# Find the critical points of f(x)=((x+1))/(x-3).

Finding the critical points of a function and the interval where it increases and decreases.
I am having trouble finding the critical points of
$f\left(x\right)=\left(x+1\right)/x-3$
I found the derivative to be ${f}^{\prime }\left(x\right)=-4/\left(x-3{\right)}^{2}$.
My next step was to equate my derivative to zero, but that does not seem to work as my x cancels out. Usually I would take the x-value(worked out by equating the derivative with zero) and substitute it into the original equation to get a y-value. This would then be the critical points. Is there anyone who could maybe help me out (maybe with an example or so) as I also have to find the intervals where the function is increasing and decreasing?
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nutnhonyl8
Step 1
Judging by the derivative you calculated, it appears the function is supposed to be $f\left(x\right)=\frac{x+1}{x-3}$.
Since the implicit domain of a rational function is the set of all real numbers except those that make the denominator equal to zero, the implicit domain of f is ${\text{Dom}}_{f}=\left\{x\in \mathbb{R}\mid x\ne 3\right\}=\left(-\mathrm{\infty },3\right)\cup \left(3,\mathrm{\infty }\right)$.
The derivative of f is ${f}^{\prime }\left(x\right)=-\frac{4}{\left(x-3{\right)}^{2}}<0$ for every x in its domain, which tells us that the function is decreasing on the intervals $\left(-\mathrm{\infty },3\right)$ and $\left(3,\mathrm{\infty }\right)$
Note that $f\left(x\right)=\frac{x+1}{x-3}=1+\frac{4}{x-3}$ so $\begin{array}{rl}\underset{x\to -\mathrm{\infty }}{lim}f\left(x\right)& =\underset{x\to -\mathrm{\infty }}{lim}\left(1+\frac{4}{x-3}\right)=1\\ \underset{x\to {3}^{+}}{lim}f\left(x\right)& =\underset{x\to {3}^{+}}{lim}\left(1+\frac{4}{x-3}\right)=-\mathrm{\infty }\\ \underset{x\to {3}^{+}}{lim}f\left(x\right)& =\underset{x\to {3}^{+}}{lim}\left(1+\frac{4}{x-3}\right)=\mathrm{\infty }\\ \underset{x\to \mathrm{\infty }}{lim}f\left(x\right)& =\underset{x\to \mathrm{\infty }}{lim}\left(1+\frac{4}{x-3}\right)=1\end{array}$
Step 2
Thus, when $x\in \left(-\mathrm{\infty },3\right)$, $f\left(x\right)\in \left(-\mathrm{\infty },1\right)$, and when $x\in \left(3,\mathrm{\infty }\right)$, $f\left(x\right)\in \left(1,\mathrm{\infty }\right)$. Since the function assumes larger values in the interval $\left(3,\mathrm{\infty }\right)$ than it does in the interval $\left(-\mathrm{\infty },3\right)$ than it does in the interval $\left(-\mathrm{\infty },3\right)$, it does not decrease over the union of the two intervals in which it is decreasing.
If you define a critical point of a function f to be a point where ${f}^{\prime }\left(x\right)=0$, then the function $f:\left(-\mathrm{\infty },3\right)\cup \left(3,\mathrm{\infty }\right)\to \mathbb{R}$ defined by
$f\left(x\right)=\frac{x+1}{x-3}$ does not have a critical point since ${f}^{\prime }\left(x\right)<0$ for every x in its domain.
If you use the alternative definition that a critical point of a function f is a point in its domain where ${f}^{\prime }\left(x\right)=0$ or f′(x) does not exist, then the function $f:\left(-\mathrm{\infty },3\right)\cup \left(3,\mathrm{\infty }\right)\to \mathbb{R}$ defined by $f\left(x\right)=\frac{x+1}{x-3}$ still does not have a critical point since the derivative is defined at every point of its domain.

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