Question

Let $f\left(x\right)$ be a polynomial of degree four having extreme values at $x=1$ and $x=2$. If $\underset{x\to 0}{lim}(1+\frac{f\left(x\right)}{x^2})=3$, then the maximum value of $f\left(x\right)$ in $x\in [0,2]$ is

Answer 1

Let $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e$
Since, $\underset{x\to 0}{lim}(1+\frac{f\left(x\right)}{x^2})=3$
$\Rightarrow \underset{x\to 0}{lim}(1+\frac{a{x}^4+b{x}^3+c{x}^2+dx+e}{x^2})=3$

In order for limit to exist, $d,e$ must be $0$
$\Rightarrow \underset{x\to 0}{lim}(1+a{x}^2+bx+c)=3$
$\Rightarrow 1+c=3$
$\Rightarrow c=2$
$\therefore f\left(x\right)=a{x}^4+b{x}^3+2x^2$

Now $f^{\prime }\left(1\right)=0$
$\Rightarrow f^{\prime }\left(1\right)=4a+3b+4=0\cdots \left(1\right)$
$f^{\prime }\left(2\right)=0\Rightarrow 32a+12b+8=0\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right),$ we have
$a=0.5,b=-2$
$\Rightarrow f\left(x\right)=0.5x^4-2x^3+2x^2$

For minimum/maximum value
$f^{\prime }\left(x\right)=0\Rightarrow 2x^3-6x^2+4x=0\Rightarrow x(x^2-3x+2)=0\Rightarrow x(x-1)(x-2)=0\Rightarrow x=0,1,2$

Now, $f^{\prime \prime }\left(x\right)=6x^2-12x+4$
$\Rightarrow f^{\prime \prime }\left(0\right)=f^{\prime \prime }\left(2\right)=4>0$
and $f^{\prime \prime }\left(1\right)=-2<0$
Hence, the maximum value of $f\left(x\right)$ in the given interval is $\frac{1}{2}.$