If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$

Question

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.

Answer 1

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; $x=\pm\sqrt{2\pm\sqrt3}$.

Answer 2

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.

Answer 3

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?

Answer 4

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): $(2x^4+ 2)(x^2- 2)- (x^4+ 1)(x^2- 1)= 2x^6- 4x^4+ 2x^2- 4- x^6+ x^4- x^2- 1= -3x^4+ x^3- 5= 0$.

Answer 5

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 $x^2$.

Answer 6

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 $$\dfrac{\pm\sqrt 6\pm\sqrt{42}}{6}.$$