Question

A cubic function f(x) vanishes at $x=0$ & has relative minimum/maximum at $x=-1$ and $x=\frac{1}{3}.$ If ${\int }_{-1}^{1}f\left(x\right)dx=\frac{14}{3},$ then find the cubic $f\left(x\right)$

• A. $f\left(x\right)=x^3-x^2-x+2$
• B. $f\left(x\right)=x^3+x^2-x+2$
• C. $f\left(x\right)=x^3-x^2+x-2$
• D. None of these

Answer 1

The correct option is D None of these

Let the polynomial be $f\left(x\right)=a{x}^3+b{x}^2+cx+d$

$f\left(0\right)=0$ implies $d=0$.

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

Now ${\frac{d\left(fx\right)}{dx}}_{x=-1,\frac{1}{3}}=0$

Hence

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

$f^{\prime }(-1)=0$ implies

$3a-2b+c=0$

$f^{\prime }\left(\frac{1}{3}\right)=0$

This gives us

$\frac{a}{3}+\frac{2b}{3}+c=0$

$a+2b+3c=0$

Hence we get two equations,

$3a-2b+c=0$ and

$a+2b+3c=0$

Adding both the equations, yields

$4a+4c=0$
$c=-a$
Therefore $b=a$

Hence, $f\left(x\right)=a{x}^3+a{x}^2-ax$

Now

${\int }_{-1}^{1}f\left(x\right).dx$

$={\int }_0^{1}f\left(x\right)+f(-x).dx$

$=a{\int }_0^1(x^3+x^2-x)+(-x^3+x^2+x).dx$

$=2a{\int }_0^1{x}^2.dx$

$=2a[\frac{x}{3}{]}_0^1$

$=\frac{2a}{3}$

$=\frac{14}{3}$

Hence, $a=7$

$\therefore f\left(x\right)=7x^3+7x^2-7x$.