Question

Determine if the Mean Value Theorem for Integral applies to the function f(X) = x^3 - 16x on the interval [-1,1]. If so, find the X-coordinates of the point(s) guaranteed to exist by the theorem.

Answer 1

Given:

Given that the function is

$f(x)=x^3-16x$

The interval is [-1,1].

Required:

To determine the mean value theorem for the given function.

Explanation:

Since the function is continuous on [-1,1], the theorem does apply.

So, by the Mean Value Theorem for integrals, there is a number, c, in the interval (-1,1) such that

$f(c)(1-(-1))=\int_{-1}^1f(x)dx$$\begin{gathered} 2f(c)=\int_{-1}^1(x^3-16x)dx \\ \\ 2f(c)=[\frac{1}{4}x^4-\frac{16x^2}{2}]_{-1}^1 \\ \\ 2f(c)=[\frac{1}{4}-\frac{1}{4}]-[8-8] \\ \\ 2f(c)=0 \end{gathered}$$\begin{gathered} 2(c^3-16c)=0 \\ \\ c^2-16=0 \\ \\ c=0,4,-4 \end{gathered}$

The only value of c that is in the interval (-1,1) is

$c=0$

Final Answer:

The mean value theorem is applies the given function.