composition functions - treatment 6

[logistic_guy]

here is the question

Find the compositions \(\displaystyle f \circ g\) and \(\displaystyle g \circ f\), and identify their respective domains.

6. \(\displaystyle f(x) = \frac{1}{x^2 - 1}, \ \ g(x) = x^2 - 2\)


my attemb
\(\displaystyle f \circ g = f(g(x)) = \frac{1}{(x^2 - 2)^2 - 1} = \frac{1}{x^4 - 4x^2 + 3}\)
Domain: \(\displaystyle (-\infty,-\sqrt{3}) U (-\sqrt{3}, -1) U (-1, 1) U (1, \sqrt{3}) U (\sqrt{3},\infty)\)

\(\displaystyle g \circ f = g(f(x)) = (\frac{1}{x^2 - 1})^2 - 2 = \frac{1}{x^4 - 2x^2 + 1} - 2\)
Domain: \(\displaystyle (-\infty,-1) U (-1,1) U (1,\infty)\)

is my analize correct?

 

[topsquark]

here is the question

Find the compositions \(\displaystyle f \circ g\) and \(\displaystyle g \circ f\), and identify their respective domains.

6. \(\displaystyle f(x) = \frac{1}{x^2 - 1}, \ \ g(x) = x^2 - 2\)


my attemb
\(\displaystyle f \circ g = f(g(x)) = \frac{1}{(x^2 - 2)^2 - 1} = \frac{1}{x^4 - 4x^2 + 3}\)
Domain: \(\displaystyle (-\infty,-\sqrt{3}) U (-\sqrt{3}, -1) U (-1, 1) U (1, \sqrt{3}) U (\sqrt{3},\infty)\)

\(\displaystyle g \circ f = g(f(x)) = (\frac{1}{x^2 - 1})^2 - 2 = \frac{1}{x^4 - 2x^2 + 1} - 2\)
Domain: \(\displaystyle (-\infty,-1) U (-1,1) U (1,\infty)\)

is my analize correct?

Did you graph these?

-Dan

 

[logistic_guy]

Did you graph these?

-Dan

that's for \(\displaystyle f \circ g\)



that's for \(\displaystyle g \circ f\)

 

[topsquark]

that's for \(\displaystyle f \circ g\)

View attachment 38972

that's for \(\displaystyle g \circ f\)

View attachment 38973

What do these tell you about your domains that you posted?

-Dan

 

[logistic_guy]

What do these tell you about your domains that you posted?

-Dan

i can't see the values of my domain exactly in the graph, but it look like they fit correctly where the graph is discontinuous

 

[topsquark]

i can't see the values of my domain exactly in the graph, but it look like they fit correctly where the graph is discontinuous

Then you have the correct answer.

-Dan

 

[Dr.Peterson]

i can't see the values of my domain exactly in the graph, but it look like they fit correctly where the graph is discontinuous

You can make the locations of the asymptotes more visible by adjusting the scale, and drawing in your claimed asymptotes. They do look correct:

​

​

I should mention, though, that graphing software doesn't always show every feature, particularly "holes", so it is not a thorough check. In this case, if you were wrong it would be visible.

 

[logistic_guy]

Then you have the correct answer.

-Dan

thank

You can make the locations of the asymptotes more visible by adjusting the scale, and drawing in your claimed asymptotes. They do look correct:

View attachment 38977​

View attachment 38978​

I should mention, though, that graphing software doesn't always show every feature, particularly "holes", so it is not a thorough check. In this case, if you were wrong it would be visible.

adjusting the scales give me new idea. i do scale in my other post to demonstarate the idea

thank Dr. very much for showing the graph