# Finding the critical points of a function and the interval where it increases and decreases.

I am having trouble finding the critical points of $$f(x)= (x+1)/{x-3}$$ I found the derivative to be $$f'(x)= -4/(x-3)^2$$ 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? Thank you very much.

• It looks like you meant to write $f(x) = \frac{x + 1}{x - 3}$, in which case you are missing needed parentheses in the denominator. As explained in this MathJax tutorial, you could type f(x) = \frac{x + 1}{x - 3} when you are in math mode to produce $f(x) = \frac{x + 1}{x - 3}$. Commented Sep 6, 2018 at 7:58

Judging by the derivative you calculated, it appears the function is supposed to be $$f(x) = \frac{x + 1}{x - 3}$$ 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 $$\text{Dom}_f = \{x \in \mathbb{R} \mid x \neq 3\} = (-\infty, 3) \cup (3, \infty)$$ The derivative of $f$ is $$f'(x) = -\frac{4}{(x - 3)^2} < 0$$ for every $x$ in its domain, which tells us that the function is decreasing on the intervals $(-\infty, 3)$ and $(3, \infty)$.

Note that $$f(x) = \frac{x + 1}{x - 3} = 1 + \frac{4}{x - 3}$$ so \begin{align*} \lim_{x \to -\infty} f(x) & = \lim_{x \to -\infty} \left(1 + \frac{4}{x - 3}\right) = 1\\ \lim_{x \to 3^+} f(x) & = \lim_{x \to 3^+} \left(1 + \frac{4}{x - 3}\right) = -\infty\\ \lim_{x \to 3^+} f(x) & = \lim_{x \to 3^+} \left(1 + \frac{4}{x - 3}\right) = \infty\\ \lim_{x \to \infty} f(x) & = \lim_{x \to \infty} \left(1 + \frac{4}{x - 3}\right) = 1 \end{align*} Thus, when $x \in (-\infty, 3)$, $f(x) \in (-\infty, 1)$, and when $x \in (3, \infty)$, $f(x) \in (1, \infty)$. Since the function assumes larger values in the interval $(3, \infty)$ than it does in the interval $(-\infty, 3)$, 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 $f'(x) = 0$, then the function $f: (-\infty, 3) \cup (3, \infty) \to \mathbb{R}$ defined by $$f(x) = \frac{x + 1}{x - 3}$$ does not have a critical point since $f'(x) < 0$ 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 $f'(x) = 0$ or $f'(x)$ does not exist, then the function $f: (-\infty, 3) \cup (3, \infty) \to \mathbb{R}$ defined by $$f(x) = \frac{x + 1}{x - 3}$$ still does not have a critical point since the derivative is defined at every point of its domain.

Hint: Critical points can also occur where your derivative is undefined (where the denominator equals 0)!

• A critical point of a function $f$ is usually defined as a point where $f'(x) = 0$. It seems you are using the definition that a critical point of a function $f$ is a point in its domain where $f'(x) = 0$ or $f'(x)$ does not exist. However, $3$ is not in the domain of $f(x) = \frac{x + 1}{x - 3}$. Commented Sep 6, 2018 at 11:07