Answer
$$
f(x)= x^{\frac{2}{3}}-x^{\frac{5}{3}}
$$
(a)
The critical numbers : $x=\frac{2}{5} $ and $x=0$ .
(b)
The function is increasing , on interval $ (0, \frac{2} {5} ). $
(c)
The function is decreasing on intervals $(-\infty , 0) $ and $ ( \frac{2} {5} , \infty ) .$
Work Step by Step
$$
f(x)= x^{\frac{2}{3}}-x^{\frac{5}{3}}
$$
We find $f^{\prime}(x) $ first, using the product rule and the chain rule.
$$
\begin{aligned}
f^{\prime}(x) &=\frac{2}{3} x^{\frac{-1}{3}}- \frac{5}{3}x^{\frac{2}{3}} \\
&=\frac{2-5x}{3x^{\frac{1}{3}}} \\
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime} =0 $. Now, solve the equation $f^{\prime} =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =\frac{2-5x}{3x^{\frac{1}{3}}} &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
2-5x &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
5x &=2
\\
\Rightarrow\quad\quad\quad\quad\quad\\
x &=\frac{2}{5}
\end{aligned}
$$
Next, we find any values of $x$ where $f^{\prime}(x) $ fails to exist. This occurs whenever the denominator of $f^{\prime}(x) $ is $0$, here we observe that the denominator $3x^{\frac{1}{3}}$ equal to 0, when $x=0$ So,
(a)
The critical numbers : $x=\frac{2}{5} $ and $x=0$ .
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. Now, check the sign of $f^{\prime}(x)$. in the intervals
$$
(-\infty, 0 ), \quad (0 , \frac{2} {5} ) \quad \text {and }\quad ( \frac{2} {5} , \infty) .
$$
(1)
Test a number in the interval $(-\infty, 0) $ say $-1$:
$$
\begin{aligned}
f^{\prime}(-1) &=\frac{2-5(-1)}{3(-1)^{\frac{1}{3}}} \\
&=-\frac{7}{3} \\
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty, 0)$.
(2)
Test a number in the interval $(0 , \frac{2} {5} ) $ say $ \frac{1} {5}$:
$$
\begin{aligned}
f^{\prime}( \frac{1}{5}) &=\frac{2-5( \frac{1}{5})}{3( \frac{1}{5})^{\frac{1}{3}}} \\
&=\frac{5^{\frac{1}{3}}}{3} \\
&\approx 0.569 \\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(0 , \frac{2} {5} ) $.
(3)
Test a number in the interval $( \frac{2} {5} , \infty ) $ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=\frac{2-5( 1)}{3( 1)^{\frac{1}{3}}} \\
&=-1 \\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $ ( \frac{2} {5} , \infty ) $.
So,
b)
The function is increasing , on interval $ (0, \frac{2} {5} ). $
(c)
The function is decreasing on intervals $(-\infty , 0) $ and $ ( \frac{2} {5} , \infty ) .$
