Question

The maximum value of $f\left(x\right)=2b{x}^2-x^4-3b$ is $g\left(b\right),$ where $b>0.$ If b varies then the minimum value of $g\left(b\right)$ is

Answer 1

$f\left(x\right)={-x}^4+2{bx}^2-3b;b>0$

$f\left(x\right)=-[x^4-{2bx}^2]-3b$

$=-[{\left(x^2\right)}^2-2b.x^2+b^2-b^2]-3b$

$=-[{\left(x^2\right)}^2-2.b.x^2+b^2]+b^2-3b$

$=-{(x^2-b)}^2+b^2-3b$

$=(b^2-3b)-{(x^2-b)}^2$

We know the square of a term will always be greater than zero $\Rightarrow {(x^2-b)}^2\ge 0$