Let f x be a polynomial of degree 5 such that x = ± 1 are its critical points. If l i m x → 0 2 +…

$f\left(x\right)=a{x}^5+b{x}^4+c{x}^3$

Consider, $\underset{x\to 0}{\lim }\left(2+\frac{a{x}^5+b{x}^4+c{x}^3}{x^3}\right)=4$

$\Rightarrow 2+c=4$

$\Rightarrow c=2$

Differentiating $f\left(x\right)$ with respect to $x$

$f^{\text{'}}\left(x\right)=5a{x}^4+4b{x}^3+6x^2$

$=x^2\left(5a{x}^2+4bx+6\right)$

Critical points of the function are given as $\pm 1$.

$f^{\text{'}}\left(1\right)=0\Rightarrow 5a+4b+6=0$

$f^{\text{'}}\left(-1\right)=0\Rightarrow 5a-4b+6=0$
Solving both we get, $b=0$, $a=-\frac{6}{5}$
So, $f\left(x\right)=\frac{-6}{5}{x}^5+2x^3$
And $f^{\text{'}}\left(x\right)=-6x^4+6x^2$

$=-6x^2\left(x+1\right)\left(x-1\right)$

Hence, local minima at $x=-1$ and local maxima at $x=1$.