Answer
$$
y=-3x+6
$$
(a)
The critical numbers : There are no critical numbers.
(b)
No interval that the function is increasing , since $ f^{\prime}(x)$ is negative for all $x$.
(c)
The function is decreasing on interval $(-\infty, \infty ) .
Work Step by Step
$$
y=-3x+6
$$
First, find the points where the derivative $f^{\prime }$ is $0$.
Here
$$
\begin{aligned}
f^{\prime}(x) &=-3 \\
\end{aligned}
$$
Solve the equation $f^{\prime}(x) =0 $ to get
$$
\begin{aligned}
& f^{\prime}(x) =-3 \ne 0\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
&\text{ [Then, there are no critical numbers ]}
\end{aligned}
$$
(a)
The critical numbers : There are no critical numbers.
Now, we can use the first derivative test.
Check the sign of $f^{\prime}(x)$.
$$
\begin{aligned}
f^{\prime}(x) &=-3 , \text {for all } x\\
\end{aligned}
$$
WE see that $ f^{\prime}(x)$ is negative for all $x$, so $f(x)$ is decreasing on $(-\infty, \infty )$
So,
(b)
No interval that the function is increasing , since $ f^{\prime}(x)$ is negative for all $x$.
(c)
The function is decreasing on interval $(-\infty, \infty ) .
