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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.

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  • $\begingroup$ 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 $f(x) = \frac{x + 1}{x - 3}$. $\endgroup$ – N. F. Taussig Sep 6 '18 at 7:58
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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.

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

Also, see if you can notice something about the sign of your derivative which can help you determine the increasing/decreasing intervals :)

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    $\begingroup$ A function cannot have a critical point where it is undefined. $\endgroup$ – N. F. Taussig Sep 6 '18 at 8:08
  • $\begingroup$ True, but when the derivative is undefined that still can be a candidate for a critical point (which is what i was saying). $\endgroup$ – Zach Sep 6 '18 at 10:50
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    $\begingroup$ 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 $f(x) = \frac{x + 1}{x - 3}$. $\endgroup$ – N. F. Taussig Sep 6 '18 at 11:07
  • $\begingroup$ Thank you very much. If the denominator is equal to 0, then x=3. How would I get the y-value for the critical point? If I substitute x=3 into the original equation it would also be undefined. $\endgroup$ – Nicci Sep 6 '18 at 12:07

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