Question

Find the inverse function for: f(x)=(√(1-x^2))/(x+1)

Answer 1

I know the answer but I don't know how to get it.

Answer 2

change f(x) to x and x to y. then isolate y in one side.

Answer 3

Yeah I know that much but then I get stuck on \[(x(y+1))^2=1-y^2 \]

Answer 4

\(\Large \rm\color{teal}{is~that~function~like~this\\
f(x)=\frac{\sqrt{1-x^2}}{x+1}}\)

Answer 5

??

Answer 6

I'm gonna assume that's the function and continue
First thing you need to do is to check for one-to-one property if it is satisfied
in this case it is just looking at the function
there is no need for restriction
but better take a look at the graph
https://www.desmos.com/calculator
surely the horizontal line test passes

Answer 7

https://www.desmos.com/calculator/kgr0o4rk2k

Answer 8

Okay now the main part
\(\large\bf {f(x)=\frac{\sqrt{1-x^2}}{x+1}}\)
set is like this
\(\large\bf {y=\frac{\sqrt{1-x^2}}{x+1}}\)
change x and y places
\(\large\bf {x=\frac{\sqrt{1-y^2}}{y+1}}\)
Solve for y
\(\large\bf {x(y+1)=\sqrt{1-y^2}}\)
square both sides
\(\large\bf {x^2(y+1)^2=1-y^2}\)

\(\large\bf {x^2(y+1)^2-\color{red}{(1-y^2)}=(1-y^2)-\color{red}{(1-y^2)}}\)
\(\large\bf {x^2(y+1)^2-\color{red}{(1-y^2)}=0}\)
\(\large\bf {x^2(y+1)^2-(1-y)(1+y)=0}\)
\(\large\bf {[x^2(y+1)-(1-y)](1+y)=0}\)
\(\large\bf {[x^2y+x^2-1+y)](1+y)=0}\)
we don't care about 1+y=0
our focus on the one that has x
\(\large\bf {[(x^2+1)y+x^2-1)]=0}\)
\(\large\bf {(x^2+1)y=1-x^2}\)
\(\large\bf {y=\frac{1-x^2}{1+x^2}}\)
then
\(\large\bf {f^{-1}(x)=\frac{1-x^2}{1+x^2}}\)

Answer 9

Now check let's see if that's true.
we don't want errors

Answer 10

hey you there?

Answer 11

yeah sorry.. that's the right answer. Thank you sooo much!!

Answer 12

Okay! how do you know it is right?
we have to check \(\large\bf {f(f^{-1}(x))=x}\)
the best think to do is choose a value
and evaluate we don't want to spend alot of time simplifying the expression
\(\large\bf {f(f^{-1}(2))=2}\) we need to verify this

Answer 13

i checked and it is true^_^

Answer 14

I'll get back to you in about an hour.. i got to go somewhere..
thank you so much anyways.

Answer 15

Alright! you are welcome
make sure to check the steps^_^