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.

$f\left(x\right)=x^{\frac{4}{3}}-x^{\frac{1}{3}}$

The given function is$f\left(x\right)=x^{\frac{4}{3}}-x^{\frac{1}{3}}.$

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

$$\begin{gathered} f\left(x\right)=0 \\ {x}^{\frac{4}{3}}-x^{\frac{1}{3}}=0 \\ {x}^{\frac{1}{3}}(x-1)=0 \\ x=0\&1 \end{gathered}$$

So the roots of the function are$x=0\&1.$

Determine the first derivative o the function.

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}\left(x^{\frac{4}{3}}-x^{\frac{1}{3}}\right) \\ =\frac{4}{3}{x}^{\frac{4}{3}-1}-\frac{1}{3}{x}^{\frac{1}{3}-1} \\ =\frac{4}{3}{x}^{\frac{1}{3}}-\frac{1}{3}{x}^{\frac{-2}{3}} \end{gathered}$$

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

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ \frac{4}{3}{x}^{\frac{1}{3}}-\frac{1}{3}{x}^{\frac{-2}{3}}=0 \\ 4x^{\frac{1}{3}}-x^{-\frac{2}{3}}=0 \\ 4x^{\frac{1}{3}-\left(\frac{-2}{3}\right)}-1=0 \\ 4x-1=0 \\ x=\frac{1}{4} \end{gathered}$$

So critical point of the function is at$x=\frac{1}{4}.$

Substitute $x=\frac{1}{4}$in the function.

$$\begin{gathered} f\left(x\right)=x^{\frac{4}{3}}-x^{\frac{1}{3}} \\ f\left(\frac{1}{4}\right)={\left(\frac{1}{4}\right)}^{\frac{4}{3}}-{\left(\frac{1}{4}\right)}^{\frac{1}{3}} \\ =4^{\frac{-4}{3}}-4^{\frac{-1}{3}} \\ =4^{\frac{-1}{3}}\left(\frac{1}{4}-1\right) \\ =4^{\frac{-1}{3}}\left(\frac{3}{4}\right) \\ =\frac{3}{4^{{\displaystyle \frac{4}{3}}}} \end{gathered}$$

So the function has local minima at $x=\frac{1}{4}.$

the function has global minima also at$x=\frac{1}{4}$because the function value increases for $x>\frac{1}{4}\mathrm{or}x<\frac{1}{4}.$

Determine the second derivative of the function.

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{4}{3}{x}^{\frac{1}{3}}-\frac{1}{3}{x}^{\frac{-2}{3}}\right) \\ =\frac{1}{3}·\frac{4}{3}{x}^{\frac{1}{3}-1}-\frac{-2}{3}·\frac{1}{3}{x}^{\frac{-2}{3}-1} \\ =\frac{4}{9}{x}^{\frac{-2}{3}}+\frac{2}{9}{x}^{\frac{-5}{3}} \end{gathered}$$

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

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=0 \\ \frac{4}{9}{x}^{\frac{-2}{3}}+\frac{2}{9}{x}^{\frac{-5}{3}}=0 \\ 2x^{\frac{-2}{3}}+x^{\frac{-5}{3}}=0 \\ {x}^{\frac{-2}{3}}\left(2-x^{-1}\right)=0 \\ 2-\frac{1}{x}=0 \\ x=\frac{1}{2} \end{gathered}$$

So function has an inflection point at$x=\frac{1}{2}.$