Let f(x) be a polynomial of degree 5, and have extrema at x=1 and x=-1.

Question

Let $f(x)$ be a polynomial of degree 5, and have extrema at $x=1$ and $x=-1$. If $\underset{x	o 0}{lim}(\frac{f(x)}{{x}^{3}})=-5$, then $f(2)-f(-2)$ is equal to:

Accepted Answer

Given $f(x)$ is a polynomial of degree $5$

Also, ${f}^{'}(1)=0$; ${f}^{'}(-1)=0$.

Also $\because \underset{x	o 0}{lim}\frac{f(x)}{{x}^{3}}=-5$ (fixed and finite)

$	herefore f(0)=0$; ${f}^{'}(0)=0$; ${f}^{''}(0)=0$

$\frac{{f}^{'''}(0)}{6}=-5	o {f}^{'''}(0)=-30$

Hence, let

${f}^{'}(x)=(ax+b)(x-0)(x-1)(x+1)$

$	o {f}^{'}(x)=a{x}^{4}+b{x}^{3}-a{x}^{2}-bx$

${f}^{''}(x)=4a{x}^{3}+3b{x}^{2}-2ax-b$

$	herefore {f}^{''}(0)=0	o b=0$

${f}^{'''}(x)=12a{x}^{2}+6bx-2a$

$	herefore {f}^{'''}(0)=-2a=-30$

$	o a=15$

$	herefore {f}^{'}(x)=15x.x.(x-1)(x+1)$

$	o {f}^{'}(x)=15{x}^{4}-15{x}^{2}$

$	o f(x)=3{x}^{5}-5{x}^{3}+C$

$	herefore f(0)=0	o C=0$

$	herefore f(x)=3{x}^{5}-5{x}^{3}$

$	herefore f(2)-f(-2)=112$

Solutions · 1