Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$f(x) = \frac{1}{x^2} + 1$$

$$g(x) = \frac{1}{x - 1}$$

Do I just insert the $g(x)$ function into the $f(x)$ function giving me this terrible looking thing?:


Thanks for any help, I really just need confirmation I'm going in the right direction.

share|improve this question
What you have written is $(g \circ f)(x)$. You need to do the opposite to get $f \circ g$. –  Brandon Carter Oct 18 '11 at 16:06
BTW, that terrible looking thing is not so terrible since it simplifies to $x^2$. –  Javier Oct 18 '11 at 16:14
@JavierBadia: Careful! The function $g\circ f$ is undefined at $x=0$, but the function $x^2$ is defined at $0$. So really you need to write $g\circ f(x) = x^2,\ x\neq 0$ to get a true equality of functions. –  Arturo Magidin Oct 18 '11 at 16:29

3 Answers 3

up vote 8 down vote accepted

Actually, it would be the other way around. $(f \circ g)(x) = f(g(x))$, and $f(g(x)) = \frac{1}{g(x)^2}+1 = \frac{1}{\frac{1}{(x-1)^2}}+1$. This last expression is almost equal to $(x-1)^2+1$. The difference is that in the former, $ f \circ g$ is undefined at $x = 1$, while in the latter, it is. The correct function would be $(f \circ g )(x) = (x-1)^2 + 1, \forall x \neq 1$.

What you did is actually $(g\circ f)(x) = g(f(x)) = \frac{1}{f(x)-1} = \frac{1}{\frac{1}{x^2}+1-1} = \frac{1}{\frac{1}{x^2}}$. The same note about the function not existing at $x = 0$ applies; therefore, $(g \circ f)(x) = x^2, \forall x \neq 0$.

share|improve this answer
See my comment above: $g\circ f$ is undefined at $0$, but $y=x^2$ is defined everywhere. You need to be careful with the domains when you "simplify" an expression, given the convention that a function given by a formula is given its natural domain unless otherwise specified. –  Arturo Magidin Oct 18 '11 at 16:30
@Arturo: You're right, I'll add that in the answer. –  Javier Oct 18 '11 at 16:39
Great, thanks for the explanation. Using the correct problem sure does make a difference. –  erimar77 Oct 18 '11 at 18:42
@erimar77: If the answer was what you wanted, you can mark it as accepted. –  Javier Oct 18 '11 at 18:58

When you compute (f o g)(x) you take f(g(x)). In other words, g(x) becomes the variable x in f(x). So, with your example, f(g(x))=((1/(g(x)^2)+1). I think you can do the rest.

share|improve this answer

$g(x)=1/(x-1)$ => $x=1+1/g(x)$ => $f(x)=1/(1+1/g(x))^2+1=1/(1+2/g(x)+1/g^2(x))+1=g^2(x)/(g(x)+1)^2+1$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.