Let f : \mathbb{R} \[ \to \] \mathbb{R} be a polynomial function of degree four having extreme values at x = 4 and x = 5. If x→0limx2f(x)=5, then f(2) is equal to :
Solutions · 1
Official solution
\[x→0limx2f(x)=5x→0limx2ax4+bx3+cx2+dx+e)=5c=5 and d=e=0f(x)=ax4+bx3+5x2f′(x)=4ax3+3bx2+10x=x(4ax2+3bx+10)\]
has extremes at 4 and so f′(4)=0&f′(5)=0
so a=81&b=2−3
so f(2)=81×24−23×23+5×22=2−12+20=10