Answer
$$
y(x)=x-4\ln(3x-9)
$$
(a)
The critical numbers : $x=7$.
(b)
The function is increasing on interval $ (7, \infty), $
(c)
The function is decreasing on interval $(3, 7 ). $
Work Step by Step
$$
y(x)=x-4\ln(3x-9)
$$
We find $y^{\prime}(x) $ first, using the power rule and the chain rule.
$$
\begin{aligned}
y^{\prime}(x) &=1-\frac{4.(3)}{3x-9}\\
&=1-\frac{12}{3x-9}\\
&=\frac{3x-9-12}{3x-9}\\
&=\frac{3x-21}{3x-9}\\
&=\frac{3(x-7)}{3(x-3)}\\
&=\frac{x-7}{x-3}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime}(x)=0 $ , here $f^{\prime}(x)=0 $ when $x=7. $ 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:
$$
\begin{aligned}
x-3 &=0\\
x &=3\\
\end{aligned}
$$
Since $f^{\prime}(3) $ does not exist but $f(3) $ is undefined, Since is not in the domain of $f$, where the domain of $f$ is $(3, \infty ) $. So $x=7 $ is the only critical number.
(a)
The critical numbers : $x=7$.
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. The number $7$ divides the number line into two intervals: $(3, 7)$ and $(7, \infty)$.
use a test point in each of these intervals as follows:
(1) Test a number in the interval $(3, 7)$ say $4$:
$$
\begin{aligned}
f^{\prime}(4) &=\frac{(4)-7}{(4)-3} \\
&=-3\\
&\lt 0
\end{aligned}
$$
we see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(3, 7).$
(2) Test a number in the interval $(7,\infty)$ say $8$:
$$
\begin{aligned}
f^{\prime}(8) &=\frac{(8)-7}{(8)-3} \\
&=\frac{1}{5}\\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(7, \infty ).$
So,
(b) The function is increasing on interval $ (7, \infty), $
(c) The function is decreasing on interval $(3, 7 ). $
