Remember that
If f is continuous over [a,b] and differentiable over (a,b), then there exists c∈(a,b) such that
[tex]f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}[/tex]
In this problem, we have the function
[tex]f(x)=\frac{9}{x^3}[/tex]
over the interval [1,3]
so
f(a)=f(1)=9/(1)^3=9
f(b)=f(3)=9/(3)^3=1/3
substitute
[tex]f^{\prime}(c)=\frac{\frac{1}{3}-9}{3-1}=\frac{-\frac{26}{3}}{2}=-\frac{26}{6}=-\frac{13}{3}[/tex]
Find out the first derivative f'(x)
[tex]f^{\prime}(x)=-\frac{27}{x^4}[/tex]
Equate the first derivative to -13/3
[tex]\begin{gathered} -\frac{27}{x^4}=-\frac{13}{3} \\ \\ x^4=\frac{27*3}{13} \\ \\ x=1.58 \end{gathered}[/tex]
therefore
The value of c is 1.58 (rounded to two decimal places)
