f(x)=ax5+bx4+cx3
Consider, x→0lim(2+x3ax5+bx4+cx3)=4
⇒2+c=4
⇒c=2
Differentiating f(x) with respect to x
f′(x)=5ax4+4bx3+6x2
=x2(5ax2+4bx+6)
Critical points of the function are given as ±1.
f′(1)=0⇒5a+4b+6=0
f′(−1)=0⇒5a−4b+6=0
Solving both we get, b=0, a=−56
So, f(x)=5−6x5+2x3
And f′(x)=−6x4+6x2
=−6x2(x+1)(x−1)
Hence, local minima at x=−1 and local maxima at x=1.