I graduated several years ago from Computer Science. In order to refresh my mathematical knowledge, I bought the book "How to Think Like a Mathematician" and currently I'm working through the chapters. The author of the book provided the solution without the solution process. In most cases I don't know how to begin. Therefore, I need your hints for how to begin the solution process. So, for the following function I've to find out the domain. For training purposes I myself want to determine also the range.

$$f(x) = \frac{x}{x^2-5x+3}$$

Thanks in advance!

  • $\begingroup$ Regarding the domain: The goal is to find an x which can't fulfill the function. This I know. But how to find this out in a efficient way? $\endgroup$ – convexfx Sep 13 '16 at 20:40

For the domain of the function you have to check which values $x \in \mathbb{R}$ you can "put in" the function in order to get a result or better: such that the function is defined. The function is not defined if the denominator is zero. So you have to find the roots of the function $x^2 - 5x +3$.

For the range you have to check which values your function can "adopt". For this it might be helpful to plot the function or to check the limit of the function at some important points such as the zeros of the function in the denominator or at infinity. Maybe the picture gives you a hint. enter image description here

  • $\begingroup$ Thank you so much! Too many years since my graduation! So, in order to $\endgroup$ – convexfx Sep 13 '16 at 21:04
  • $\begingroup$ it turns out that I have to use quadratic equation. So the solution is $x \in R | x \neq \frac{-5 -\sqrt 13}{2} and x \neq \frac{-5 +\sqrt 13}{2}$. This was the first and last time I wrote with the iPad. It's exhausting to write with it this kind of text. $\endgroup$ – convexfx Sep 13 '16 at 21:21
  • $\begingroup$ Your solution of the denominator function is correct. A nice way to write the domain could be $D = \mathbb{R} \setminus\{ \frac{-5-\sqrt{13}}{2}, \frac{-5+\sqrt{13}}{2} \}$. What do you have for the range of your function? $\endgroup$ – JDoe Sep 14 '16 at 7:42
  • $\begingroup$ Hi JDoe, meanwhile I realized that I could to solve the function also by completing the square. But I don't get the same result as for the quadratic equation. Here is my solution process: 1. Bring function into complete square form -> $(x-a)^2$ $(x- \frac{5}{2})^2-( \frac{5}[2})^2+3$ 2. Combine constants $ = (x-\frac{5}{2})^2-3,25$ 3. Take root $ = $(x-\frac{5}{2}) - \sqrt 3,25 4. Rearrange and solve $x = 2,5 - \sqrt 3,25 or x = 2,5 + \sqrt 3,25$ So, I don't get the same result via quadratic equation. What am I doing wrong? $\endgroup$ – convexfx Sep 16 '16 at 22:15

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