# Video: The Mean Value Theorem

For the function 𝑓(𝑥) = 𝑥³ − 4𝑥, find all the possible values of 𝑐 that satisfy the mean value theorem over the closed interval [−2, 2].

02:09

### Video Transcript

For the function 𝑓 of 𝑥 equals 𝑥 cubed minus four 𝑥, find all the possible values of 𝑐 that satisfy the mean value theorem over the closed interval negative two to two.

Remember, the mean value theorem says that if 𝑓 is a function which is continuous over some closed interval 𝑎 to 𝑏 and differentiable at every point of some open interval 𝑎 to 𝑏. Then there exists a point 𝑐 in that open interval such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎. Our 𝑓 of 𝑥 is equal to 𝑥 cubed minus four 𝑥. And our closed interval is from negative two to two. We’re looking to find the value of 𝑐 such that the derivative of our function evaluated at 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎. That’s 𝑓 of two minus 𝑓 of negative two over two minus negative two. So we’ll do two things. We’ll evaluate this quotient. And we’ll also find an expression for the derivative of our function. 𝑓 of two is two cubed minus four times two. And 𝑓 of negative two is negative two cubed minus four times negative two. And actually, this gives us a value of zero. So in order to find the values for 𝑐 such that 𝑓 prime of 𝑐 is equal to zero, let’s find the derivative of our function.

The derivative of 𝑥 cubed is three 𝑥 squared. And the derivative of negative four 𝑥 is negative four. So 𝑓 prime of 𝑥 is three 𝑥 squared minus four. And we can say that 𝑓 prime of 𝑐 is equal to three 𝑐 squared minus four. So let’s set this equal to zero and solve for 𝑐. We begin by adding four to both sides of our equation to obtain that three 𝑐 squared is equal to four. Next, we divide by three. And we see that 𝑐 squared is equal to four-thirds. And finally, we take the square root of both sides, remembering to take by the positive and negative square root of four-thirds. To obtain that 𝑐 is equal to plus or minus the square root of four over three. And which we can say is equal to two over root three and negative two over root three. Notice also that these values of 𝑐 are indeed in the closed interval negative two to two as required by the mean value theorem.