Question

Prove that $f^{\prime }\left(1\right)+f(-1)=-f\left(0\right)$, if $f\left(x\right)=x^5+x^3-2x-3$

Answer 1

$f\left(x\right)=x^5+x^3-2x-3$

So $f^{\prime }\left(x\right)=5x^4+3x^2-2$

$\therefore f^{\prime }\left(1\right)=5(1{)}^4+3(1{)}^2-2$

$=5+3-2$

$=6$

$Alsof\left(1\right)=(1{)}^5+(1{)}^3-2\left(1\right)-3$

$=1+1-2-3$

$=2-2-3$

$=-3$

$\therefore f\left(0\right)=-3$

and $f(-1)=(-1{)}^5+(-1{)}^3-2(-1)-3$

$=-1-1+2-3$

$\Rightarrow -3$

So $f^{\prime }\left(1\right)+f(-1)=6-3=3=-f\left(0\right)$.