Answer
$$
y= x^{\frac{1}{3}}+x^{\frac{4}{3}}
$$
(a)
The critical numbers : $x=\frac{-1}{4} $ and $x=0$ .
(b)
The function is increasing on interval $ (\frac{-1} {4} , \infty ) $.
(c)
The function is decreasing on interval $(-\infty , \frac{-1} {4} ) .$
Work Step by Step
$$
y= x^{\frac{1}{3}}+x^{\frac{4}{3}}
$$
We find $y^{\prime}(x) $ first, using the product rule and the chain rule.
$$
\begin{aligned}
y^{\prime}(x) &=\frac{1}{3}x^{\frac{-2}{3}}+\frac{4}{3}x^{\frac{1}{3}}\\
&=\frac{1+4x}{3x^{\frac{2}{3}}} \\
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $y^{\prime} =0 $. Now, solve the equation $y^{\prime} =0 $ to get
$$
\begin{aligned}
y^{\prime}(x) =\frac{1+4x}{3x^{\frac{2}{3}}} &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
1+4x &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
4x &=-1
\\
\Rightarrow\quad\quad\quad\quad\quad\\
x &=\frac{-1}{4}
\end{aligned}
$$
Next, we find any values of $x$ where $y^{\prime}(x) $ fails to exist. This occurs whenever the denominator of $y^{\prime}(x) $ is $0$, here we observe that the denominator $3x^{\frac{2}{3}}$ equal to 0, when $x=0$ So,
(a)
The critical numbers : $x=\frac{-1}{4} $ and $x=0$ .
We can still apply the first derivative test, however, to find where $y$ is increasing and decreasing. Now, check the sign of $y^{\prime}(x)$. in the intervals
$$
(-\infty, \frac{-1} {4} ), \quad (\frac{-1} {4} ,0 ) \quad \text {and }\quad ( 0 , \infty) .
$$
(1)
Test a number in the interval $(-\infty, \frac{-1} {4} ) $ say $-1 $:
$$
\begin{aligned}
y^{\prime}(-1) &=\frac{1+4(-1)}{3(-1)^{\frac{2}{3}}}\\
&=-1 \\
& \lt 0
\end{aligned}
$$
we see that $y^{\prime}(x)$ is negative in that interval, so $y(x)$ is decreasing on $(-\infty, \frac{-1} {4} )$.
(2)
Test a number in the interval $(\frac{-1} {4} , 0 ) $ say $ \frac{-1} {8}$:
$$
\begin{aligned}
y^{\prime}(\frac{-1} {8}) &=\frac{1+4(\frac{-1}{8})}{3(\frac{-1}{8})^{\frac{2}{3}}}\\
&=\frac{2}{3}\\
&\gt 0
\end{aligned}
$$
we see that $y^{\prime}(x)$ is positive in that interval, so $y(x)$ is increasing on $( \frac{-1} {4} , 0 ) $.
(3)
Test a number in the interval $( 0 , \infty ) $ say $1$:
$$
\begin{aligned}
y^{\prime}(1) &=\frac{1+4(1)}{3(1)^{\frac{2}{3}}}\\
&=\frac{5}{3} \\
&\gt 0
\end{aligned}
$$
we see that $y^{\prime}(x)$ is positive in that interval, so $y(x)$ is increasing on $ ( 0 , \infty ) $.
So,
b)
The function is increasing , on intervals $ (\frac{-1} {4} ,0 ) $ and $ (0, \infty ) $. Since $y(0)$ is defined then, the function is increasing on interval $ (\frac{-1} {4} , \infty ) $.
(c)
The function is decreasing on interval $(-\infty , \frac{-1} {4} ) .$
