91Ó°ÊÓ

We have been given the function

$f\left(x\right)=x^{4/3}$

We have to find the derivative of the given function

$f\left(x\right)=x^{4/3}$

Therefore,

$\begin{array}{rl}{f}^{\mathrm{′}}\left(x\right)& =\frac{4}{3}{x}^{\frac{4}{3}−1}\\ & =\frac{4}{3}{x}^{\frac{1}{3}}\end{array}$

Since the derivative is always defined and continuous, the critical points of the function are just the places where $f\text{'}\left(x\right)=0$; that is,

localid="1648629917878" $\begin{array}{rl}\frac{4}{3}{x}^{\frac{1}{3}}& =0\\ x^{\frac{1}{3}}& =0\\ x& =0\end{array}$

These critical points divide the real number line into two parts$(−\mathrm{∞},0),(0,\mathrm{∞})$

Here, for testing take the sign of $f\text{'}\left(x\right)$at one point in that interval.

Let the value of $x$berole="math" localid="1648629662005" $x=−1,x=1$

$\begin{array}{rl}{f}^{\mathrm{′}}(−1)& =\frac{4}{3}(−1{)}^{\frac{1}{3}}\\ & =−\frac{4}{3}<0\\ f^{\mathrm{′}}\left(1\right)& =\frac{4}{3}(1{)}^{\frac{1}{3}}\\ & =\frac{4}{3}>0\end{array}$

Let us draw the sign chart,

Therefore, $f\text{'}$changes from negative to positive at $x=0$; that is $f$changes from decreasing to increasing at $x=0$. The function is concave up at $(-∞,∞)$.