Question

Find the intervals in which the function f given by $f\left(x\right)=x^3+\frac{1}{x^3},x\ne 0$ is
increasing

Find the intervals in which the function f given by $f\left(x\right)=x^3+\frac{1}{x^3},x\ne 0$ is
decreasing

Answer 1

Given, $f\left(x\right)=x^3+\frac{1}{x^3}$
On differentiating w.r.t. x, we get $f^{\prime }\left(x\right)=3x^2-\frac{3}{x^4}$
(a) f(x) is increasing, if $f^{\prime }\left(x\right)\ge 0$
$\Rightarrow 3x^2-\frac{3}{x^4}\ge 0\Rightarrow x^6\ge 1\Rightarrow (x^2{)}^3\ge 1\Rightarrow x^2\ge \Rightarrow xϵ(-\infty ,-1)\cup (1,\infty )$
Hence, f is increasing in $(-\infty ,-1)\cup (1,\infty )orx<-1andx>1.$

Given, $f\left(x\right)=x^3+\frac{1}{x^3}$
On differentiating w.r.t. x, we get $f^{\prime }\left(x\right)=3x^2-\frac{3}{x^4}$
(b) f(x) is decreasing, if $f^{\prime }\left(x\right)\le 0$
$\Rightarrow 3x^2-\frac{3}{x^4}\le 0\Rightarrow x^6\le 1\Rightarrow (X^2{)}^3\le 1\Rightarrow x^2\le 1\Rightarrow -1\le x\le 1$
$\therefore$ f is decreasing in -1 < x < 1.