Domain, range issues.

[NIN9S]

Honestly, I'm a college student who has been pushed through school, & now trying to earn a degree & I'm in an intermediate math course so I'm not sure this falls in this category.

That said, I have a problem with a skipped step & my online homework won't show me so I'm stumped. I'll try my best to write it out, but it gave me a new one to try with new numbers.



y = 8/x - 10

Is it a function? Yes
What is the domain?
What is thre range?

It is 8 over x minus 10 if that helps clarify. I have problems with any fraction type problems & don't know what to do, the online course skipped that portion & just gave me the answer (of course I get no credit once I get help from it, plus I need to know this for myself for real tests in class anyway). Any help is greatly appreciated.

Edit: After looking over the examples online I was able to guess domain = (-inf,10) U (10,inf) and range was (-inf,0) U (0,inf), but I don't know how it got to 10. Thanks for the patience everyone...(the "inf" stands for the infinity symbol). A complete draw-out of the problem would be nice though so I better understand the process.

 

[pappus]

...

y = 8/x - 10

Is it a function? Yes
What is the domain?
What is thre range?
...
Edit: After looking over the examples online I was able to guess (<-- you are doing Math, not guessing!) domain = (-inf,10) U (10,inf) and range was (-inf,0) U (0,inf), but I don't know how it got to 10. Thanks for the patience everyone...(the "inf" stands for the infinity symbol). A complete draw-out of the problem would be nice though so I better understand the process.

1. First assumption: x and y are real numbers. In Math: \(\displaystyle x, y \in \mathbb{R}\)

2. Determine the domain. Since x and y are real numbers there could only be some restriction so that x or y don't belong to the complete set \(\displaystyle \mathbb{R}\).

The restrictions could be:
a) the denominator of a fraction must unequal zero
b) the radicand of a square-root must be positive or zero
c) the argument of a logarithm function has to be positive

Your example contains a fraction with the variable in the denominator. So you have to make sure that the denominator is never zero. In Math: \(\displaystyle x \in \mathbb{R} \setminus \{0\}\)
\(\displaystyle x\in\{(-\infty,0) \cup (0,\infty)\}\) is of course the same set of valid numbers for x.

This set of valid numbers for x is called the domain of the function.

3. Now examine the values for y:

If x approaches \(\displaystyle -\infty\) or \(\displaystyle \infty\) then \(\displaystyle \frac8x\) approaches zero. That means y approaches \(\displaystyle -10\).
If x < 0 and it approaches 0 then \(\displaystyle \frac8x\) approaches \(\displaystyle -\infty\) (keep in mind: x is negative!). So the y-values are running from \(\displaystyle -\infty\) to -10.

If x > 0 and it approaches 0 then \(\displaystyle \frac8x\) approaches \(\displaystyle \infty\) (keep in mind: x is positive!). So the y-values are running from \(\displaystyle \infty\) to -10.

Collect these results: \(\displaystyle y \in \mathbb{R}\setminus\{-10\}\) or which is the same:

\(\displaystyle y\in\{(-\infty,-10) \cup (-10,\infty)\}\)

The set of valid values for y is called range of the function.

 

[lookagain]

y = 8/x - 10

Is it a function? Yes
What is the domain?
What is thre range?

\(\displaystyle > > \)It is 8 over x minus 10 if that helps clarify. \(\displaystyle < < \)



Edit: After looking over the examples online I was able to guess domain = (-inf,10) U (10,inf) and range was (-inf,0) U (0,inf), but I don't know how it got to 10. Thanks for the patience everyone...(the "inf" stands for the infinity symbol). A complete draw-out of the problem would be nice though so I better understand the process.

No, that doesn't help to clarify it, because that description is ambiguous.
If you had instead stated, "It is 8 over the quantity of x minus 10,"
then that would have helped to clarify it.

But from your edit, it appears that you meant for it to be y = 8/(x - 10).

And then, pappus, that would make your work not apply because it's a different enough type
of function.

 

[pappus]

...
And then, pappus, that would make your work not apply because it's a different enough type
of function.

Ofcourse, but I always loved to work for the waste-bin!