Answer
$$
f(x)=\frac{x+3}{x-4}
$$
(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 $x$ in the domain of $f$.
(c)
The function is decreasing on interval $(-\infty, 4)$ and $(4, \infty)$.
Work Step by Step
$$
f(x)=\frac{x+3}{x-4}
$$
Notice that the function $f$ is undefined when $x=4$ , so $4$ is not in the domain of $f$. To determine any critical numbers, first use the quotient rule to find
$$
\begin{aligned}
f^{\prime}(x) &=\frac{(1)(x-4)-(1)(x+3)}{(x-4)^{2}} \\
&=\frac{x-4-x-3}{(x-4)^{2}} \\
&=\frac{-7}{(x-4)^{2}}
\end{aligned}
$$
This derivative is never $0$, but it fails to exist at $x=4$ , where the function is undefined. Since $4$ is not in the domain of $f$, there are no critical numbers for $f$.
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. The number $4$ (where $f $is undefined) divides the number line into two intervals: $(-\infty, 4)$ and $(4, \infty)$.
use a test point in each of these intervals to find that $f^{\prime}(x) \lt 0 $ for all $x$ except $4$ . (This can also be determined by
observing that $f^{\prime}(x) $ is the quotient of $-7$, which is negative, and $(x-4)^{2}$ , which is always positive or $0$.) This means that the function $f$ is decreasing on both $(-\infty, 4)$ and $(4, \infty)$.
Thus
(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 $x$ in the domain of $f$.
(c)
The function is decreasing on interval $(-\infty, 4)$ and $(4, \infty)$.
