Question

Let $f\left(x\right)=a{x}^5+b{x}^3+c\sin x+d{\tan }^{3}x-1$ .

If $f(-\alpha )=-5$, then $f\left(\alpha \right)$ equals

Answer 1

$f(-\alpha )$
$=-a{\alpha }^5-b{\alpha }^3-c\sin \alpha -d\tan {\alpha }^3-1$
$=-(a{\alpha }^5+b{\alpha }^3+c\sin \alpha +d\tan {\alpha }^3+1)$

$=-f\left(\alpha \right)-2$ {We can see this step after subsituting the value of $f\left(x\right)$ at $\alpha$ and simplifying we get the last step]

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