Question

A polynomial function $f\left(x\right)$ is such that $f\left(2x\right)=f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right)$, then find the value of $f\left(3\right)$.

Answer 1

Let $f\left(x\right)$ be a 'n' degree polynomial

Degree of $f\left(2x\right)=n$

Degree of $f^{\prime }\left(x\right)=n-1$

Degree of$f^{\prime \prime }\left(x\right)=n-2$

Then using given condition

$n=n-1+n-2$

$\Rightarrow n=3$

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

$\Rightarrow f\left(2x\right)=8a{x}^3+4b{x}^2+2cx+d$

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

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

Putting in above conditions we get

$(8a-18a^2)x^3+(4b-18ab)x^2+(2c-4b^2-6ac)x+(d-2bc)=0$

Equating the coefficients of x with zero we get:

$8a-18a^2=0$

$\Rightarrow a=\frac{4}{9}$

$4b-18ab=0$

$\Rightarrow b=0$, as a is non-zero

$2c-4b^2-6ac=0$

$\Rightarrow c=0$

$d-2bc=0$

$\Rightarrow d=0$

So $f\left(x\right)=\frac{4x^3}{9}$

Hence $f\left(3\right)=12$