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I tried questions like these by putting the values of range and domain in the linear function $ax+b$ type;they were closed intervals. Now I don't know how to proceed further. That approach is not working here. If anyone can suggest that would be great help thanks.

Norse
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Amrit
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    A linear function on a finite domain will always have a finite range. You need something that has vertical asymptotes going to $+\infty$ and $-\infty$ as $x$ approaches $0$ and $4$. The tangent function has vertical asymptotes, so you can transform that into something that will work. You can also work with combinations of transformations of the function $\frac{1}{x}$. – kccu Jan 21 '20 at 15:55
  • But what if the range is finite and some numbers are included some are not how can we draw graphs for just random ranges and domains – Amrit Jan 21 '20 at 16:01
  • To clarify, do you need to write down a function or just sketch it? What do you mean by "some numbers are included some are not"? – kccu Jan 21 '20 at 16:05
  • What of domain is like [0,4] here numbers are included and range [-2,5) here they are not – Amrit Jan 21 '20 at 16:07
  • $f(x)=\frac{1}{4-x}-\frac{1}{x}$ works. – Thomas Andrews Jan 21 '20 at 16:12
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    Oh, well that's a little trickier if the domain is closed but the range is half-open. This is getting a little advanced, but basically no continuous function will work in that case. You'd have to get creative with a piece-wise defined function. I don't know what level of math you're in, so that might be beyond the scope of what you'd be expected to do. – kccu Jan 21 '20 at 16:14
  • I am doing pre-calculus and there are no such questions in my book. – Amrit Jan 21 '20 at 16:20

2 Answers2

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Take the function $\tan\left(\dfrac{\pi}{4}x-\dfrac{\pi}{2}\right)$ for $x \in (0,4)$.

I know that $\tan(x)$ has range $(-\infty, \infty)$ for $x \in \left(\frac{-\pi}{2},\frac{\pi}{2}\right)$. So if I get an arbitrary interval for the domain, say $(a,b)$ where $a \neq b$, then I just define a linear function $$f(x)=\frac{\pi}{b-a}(x-b)+\frac{\pi}{2}$$ We see that $f(a)=\frac{-\pi}{2}$ and $f(b)=\frac{\pi}{2}$ hence the function $\tan(f(x))$ has range $(-\infty, \infty)$ on $(a,b)$

Norse
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  • But how do we know that this the function how is this relation derived – Amrit Jan 21 '20 at 15:57
  • This linear function is a standard result ?can it be applied only for these type of intervals? – Amrit Jan 21 '20 at 16:09
  • What other kinds of intervals are there? This above idea only holds for open intervals $(a,b)$. If you are thinking about closed intervals, then there are no continuous functions which are unbounded on a closed and bounded interval. Hence you would have to look at functions which have unbounded domain – Norse Jan 21 '20 at 16:16
  • Ok thank u. Can u suggest an approach for solving above closed and half open intervals? – Amrit Jan 21 '20 at 16:22
  • Can you specify the question. Are you asking for a function on a closed interval with range $(-\infty,\infty)$? If that is the case, then take the function $f(x)=x$ for $x \in \Bbb R$. $\Bbb R$ is closed and $f$ has the wanted range. – Norse Jan 21 '20 at 16:26
  • What would be the function if range is [-2,5) and domain is [0,4] – Amrit Jan 21 '20 at 16:29
  • Then you can take the function $$f=\begin{cases} 7x-2, \quad &\text{for } 0 \le x < 1 \ 2, \quad &\text{for } 1 \le x \le 4 \end{cases}$$ – Norse Jan 21 '20 at 16:32
  • To make functions like this it comes with practice? – Amrit Jan 21 '20 at 16:35
  • I think so, but I don't know – Norse Jan 21 '20 at 16:37
  • Is this right? If F=6x-2 , 0<_x<1and 2, 1<_x<_4 – Amrit Jan 21 '20 at 16:38
  • That would be right if the domain is $(0,4)$ and range is $(-2,4)$ – Norse Jan 22 '20 at 14:03
  • Thank you for helping – Amrit Jan 23 '20 at 15:22
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Alternatively, take the function $\dfrac1{4-x}-\dfrac1x$ for $x\in(0,4)$.

Note that it approaches $-\infty$ as $x\to0$, it approaches $\infty$ as $x\to4$, and it is continuous.

J. W. Tanner
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  • How is that relation derived does this come with practice or there is a trick or method for this? – Amrit Jan 21 '20 at 16:02
  • As @kccu commented above, with practice you'll come to see that combinations of transformations of the reciprocal function or transformations of the tangent function can be used to get functions with vertical asymptotes you wish (in this case $y=0$ and $y=4$) – J. W. Tanner Jan 21 '20 at 16:08
  • What to do when we have range like say [-4,3) and range like [0,2] here everything is finite? – Amrit Jan 21 '20 at 16:16
  • Indeed, given any continuous function $f:(0,4]$ with $f(x)\to \infty$ as $x\to 0$ will give $g(x)=f(4-x)-f(x)$ is continuous on $(0,4)$ and range $(-\infty,+\infty).$ It's just that $f(x)=\frac{1}{x}$ is the most obvious such function.\ – Thomas Andrews Jan 21 '20 at 16:16
  • Thank u for helping – Amrit Jan 21 '20 at 16:55