Question

Verify mean value theorem for the function $f\left(x\right)=x^3-5x^2-3x$, in the interval $[a,b]$, where $a=1$ and $b=3$. Find all $cϵ(1,3)$ for which $f^1\left(c\right)=0$.

Answer 1

The given function is $f\left(x\right)=x^3-5x^2-3x$
$f$ being a polynomial function, so it is continuous in $[1,3]$ and is differentiable in $(1,3)$ whose derivative is $3x^2-10x-3$.
$f\left(1\right)=1^3-5\cdot 1^2-3\cdot 1=-7,f\left(3\right)=3^3-5\cdot 3^2-3\cdot 3=-27$
$\therefore \frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{-27-(-7)}{3-1}=-10$
Mean Value Theorem states that there exist a point $c\in (1,3)$ such that $f^{\prime }\left(c\right)=-10$
$\Rightarrow 3c^2-10c-3=-10$
$\Rightarrow 3c^2-10c+7=0$
$\Rightarrow 3c^2-3c-7c+7=0$
$\Rightarrow 3c(c-1)-7(c-1)=0$
$\Rightarrow (c-1)(3c-7)=0$
$\Rightarrow c=1,\frac{7}{3}$, where $c=\frac{7}{3}\in (1,3)$
Hence, Mean Value Theorem is verified for the given function and $c=\frac{7}{3}\in (1,3)$ is the point for which $f^{\prime }\left(c\right)=0$