Answer
$$
f(x)=(x+1)^{\frac{4}{5}}
$$
(a)
The critical numbers : $x=-1$.
(b)
The function is increasing on interval $ (-1, \infty). $
(c)
The function is decreasing on interval $(-\infty, -1 ). $
Work Step by Step
$$
f(x)=(x+1)^{\frac{4}{5}}
$$
We find $f^{\prime}(x) $ first, using the power rule and the chain rule.
$$
\begin{aligned}
f^{\prime}(x) &=\frac{4}{5}(x+1)^{\frac{-1}{5}}\\
&=\frac{4}{5(x+1)^{\frac{1}{5}}}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime}(x)=0 $ , but here $f^{\prime}(x) $ is never $0$. Next, we find any values of $x$ where $f^{\prime}(x) $ fails to exist. This occurs whenever the denominator of $f^{\prime}(x) $ is $0$, so set the denominator equal to 0 and solve:
$$
\begin{aligned}
5(x+1)^{\frac{1}{5}} &=0\\
[(x+1)^{\frac{1}{5}}]^{5}&=(0)^{5}\\
x+1&= 0\\
x &= -1\\
\end{aligned}
$$
Since $f^{\prime}(-1) $ does not exist but $f(-1) $ is defined, $x=-1 $ is a critical number, the only critical number.
(a)
The critical numbers : $x=-1$.
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. The number $-1$ divides the number line into two intervals: $(-\infty, -1)$ and $(-1, \infty)$.
use a test point in each of these intervals as follows:
(1) Test a number in the interval $(-\infty, -1)$ say $-2$:
$$
\begin{aligned}
f^{\prime}(-2) &=\frac{4}{5((-2)+1)^{\frac{1}{5}}} \\
&=-\frac{4}{5}\\
& \lt 0
\end{aligned}
$$
we see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty, -1).$
(2) Test a number in the interval $(-1, \infty)$ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=\frac{4}{5((0)+1)^{\frac{1}{5}}} \\
&=\frac{4}{5((1)^{\frac{1}{5}}} \\
&=\frac{4}{5}\\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-1, \infty ).$
So,
(b) The function is increasing on interval $ (-1, \infty), $
(c) The function is decreasing on interval $(-\infty, -1 ). $
