If 5f(x)+4f((1)/(x))=x^2-2, 0 and y=9x^2f(x), then y is strictly increasing in :

Question

If $5f(x)+4f(\frac{1}{x})={x}^{2}-2,\forallx
e 0$ and $y=9{x}^{2}f(x)$, then $y$ is strictly increasing in :

Accepted Answer

$5f(x)+4f(1/x)={x}^{2}-2$ ........(1)

Replace $x$ by $1/x$

$5f(1/x)+4f(x)=\frac{1}{{x}^{2}}-2$ ..........(2)

Multiply equation (1) by 5 and multiply equation (2) by 4 and then subtract equation (2) from (1)

$\begin{gathered} 25f(x)-16f(x)=5{x}^{2}-10-\frac{4}{{x}^{2}}+8 \ 9f(x)=5{x}^{2}-\frac{4}{{x}^{2}}-2 \ 9f(x)=\frac{5{x}^{4}-4-2{x}^{2}}{{x}^{2}} \end{gathered}$

$\begin{gathered} y=9{x}^{2}f(x) \ y=5{x}^{4}-2{x}^{2}-4 \ {y}^{'}=20{x}^{3}-4x \ {y}^{'}>0 \ 20{x}^{3}-4x>0 \ 5{x}^{3}-x>0 \ x(5{x}^{2}-1)>0 \end{gathered}$

$x\in (-\frac{1}{\sqrt{5}},0)\cup (\frac{1}{\sqrt{5}},\infty)$

Solutions · 1