Question

find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval.

Answer 1

Remember that

If f is continuous over [a,b] and differentiable over (a,b), then there exists c∈(a,b) such that

$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$

In this problem, we have the function

$f(x)=\frac{9}{x^3}$

over the interval [1,3]

so

f(a)=f(1)=9/(1)^3=9

f(b)=f(3)=9/(3)^3=1/3

substitute

$f^{\prime}(c)=\frac{\frac{1}{3}-9}{3-1}=\frac{-\frac{26}{3}}{2}=-\frac{26}{6}=-\frac{13}{3}$

Find out the first derivative f'(x)

$f^{\prime}(x)=-\frac{27}{x^4}$

Equate the first derivative to -13/3

$\begin{gathered} -\frac{27}{x^4}=-\frac{13}{3} \\ \\ x^4=\frac{27*3}{13} \\ \\ x=1.58 \end{gathered}$

therefore

The value of c is 1.58 (rounded to two decimal places)