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.
 A: 
Your calculation is fine, except the interpretation. Note that,
$$h(-\sqrt{2-\sqrt3}) = h(\sqrt{2+\sqrt3}) = 2\sqrt2$$
As seen from the plot, $2\sqrt2 $ at $-\sqrt{2-\sqrt3}$ and $\sqrt{2+\sqrt3}$ are the two local minima, while $-2\sqrt2 $ are the local maxima.
A: 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
$$\dfrac{\pm\sqrt 6\pm\sqrt{42}}{6}.$$
A: Note that 
$$\frac{f(x)}{g(x)}=\frac{(x-1/x)^2+2}{x-1/x}=x-\frac1x+2\frac{1}{x-\frac1x}.
$$
Letting $u=x-1/x$ we take the derivative of $u+2/u$ to get $u'-\frac{2}{u^2}u'=u'(1-2/u^2)=0$.  As $u'$ is always positive we must have $2=u^2=(x-1/x)^2$ which is easily solved; $x=\pm\sqrt{2\pm\sqrt3}$.
A: If you are dealing with $h(x)=f(x)/g(x)$ then:
$$h'(x_0)=0=\frac{f'(x_0)g(x_0)-f(x_0)g'(x_0)}{[g(x_0)]^2}\to \frac{f'(x_0)}{g'(x_0)}=\frac{f(x_0)}{g(x_0)}   \quad \quad (1)$$
but, $f(x)=[g(x)]^2+2$, so
$$f'(x)=2g'(x)g(x)\to \frac{f'(x)}{g'(x)}=2g(x)\quad \quad (2)$$
From $(1)$ and $(2)$,
$$f(x_0)=2[g(x_0)]^2=[g(x_0)]^2+2$$
then,
$$g(x_0)=\pm \sqrt{2}=x_0-\frac{1}{x_0}$$
Can you finish?
A: Assuming that h(x)= f(x)/g(x), I would not actually calculate h itself.  Instead, use the "quotient rule": $\frac{h'(x)= f'(x)g(x)- f(x)g'(x)}{g^2(x)}$.  That will be 0 if and only if the numerator is 0: $f'(x)g(x)- f(x)g'(x)= 0$.  
$f(x)= x^2+ x^{-2}$ so $f'(x)= 2x+ 2x^{-3}$, $g(x)= x- x^{-1}$ so $g'(x)= 1- x^{-2}$.
So $f'(x)g(x)- f(x)g'(x)= (2x+ 2x^{-3})(x- x^{-1})- (x^2+ x^{-2})(1- x^{-2})= 0$
Seeing that "$x^{-3}$ and $x^{-1}$ I would now multiply both sides by $x^4$ (of course x cannot be 0):
$(2x^4+ 2)(x^2- 2)- (x^4+ 1)(x^2- 1)= 2x^6- 4x^4+ 2x^2- 4- x^6+ x^4- x^2- 1= -3x^4+ x^3- 5= 0$.
A: Assuming that 
$$h(x)=\frac{f(x)}{g(x)}\implies h'(x)=\frac{\left(x^2+1\right) \left(x^4-4 x^2+1\right)}{x^2 \left(x^2-1\right)^2}$$ Assuming $x\neq 0$ and $x\neq 1$ you are left with
$$x^4-4x^2+1=0 \implies (x^2)^2-4(x^2)+1=0$$ which is a quadratic in $x^2$.
