Question

Verify Mean Value Theorem if $f\left(x\right)=x^3-5x^2-3x$ in the interval $[1,-3]$

Answer 1

Given : $f\left(x\right)=x^3-5x^2-3x$

It is a polynomial.

Therefore it is continuous in the interval $[1,3]$ and derivative in the interval $(1,3)$.

$f\left(x\right)=x^3-5x^2-3x$

$f^{\prime }\left(x\right)=3x^2-10x-3$

Let $a=1,b=3$

$f\text{'}\left(c\right)=3c^2-10c-3$

$f\left(1\right)=(1{)}^3-5(1{)}^2-3\left(1\right)$

$=1-5-3=-7$

$f\left(3\right)=(3{)}^3-5(3{)}^2-3\left(3\right)$

$=27-45-9=-27$

By Mean value theorem $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$⟹3c^2-10c-3=\frac{-27-(-7)}{3-1}=-\frac{20}{2}=-10$

$3c^2-10c-3=-10$

$3c^2-10c-3+10=0$

$3c^2-10c+7=0$

$3c^2-7c-3c+7=0$

$3c^2-3c-7c+7=0$

$3c(c-1)-7(c-1)=0$

$(3c-7)(c-1)=0$

$c=\frac{7}{3}orc=1$

Now $c=\frac{7}{3}\in (1,3)$

$c=\frac{7}{3}$

$f\text{'}\left(c\right)=0⟹3c^2-10c-3=0$

$c=\frac{10\pm \sqrt{100+36}}{6}$

$c=\frac{10\pm \sqrt{136}}{6}$

$=\frac{10\pm \sqrt{2}34}{6}$

$=\frac{5\pm \sqrt{34}}{3}$

$=3.61,-0.28$

None of these values $\in (1,3)$

Hence, Mean value theorem is not verified.