Question

Determine for what values of $x$, the function $f\left(x\right)=x^3+\frac{1}{x^3}(x\ne 0)$ is strictly increasing or strictly decreasing.

Answer 1

REF.Image.

given

$f\left(x\right)=x^3+\frac{1}{x^3}$

$=(x+\frac{1}{x}{)}^3-3\times x\times \frac{1}{x}(x+\frac{1}{x})$

$f\left(x\right)=(x+\frac{1}{x}{)}^3-3(x+\frac{1}{x})$

$f^{\prime }=3(x+\frac{1}{x}{)}^2(1-\frac{1}{x^2})-3(1-\frac{1}{x^2})$

$=3(1-\frac{1}{x^2})[x^2+\frac{1}{x^2}+2-1]$

$=3(1-\frac{1}{x^2})(x^2+\frac{1}{x^2}+1)$

Hence, f(x)is increasing for $xϵ[-\infty ,-1]\cup [1,\infty ]$

f(x)is strictly decreasing for $xϵ(-1,0)\cup (0,1)$