Answer:
c = 0 or c = -4
Step-by-step explanation:
According to the Mean Value Theorem
[tex]f^{'}(c) = \frac{f(0) - f(-6)}{0 - (-6)} = \frac{0 - (-6)^3 - 6*(-6)^2 - 6*(-6)}{6} = 36 - 36 + 6 = 6[/tex]
By taking the first derivative of f
[tex]f^{'}(x) = 3x^2 + 12x + 6[/tex]
Since[tex]f^{'}(c) = 6[/tex]
We can solve for c
[tex]3c^2 + 12c + 6 = 6[/tex]
[tex]3c^2 + 12c = 0[/tex]
[tex]c(c + 4) = 0[/tex]
c = 0 or c = -4
