<s>[INST] If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$$
$$h(x)=\dfrac{x^2+x^2-2+2}{x-\dfrac{1}{x}}$$
$$h(x)=x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}$$
Case $1$: $x-\dfrac{1}{x}>0$
$$\dfrac{x^2-1}{x}>0$$
$$x\in(-1,0) \cup (1,\infty)$$
$$AM\ge GM$$
$$\dfrac{x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}}{2}>\sqrt{2}$$
$$h(x)\ge 2\sqrt{2}$$
$$x-\dfrac{1}{x}=\dfrac{2}{x-\dfrac{1}{x}}$$
$$x^2+\dfrac{1}{x^2}-2=2$$
$$x^2+\dfrac{1}{x^2}=4$$
$$x^4-4x^2+1=0$$
$$x^2=2\pm\sqrt{3}$$
Only $x=\sqrt{2+\sqrt{3}},-\sqrt{2-\sqrt{3}}$ are the valid solutions.
Case $2$: $x-\dfrac{1}{x}<0$
$$x\in(-\infty,-1) \cup (0,1)$$
$$h(x)=-\left(\dfrac{1}{x}-x+\dfrac{2}{\dfrac{1}{x}-x}\right)$$
By $AM\ge GM$, $h(x)\ge-2\sqrt{2}$
We will get this minimum value at $-\sqrt{2+\sqrt{3}},\sqrt{2-\sqrt{3}}$
So answer should have been $-2\sqrt{2}$ but actual answer is $2\sqrt{2}$. What am I missing here.
 [/INST] With $t:=x-\dfrac1x$
$$h(x)=\left(x^2+\frac1{x^2}\right)\left(x-\frac1x\right)=(t^2-1)t=t^3-2t.$$
Then by the chain rule
$$h'(x)=0\iff (3t^2-2)\left(1-\frac1{x^2}\right)=0.$$
We have the two solutions $x=\pm1$, and the solutions of
$$x-\frac1x=\pm\sqrt{\frac23},$$
which are
$$\dfrac{\pm\sqrt 6\pm\sqrt{42}}{6}.$$
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