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I have this exercise $$z={2x\over y+5}$$, and I am supposed to obtain the domain and range. I understand that the domain is all the pair of $(x,y)$ except $y=-5$ , then the exercise said that the range is $z=R$

I dont understand why z accept all the values, suppose that you want to plot the point $(1,-5)$ you wont be able to plot that point because $y=-5$ its not accepted by the domain so it cant output a value for z (range)

Please help me undestand this or how i get the range for rational functions

Thank you!

Math Lover
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Gabriel B.
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3 Answers3

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Yes, if you say $y=-4$ you get $z=2x$ which is a linear function and that one has range the $\mathbb{R}$.

nonuser
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  • yes i'm aware of that line, but i think about a rational function...let´s say y=1/(x+5) the domain are all the real numbers por x except x=-5 and that number affect the range somehow so it wont have graph in tnat point. – Gabriel B. Jan 11 '18 at 18:55
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Domain is the set of all inputs a function can take. As you correctly note, the only thing that makes your $f$ undefined is any input with $y=-5$, hence the domain of $f$ is the entire $\mathbb{R}^2$ with the exception of the line $y=-5$, i.e. the set $$ dom(f) = \{(x,y)|y \ne -5\}. $$

As for the range, it is the set of all points to which $f$ can map something. Note that if $r \in \mathbb{R}$, then $f(r/2, -4)=r$, so the range of $f$ is the entire real line.

UPDATE

For example, to clarify range, suppose you want to check if $\pi$ is in the range of $f$. Note that $$ f(\pi/2, -4) = \frac{2 \cdot (\pi/2)}{(-4)+5} = \pi, $$ so indeed $\pi$ is in the range of $f$. So would be any other real number.

gt6989b
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  • thanks!, for the domain it's all clear! I need help with the Range, you saaid that "is the set of all point to wich f can map something" how about this point ( 10,-5) it map something? – Gabriel B. Jan 11 '18 at 18:57
  • @GabrielB. Remember your function $f$ takes 2D-points (from $\mathbb{R}^2$) and produces 1D points (from $R$). So you cannot map anything to $(10,-5)$ under $f$. Also, $(10,-5)$ is not in the domain of $f$ either. I did update the answer with the another example to clarify range, hope it makes it clearer. – gt6989b Jan 11 '18 at 20:28
  • thank you!, just to put another thing in my head clearer, I can say that the function have domain in $r^2$ but the range only in $r$, and because $y$ it's in $r^2$ therefore it won't affect the range...because $x$ accept all the values? – Gabriel B. Jan 12 '18 at 16:26
  • @GabrielB. You can definitely say that $f$ has domain in $\mathbb{R}^2$ (but not all of it) and range in $\mathbb{R}$. The rest of your statement I did not understand – gt6989b Jan 12 '18 at 20:19
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Or, let $x=1$, then $z=\dfrac{2}{y+5}$, as a function of $y$, $z$ takes all values except $z=0$.

But, for $x=0$, $y=1$, $z=\dfrac{2\cdot 0}{1+5}=0$.

So it takes all values.

user284331
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  • thanks!, but what happened when i put a pair of any real number for X but y=-5, i think that part its my real problem to understand this – Gabriel B. Jan 11 '18 at 18:59
  • Of course you cannot input $y=-5$, but it doesn't mean that the value of $z$ must have "lost" something. – user284331 Jan 11 '18 at 19:00
  • Can you help me understand that part of "doesn't lost something", because if i want to graph the function what im going to plot on that specific line – Gabriel B. Jan 11 '18 at 19:11
  • You cannot plot any value regarding to that specific line $y=-5$. Still, the values are all of real numbers. And the graph is 3-D. – user284331 Jan 11 '18 at 19:12
  • So let's say if i have any rational multivariable function the domain won't accept some values, but the range will be always the real numbers, regarding the rules of the domain – Gabriel B. Jan 11 '18 at 19:16
  • No matter what, you cannot put $y=-5$ in to the function $z=\dfrac{2x}{y+5}$. But this doesn't matter, since we just need to determine the value of $z$. The trick is that, we arbitrary fix a variable, in my case, I fixed $x$ to be $1$, then I look at the function $z=\dfrac{2}{y+5}$, this function takes all the values except $z=0$. It seems that the value $z=0$ is lost, but it doesn't, since we have fixed $x=1$, what about we fix the other $x$? Just fix $x=0$, then back to the original function, the value is exactly zero, and this "rescues" the value $z=0$ that have lost when fixing $x=1$. – user284331 Jan 11 '18 at 19:16
  • oh! i think also the trick is that the range is all $r$ instead of $r^2$ wich automatically includes all the values from $y$ – Gabriel B. Jan 11 '18 at 19:22