of our function from 𝑥 one to 𝑥 two, we can use the formula 𝑓 of 𝑥 two minus 𝑓 of 𝑥 one over 𝑥 two minus 𝑥 one. In this example, we can say that 𝑥 one is equal to negative two and 𝑥 two is equal to two. And so, the average rate of change of our function between these values will be given by 𝑓 of two minus 𝑓 of negative two over two minus negative two.
We can find 𝑓 of two by substituting 𝑥 equals two into our function. We get two cubed minus two times two squared plus two minus three which is negative one. Similarly, 𝑓 of negative two is negative two cubed minus two times negative two squared plus negative two minus three which is negative 21. And so, our average rate of change function becomes negative one minus negative 21 all over two minus negative two. Well, that’s 20 over four which is equal to five.
So we need to find the 𝑥-coordinate of the point where the instantaneous rate of change of our function is five. So how do we work out the instantaneous rate of change of our function? Well, we find the derivative of our function with respect to 𝑥. And since we’re trying to find the 𝑥-coordinate of the point where this is equal to five, we’re going to set our derivative equal to five and solve for 𝑥. So let’s find 𝑓 prime of 𝑥.
The derivative of 𝑥 cubed is three 𝑥 squared. The derivative of negative two 𝑥 squared is two times negative two 𝑥 which is negative four 𝑥. The derivative of 𝑥 is one. And the derivative of negative three is zero. So we’re going to set 𝑓 prime of 𝑥 equal to five. Here, we have a quadratic equation in 𝑥. To solve this, we’ll begin by subtracting five from both sides of our equation. Our next step will be to factor the expression three 𝑥 squared minus four 𝑥 minus four. And when we do, we get that it’s equal to three 𝑥 plus two times 𝑥 minus two.
And since the product of three 𝑥 plus two and 𝑥 minus two is zero, that tells us that either three 𝑥 plus two is equal to zero or 𝑥 minus two is equal to zero. On our left equation, we subtract two and divide through by three. And we see that 𝑥 is equal to negative two-thirds. And in our right equation, we add two to both sides. And we get 𝑥 is equal to two.
Actually, we can instantly disregard this solution. Remember we were told that 𝑥 must be less than two and greater than negative two. So it can’t actually be equal to two. And this means the value of the 𝑥-coordinate at the point where the instantaneous rate of change of 𝑓 is the same as the average rate of change of 𝑓 — remember that was five — is negative two-thirds.
