Range of Rational Functions....2

[mathdad]

Find the range of y = (x + 2)/(x - 2).

I need the steps. According to the textbook, graphing the function leads to finding the range. This may be true for others but not for me. I am not clear on the range idea.

 

[MarkFL]

I would write:

\(\displaystyle y=\frac{x+2}{x-2}=\frac{x-2+4}{x-2}=1+\frac{4}{x-2}\)

Now, the part:

\(\displaystyle \frac{4}{x-2}\)

has the same range as:

\(\displaystyle y=\frac{1}{x}\)

And then the 1 will shift the range up one unit.

Another approach, as I mentioned in your other thread is to find the inverse function and take its domain:

\(\displaystyle x=\frac{y+2}{y-2}\)

\(\displaystyle xy-2x=y+2\)

\(\displaystyle xy-y=2+2x\)

\(\displaystyle y(x-1)=2(x+1)\)

\(\displaystyle y=\frac{2(x+1)}{x-1}\)

So, this is the inverse of the original...what's its domain?

 

[mathdad]

For y = 2(x + 1)/(x - 1), set denominator = 0.

x - 1 = 0

x - 1 + 1 = 1

x = 1

The domain is ALL REAL NUMBERS except for 1.

This, based on what you said, is the range of the original function.

Correct?

 

[MarkFL]

For y = 2(x + 1)/(x - 1), set denominator = 0.

x - 1 = 0

x - 1 + 1 = 1

x = 1

The domain is ALL REAL NUMBERS except for 1.

This, based on what you said, is the range of the original function.

Correct?

Yes, that's correct. :D

 

[mathdad]

Does this work with all functions?

 

[MarkFL]

Does this work with all functions?

Sometimes it can be difficult, if not impossible, to algebraically determine the inverse, and special care has to be taken with functions that aren't one-to-one. :D

 

[mathdad]

As I go through the textbook, I will post functions that will give me all the needed practice to find domain and range. This is very important.