Inverse Functions

[lewch45]

Can someone please help....I need to verify that f and g are inverse functions by showing that (f o g)(x) = x and (g o f)(x)=x. I understand the concept but I just can't get the math right on this specific problem.

It is: f(x)=1/x +1, g(x)=1-x/1

Can someone please help me figure this out? Thanks.

 

[stapel]

I understand the concept but I just can't get the math right on this specific problem.

We'll be glad to check your work and helop you find any errors. Please post whatever you've done so far.

Thank you!

Eliz.

 

[Gene]

Check it.
f(x) = 1/(x+1)
g(x) = (1/x)-1
****************
f(g(x)) =
f((1/x)-1) = 1/((1/x)-1)+1) =
1/(1/x) =
x
I'll leave the reverse to you

 

[lewch45]

(f o g)(x)=f[g(x)]=f[1/x-1]=1 divided by 1/x+1

This is where it gets quite confusing. I'm not sure where to go from here.

 

[Gene]

Did what I posted help or do you need more?
------------------
Gene

 

[lewch45]

I actually switched the quantities of f(x) and g(x).

Therefore: f(x)=(1-x)/1, g(x)=1/x + 1

Would this make a difference?

 

[Gene]

I'm still suspicious and unsure of your equations.
(x+1)/1=x+1 Very unusual.
1/x + 1
Do you mean 1/(x+1) or (1/x)+1?

No, if you did just swap them, it would make no difference. Please retype them after CAREFUL checking, using ()s as I have indicated so there is no doubt of what you mean.

 

[lewch45]

correct quantities

Sorry....I just checked the book and here are the exact quantities:

f(x)= (1/x) + 1

g(x)= 1 / (x-1)

I've just now totally confused myself. Can you please help me out? Thanks!

 

[pka]

If \(\displaystyle \L
f(x) = \frac{1}{x} + 1\quad \& \quad g(x) = \frac{1}{{x - 1}}\) then:

\(\displaystyle \L
f \circ g(x) = f(g(x)) = \frac{1}{{g(x)}} + 1 = \frac{1}{{\left( {\frac{1}{{x - 1}}} \right)}} + 1\) and

\(\displaystyle \L
g \circ f(x) = g(f(x)) = \frac{1}{{f(x) - 1}} = \frac{1}{{\left( {\frac{1}{x} + 1} \right) - 1}}\).