Answer
$$
f(x)=x^{4}+4 x^{3}+4x^{2}+1
$$
(a)
The critical numbers are: $-2 , -1,$ and $0$.
(b)
The function is increasing on interval $(-2, -1 ) \text { and } (0, \infty). $
(c)
The function is decreasing on interval $(-\infty, -2 ) \text { and } (-1, 0). $
Work Step by Step
$$
f(x)=x^{4}+4 x^{3}+4x^{2}+1
$$
First, find the points where the derivative $f^{\prime }$ is $0$.
Here
$$
\begin{aligned}
f^{\prime}(x) &=(4)x^{3}+4 (3)x^{2}+4 (2)x \\
&=4x^{3}+12x^{2}+8x\\
&=4x( x^{2}+3 x+2) \\
\end{aligned}
$$
Solve the equation $f^{\prime}(x) =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =4x( x^{2}+3 x+2)&=0\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
f^{\prime}(x)=4x(x+2)(x+1)&=0 \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x =-1 , \quad x =0, \quad &\text {and} \quad x =-2 \\
\end{aligned}
$$
(a)
The critical numbers are: $-2 , -1,$ and $0$.
Now, we can use the first derivative test.
Check the sign of $f^{\prime}(x)$ in the intervals
$$
(-\infty, -2 ), \quad ( -2, -1), \quad ( -1, 0),\quad \text {and } ( 0, \infty) .
$$
(1)
Test a number in the interval $(-\infty, -2)$ say $-3$:
$$
\begin{aligned}
f^{\prime}(-3) &=4(-3)( (-3)^{2}+3 (-3)+2) \\
&=-24 \\
& \lt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty, -2 )$
(2)
Test a number in the interval $( -2, -1)$ say $-\frac{3}{2}$:
$$
\begin{aligned}
f^{\prime}(-\frac{3}{2}) &=4(-\frac{3}{2})( (-\frac{3}{2})^{2}+3 (-\frac{3}{2})+2) \\
&=\frac{3}{2} \\
& \gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is decreasing on $( -2 , -1),$
(3)
Test a number in the interval $(-1,0)$ say $-\frac{1}{2}$:
$$
\begin{aligned}
f^{\prime}(-\frac{1}{2}) &=4(-\frac{1}{2})( (-\frac{1}{2})^{2}+3 (-\frac{1}{2})+2) \\
&=-\frac{3}{2} \\
& \lt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-1, 0 ) .$
(4)
Test a number in the interval $(0, \infty)$ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=4(1)( (1)^{2}+3 (1)+2) \\
&=24 \\
& \gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(0, \infty ) .$
So,
(b) The function is increasing on interval $(-2, -1 ) \text { and } (0, \infty), $
(c) The function is decreasing on interval $(-\infty, -2 ) \text { and } (-1, 0). $
