Official solution
let f(x) = ax5 + bx4 + cx3 + dx2 + ex + f
Given \[\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4\]
\[ \Rightarrow \] \[\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f} \over {{x^3}}}} \right)\] = 4
As this limit exists so d = e = f = 0
\[ \therefore \] f(x) = ax5 + bx4 + cx3
\[\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3}} \over {{x^3}}}} \right)\] = 4
\[ \Rightarrow \] 2 + c = 4
\[ \Rightarrow \] c = 2
\[ \therefore \] f(x) = ax5 + bx4 + 2x3
\[ \Rightarrow \] f'(x) = 5ax4 + 4bx3 + 6x2
As x = \pm 1 are its critical points so f'(x) = 0 at x = \pm 1.
f'(1) = 5a + 4b + 6 = 0 ....(1)
and f'(-1) = 5a - 4b + 6 = 0 .....(2)
Solving (1) and (2),
a = \[ - {6 \over 5}\] and b = 0
\[ \therefore \] f(x) = \[ - {6 \over 5}{x^5} + 2{x^3}\]
So f'(x) = -6x4 + 6x2 = 6x2(1 + x)(1 - x)
Sign scheme for f'(x)
It is clear that maxima at x = 1 and minima at x
= --1.