Use Inverse Function Property to show f and g are inverses.

[Oneiromancy]

I need to show the function f and g are inverses of each other.

Inverse Function Property:

Let f be a one-to-one function with domain A and range B. The inverse function f^-1 satisfies the following cancellation properties.

f^-1(f(x)) = x for every x in A

f(f^-1(x)) = x for every x in B

Problem:

f(x) = sqrt(4 - x^2) , 0 =< x =< 2;
g(x) = sqrt(4 - x^2), 0 =< x =< 2

_____

The inequalities threw me off, and how can the same exact two functions be an inverse?!

 

[galactus]

\(\displaystyle \L\\x=\sqrt{4-y^{2}}\)

\(\displaystyle \L\\x^{2}=4-y^{2}\)

\(\displaystyle \L\\x^{2}-4=-y^{2}\)

\(\displaystyle \L\\4-x^{2}=y^{2}\)

\(\displaystyle \L\\y=\sqrt{4-x^{2}}\)

Happy Dreams

 

[Oneiromancy]

Sorry but I don't think that answered my question.

I solved the other questions by doing a compound function f(g(x)) and g(f(x)). Whenever you solve those, it equals x (since, I suppose inverses are reflected across y = x). That's how my book does it.

 

[Mrspi]

Sorry but I don't think that answered my question.

I solved the other questions by doing a compound function f(g(x)) and g(f(x)). Whenever you solve those, it equals x (since, I suppose inverses are reflected across y = x). That's how my book does it.

You said

f(x) = sqrt(4 - x²)
and
g(x) = sqrt(4 - x²)

The domains for both of these, you said, were 0 <= x <=2

f and g are INVERSES of each other if f º g (x) = g º f(x) = x

Did you try this?

Let's look at f º g(x).......

f(x) = sqrt(4 - x<SUP>2</SUP>)

g(x) = sqrt(4 - x<SUP>2</SUP>)

to find f º g(x), replace each "x" in f(x) with the expression given for g(x):

f º g(x) = sqrt[4 - sqrt(4 - x<SUP>2</SUP>)<SUP>2</SUP>]
f º g(x) = sqrt[ 4 - (4 - x<sup>2</SUP>)]

f º g(x) = sqrt[4 - 4 + x²)]

f º g(x) = sqrt(x<SUP>2</SUP>]

f º g(x) = x

Ok...that verifies ONE part of this.

You try the other part, showing that g º f(x) = x.

 

[Oneiromancy]

Duh, thanks!