Solution
The given function is
[tex]f(x)=4x^3[/tex]With given interval
[tex]\lbrack1,2\rbrack[/tex]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:
[tex]f^{\prime}(c)=\frac{f(2)-f(1)}{2-1}[/tex]This gives
[tex]\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}[/tex]Differentiating the given function gives:
[tex]f^{\prime}(x)=12x^2[/tex]Equate f'(c) and f'(x)
This gives
[tex]x^2=\frac{28}{12}[/tex]Solve the equation for x
[tex]\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}[/tex]Therefore the values of c are
[tex]\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}[/tex]