Find all roots, local maxima and minima, and inflection points of each function f. In addition, determine whether any local extrema are also global extrema on the domain of f.

$f\left(x\right)={\tan }^{-1}{x}^2$

The given function is$f\left(x\right)={\tan }^{-1}{x}^2.$

Equate the function to zero for the root of the function.

$$\begin{gathered} f\left(x\right)=0 \\ {\tan }^{-1}{x}^2=0 \\ {x}^2=\tan 0 \\ {x}^2=0 \\ x=0 \end{gathered}$$

So the root of the function is$x=0.$

Determine the first derivative o the function.

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}\left({\tan }^{-1}{x}^2\right) \\ =\frac{2x}{1+x^2} \end{gathered}$$

The first derivative becomes zero at the critical point of the function.

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ \frac{2x}{1+x^2}=0 \\ 2x=0 \\ x=0 \end{gathered}$$

So critical point of the function is at$x=0.$

Substitute $x=0$in the function.

$$\begin{gathered} f\left(x\right)={\tan }^{-1}{x}^2 \\ f\left(0\right)={\tan }^{-1}{\left(0\right)}^2 \\ f\left(0\right)=0 \end{gathered}$$

So the function has local minima at$x=0$

the function has global minima also at$x=0$because the function value increases for$x>0orx<0.$

Determine the second derivative of the function.

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{2x}{1+x^2}\right) \\ =\frac{2\left(1+x^2\right)-2x\left(2x\right)}{{\left(1+x^2\right)}^2} \\ =\frac{2+2x^2-4x^2}{{\left(1+x^2\right)}^2} \\ =\frac{2-2x^2}{{\left(1+x^2\right)}^2} \end{gathered}$$

The second derivative becomes zero at the inflection point of the function.

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=0 \\ \frac{2-2x^2}{{\left(1+x^2\right)}^2}=0 \\ 2-2x^2=0 \\ {x}^2=1 \\ x=\pm 1 \end{gathered}$$

So function has an inflection point at$x=\pm 1.$