Find the domain and range of the following function f(x)=x-2/sqrt(10-2x)

[shamash9]

Hello, I'm sorry for registering to the the boards and immediately posting a new thread. I did use the search function to try and find a similar thread that addresses the issue I'm having but nothing jumped out at me and frankly I'm so green that I wouldn't know what to look for. This is my first math class of this level.

So here is my issue. This is the second homework question of the start of the quarter and I'm already stumped just to give you an indication of what I'm working with here.

So the question is as follows:

Find the domain and range of the following function
f(x)=\(\displaystyle \frac{x-2}{\sqrt{10-2x}}\)

Now I'm easily able to find the domain by setting the values inside the radical of the denominator >0 and evaluating. I've searched and searched and I can't find anything near as definitive for finding the range other than one site that suggested graphing it out in a calculator. I vaguely remember covering this in Algebra 2 but the fog has settled since then, even though it's only been a few months.

I'm not sure how to plot this beast in a standard ti-83. Do I enclose the numerator in parentheses? I get 2 different plots when I enclose the numerator in parentheses and when I do not.

Please help me! How does a dunce evaluate the range of the function! I know it's something so simple that I'll fall flat on my face when I figure it out.

 

[HallsofIvy]

Finding the "range" of a function is, in fact, much harder than finding the domain. As you say, in order that $$\sqrt{10- 2x}$$ be a real number, 10- 2x must be larger than or equal to 0 so x must be less than or equal to 5. Further, in order that the denominator not be 0, x must be strictly less than 5. As you say the domain is "all real numbers less than 5": $$\{x | x< 5\}$$.

But then what "y" values do we get in $$y= \frac{x- 2}{\sqrt{10- 2x}}$$? One method is, indeed, to graph the function but that can be tedious. Since it is comparatively easy to find the domain, let's try to invert the formula, swapping "domain" and "range". Multiply both sides of the equation by $$\sqrt{10- 2x}$$ to get $$y\sqrt{10- 2x}= x- 2$$. Get rid of the square root by squaring both sides: $$y^2(10- 2x)= 10y^2- 2y^2x= (x- 2)^2= x^2- 4x+ 4$$. Combine like powers of x: $$x^2+ (2y- 4)x+ 4- 10y^2= 0$$. That is now a quadratic equation in x which we can solve using the quadratic formula:
$$x= \frac{4- 2y\pm\sqrt{(2y- 4)^2- 4(4- 10y^2)}}{2}$$$$= \frac{4- 2y\pm\sqrt{2y^2- 16y+ 16- 16+ 40y^2}}{2}$$$$= \frac{4- 2y\pm\sqrt{42y^2- 16y}}{2}$$.
That will be a real number as long as $$42y^2- 16y= 2y(21y- 8)\ge 0$$.

 

[shamash9]

Wow! Thank you x100000!