Minimum and maximum..?

Let $f(x) = \frac {x^4-2x^2+3} {(x^2+1)^2}$ is at its extrema (maximum and minimum), when $f(x) = k$ ; that is $k = a$ or $b$ .

That is when $f(x) = \dfrac {x^4-2x^2+3} {(x^2+1)^2} = k \quad \Rightarrow x^4-2x^2+3 = k(x^2+1)^2$

$\Rightarrow x^4-2x^2+3 = kx^4+2kx^2 + k \quad \Rightarrow (1-k)x^4-2(1+k)x^2+3-k = 0$

Now when $f(x) = k$ , $\dfrac {df(x)}{dx} = 0\quad \Rightarrow 4(1-k)x^3 - 4(1+k)x =0$

$\Rightarrow x\left[ (1-k)x^2 - (1+k)\right] =0 \quad \Rightarrow x = 0\quad$ or $\quad x^2 = \dfrac {1+k}{1-k}$

Substituting $k = \frac {x^4-2x^2+3} {(x^2+1)^2}$ , we have:

$x^2 = \dfrac {(x^2+1)^2 + x^4-2x^2+3} {(x^2+1)^2 - x^4+2x^2-3} = \dfrac {2x^4+4} {4x^2-2} = \dfrac {x^4+2} {2x^2-1}$

$\Rightarrow 2x^4 - x^2 = x^4+2 \quad \Rightarrow x^4 - x^2 - 2 = 0$

$\Rightarrow (x^2+1)(x^2-2)=0 \quad \Rightarrow x = \pm \sqrt{2}\quad$ taking only the real roots.

We find that $f(0) = 3\quad$ and $\quad f(\pm \sqrt{2}) = \dfrac {4-4+3}{(2+1)^2} = \dfrac {1}{3}$

$\Rightarrow m = \frac {1}{3}\quad$ and $\quad n = 3\quad \Rightarrow mn = \boxed {1}$

let $f(x)=\frac{x^4-2x^2+3}{(x^2+1)^2}$

manipulating $f(x)=1-\frac{4}{x^2+1}+\frac{6}{(x^2+1)^2}$

let $g(x)=+\frac{6}{(x^2+1)^2}-\frac{4}{x^2+1}$ is maximum obviously at $x=0$ which is equal to 2

At $x=0$ , $f(x)=m=3$ which is its maximum value

Now $\frac{dg(x)}{dx}=0\quad\Rightarrow-\frac{12}{(x^2+1)^3}+\frac{4}{(x^2+1)^2}=0$

$\Rightarrow3=x^2+1\quad\Rightarrow x^2=2$

Putting $x^2=2$ $g(x)=\frac{-2}{3}$ and $f(x)=n=\frac{1}{3}$ which is its minimum value.

Therefore $mn=1$ .