Question

If the function $f\left(x\right)=x^4-2x^3+a{x}^2+bx$ on $[1,3]$ satisfies the conditions of Rolle's Theorem with $c=\frac{3}{2}$, then find $a$ and $b$.

Answer 1

Given $f\left(x\right)=x^4-2x^3+a{x}^2+bx$ on $[1,3]$

$⟹f^{\prime }\left(x\right)=4x^3-6x^2+2ax+b$

According to Rolle's theorem, if f(x) is continuous on $[a,b]$, differentiable on $(a,b)$ and $f\left(a\right)=f\left(b\right)$ then there exists some $c\in (a,b)$ such that $f^{\prime }\left(c\right)=0$

Given that the function $f\left(x\right)$ satisfies the conditions of Rolle's theorem with $c=\frac{3}{2}$

$f\left(1\right)=1^4-2(1{)}^3+a(1{)}^2+b\left(1\right)=1-2+a+b=a+b-1$

$f\left(3\right)=3^4-2(3{)}^3+a(3{)}^2+b\left(3\right)=81-54+9a+3b=9a+3b-27$

from Rolles's theorem, $f\left(1\right)=f\left(3\right)$

$⟹a+b-1=9a+3b-27$

$⟹8a+2b=26$

$⟹4a+b=13$ .......(1)

from Rolles's theorem,$f^{\prime }\left(c\right)=4c^3-6c^2+2ac+b=0$

$⟹f^{\prime }\left(\frac{3}{2}\right)=4(\frac{3}{2}{)}^3-6(\frac{3}{2}{)}^2+2a\left(\frac{3}{2}\right)+b=0$

$⟹\frac{27}{2}-\frac{27}{2}+3a+b=0$

$⟹3a+b=0$ .......(2)

(1) - (2) $⟹a=13$

substituting $a=13$ in (1)

$⟹4\left(13\right)+b=13$

$⟹b=-39$

Therefore, $a=13,b=-39$