Question

Minimum value of $x^2+\frac{1}{x^2+1}-3$ is:-

Answer 1

Minimum value of

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

$f^1=2x+\frac{2x}{-{(x^2+1)}^2}=0$.

$x(1-{(x^2+1^2)}^2)=0$.

$X=0$

$f\left(0\right)=0+\frac{1}{1}-3=-2$

Minimum value is $-2$.