Question

Let $f\left(x\right)=x^3-16x$ and let $c$ be the number that satisfies the Mean value theorem for $f$ on the interval $[-4,2]$. What is $c$ ?

• A. $-1$
• B. $2$
• C. $0$
• D. $-2$

Answer 1

The correct option is D $-2$

Given : $f\left(x\right)=x^3-16(-4,2)$

According to mean value theorem : $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{(b-a)}$

${3c}^2-16=\frac{(-24)-0}{(2-(-4\left)\right)}=-4{3c}^2=16-4=12c^2=4c=\pm 2c\ne 2$

$-2$ Lies between $(-4,2)$

Hence $c=-2$ is the correct answer.