How do you show that #f(x)=1/(1+x)# and #g(x)=(1-x)/x# are inverse functions algebraically and graphically?

1 Answer
jim p.
Aug 25, 2017

See explanation...

Explanation:

g(x) and f(x) are inverse functions if g(f(x)) = x and f(g(x)) = x.

So, algebraically, it's pretty straightforward:

#f(g(x)) = 1/(1 + (1 - x)/x)#

#= 1/((x/x) + (1 - x)/x)#

#=1/((1 + x - x)/x)#

#= 1/(1/x)#

#=x#

Which is sufficient to show that g(x) is an inverse function of x.

The converse also applies. So you shouldn't have to work out g(f(x)).

But for completeness, we'll go ahead:

#g(f(x)) = (1 - (1/(1+x)))/(1/(1+x))#

# =( (1+x)/(1+x) - (1/(1+x)))/(1/(1+x))#

#= ((1 + x - 1)/(1+x))/(1/(1+x))#

#=(x/(1+x)) * ((x+1)/1)#

#=x#

To show the inverse relationship graphically, you need a graph of each function:
Here's #f(x) = 1/(1+x)#

graph{1/(1+x) [-10, 10, -5, 5]}

...and #g(x)= (1-x)/x#
graph{(1-x)/x [-10, 10, -5, 5]}

And the way they taught me is, if the graph of the second function is just the graph of the first function FLIPPED around the diagonal line y = x, then they are inverse functions.

GOOD LUCK!

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