Question

Determine whether the Mean Value theorem can be applied to f on the closed interval [a, b]. (Select all that apply.)

Answer 1

Solution

The given function is

$f(x)=4x^3$

With given interval

$\lbrack1,2\rbrack$

The function is differentiable on the open interval (1,2) and it is continuous on the closed interval [1,2]

Therefore mean value theorem can be used

Calculating the c value iit follows:

$f^{\prime}(c)=\frac{f(2)-f(1)}{2-1}$

This gives

$\begin{gathered} f^{\prime}(c)=\frac{4(2)^3-4(1)^3}{1} \\ f^{\prime}(c)=\frac{32-4}{1} \\ f^{\prime}(c)=28 \end{gathered}$

Differentiating the given function gives:

$f^{\prime}(x)=12x^2$

Equate f'(c) and f'(x)

This gives

$x^2=\frac{28}{12}$

Solve the equation for x

$\begin{gathered} x^2=\frac{28}{12} \\ x^2=\frac{7}{3} \\ x=\pm\sqrt{\frac{7}{3}} \\ x=\sqrt{\frac{7}{3}},x=-\sqrt{\frac{7}{3}} \end{gathered}$

Therefore the values of c are

$\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}$