Question

It is given that the Rolle's theorem holds for the function $f\left(x\right)=x^3+b{x}^2+cx,xϵ[1,2]$ at the point $x=\frac{4}{3}$. Find the values of $b$ and $c$.

Answer 1

Given $f\left(x\right)=x^3+b{x}^2+cx$

As we know that

Rolle's theorem states that if $f\left(x\right)$ be continuous on $[a,b]$,differentiable on $(p,q)$ and $f\left(p\right)=f\left(q\right)$ then there exists some $r\in (p,q)$ such that $f^{\prime }\left(r\right)=0$

Given $p=1,q=2$ and $r=\frac{4}{3}$

$f\left(1\right)=(1{)}^3+b(1{)}^2+c\left(1\right)=1+b+c$

$f\left(2\right)=(2{)}^3+b(2{)}^2+c\left(2\right)=8+4b+2c$

According to Rolle's theorem

$f\left(1\right)=f\left(2\right)$

$⟹1+b+c=8+4b+2c$

$⟹c=-7-3b\cdots \left(1\right)$

$f^{\prime }\left(x\right)=3x^2+2bx+c$

According to Rolle's theorem

$f^{\prime }\left(r\right)=f^{\prime }\left(\frac{4}{3}\right)=0$

$3(\frac{4}{3}{)}^2+2b(\frac{4}{3})+c=0$

$⟹\frac{16}{3}+\frac{8b}{3}+c=0$

$⟹16+8b+3c=0$

$⟹16+8b+3(-7-3b)=0$ $(\because \text{from}(1\left)\right)$

$⟹16+8b-21-9b=0$

$⟹b=-5$

$⟹c=-7-3b=-7-3(-5)=15-7=8$

Hence $b=8,c=-5$