Question

Let $f\left(x\right)=x^3+6x^2+6x$ and let $c$ be the number that satisfies the Mean value theorem for $f$ on the interval $[-6,0]$. What is $c$ ?

• A. $-5$
• B. $-4$
• C. $-3$
• D. $-2$

Answer 1

The correct option is B $-4$

MVT states that there exists some $c$ such that $f^1\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Here $a=-6$ and $b=0$

$f\left(a\right)=f(-6)=-216+216-36=-36$

$f\left(b\right)=f\left(0\right)=0$

$f^1\left(c\right)=3c^2+12c+6$

therefore $3c^2+12c+6=\frac{36}{6}=6$

$\Rightarrow 3c^2+12c=0$

$\Rightarrow c=-4$