Question 1208753
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Find the range of y = x/(x - 5).
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        Here is another solution, which does not use the conception of limits.



<pre>
The function is y = {{{x/(x-5)}}}.


The range of this function is the domain of the inverse function.


Let's find the inverse function.


For it, swap x and y in the original formula to get  x = {{{y/(y-5)}}}  and solve for y.


You have

    x*(y-5) = y,

    xy - 5x = y,

    xy - y = 5x,

    y*(x-1) = 5x,

    y = {{{(5x)/(1-x)}}}.


Thus, the domain of the inverse function  is the set of all real numbers except of x= 1.


Hence, the range of the original function is the same set of all real numbers except of x= 1.
</pre>

Solved.



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<U>Comment from student</U>&nbsp;: &nbsp;&nbsp;The domain of the inverse function is the range of the given function. &nbsp;Yes?



<U>My response</U>&nbsp;: &nbsp;&nbsp;Yes.  &nbsp;&nbsp;For one-to-one functions it is always so.


The given function in this exercise is one-to-one, &nbsp;so this statement is applicable to it. 


It is practically the same as if you look into a mirror and see there your image . . . 



Thank you for good question.



By the way, it is the kind of knowledge which everybody must have, who learns this topic and solves such problems.



Examples of one-to-one functions


<pre>
    - all linear functions  y = ax + b,  except of horizontal;

    - all monotonic polynomial functions (like y = ax^3 + const;  y = ax^5 + const and many others);

    - many trigonometric functions  y = tan(x) + const;  y = cot(x) + const  in one period domain;

    - power functions  y = {{{a*x^b + const}}};

    - exponential functions  y = {{{a*b^x + const}}};

    - logarithmic functions  y = a*log(bx) + const;

    - fractional linear functions of the form  y = {{{(ax+b)/(cx+d)}}};

    - and all other monotonic functions . . . 
</pre>