find the two functions which are the inverse of the functions
1/(1+x^2), x is equal to and > 0 AND 1/(1+x^2), x is equal to and <0

show that the following pairs of functions are inverses by showing that
f(g(x)) = g(f(x)) = x
f(x) = sqrt(16-x^2) -4<x<0 (x also equals -4 and 0)
g(x) = - sqrt(16-x^2), 0<x<4 (x also equals 0 and 4)

thank you so much for any help given !!!

2. ## Inverse Functions

Hello iiharthero
Originally Posted by iiharthero
find the two functions which are the inverse of the functions
1/(1+x^2), x is equal to and > 0 AND 1/(1+x^2), x is equal to and <0

show that the following pairs of functions are inverses by showing that
f(g(x)) = g(f(x)) = x
f(x) = sqrt(16-x^2) -4<x<0 (x also equals -4 and 0)
g(x) = - sqrt(16-x^2), 0<x<4 (x also equals 0 and 4)

thank you so much for any help given !!!
1a) Let $y=f(x) = \frac{1}{1+x^2}, x\ge0$

Then $1+x^2 = \frac{1}{y}$

$\Rightarrow x^2 = \frac{1}{y}-1 = \frac{1-y}{y}$

$\Rightarrow x = \sqrt{\frac{1-y}{y}}$, since $x \ge 0$

So the inverse function, replacing $y$ by $x$, is $f^{-1}(x)=\sqrt{\frac{1-x}{x}}$

1b) In the same way if $g(x) = \frac{1}{1+x^2}, x \le 0$ we get
$g^{-1}(x) =-\sqrt{\frac{1-x}{x}}$
2) $f(x) = \sqrt{16-x^2}, -4 \le x \le 0$ and $g(x) = -\sqrt{16-x^2}, 0 \le x \le 4$

Note first of all that $-4\le g(x) \le 0$, so $g(x)$ is a valid input into $f$, for $0\le x\le 4$.

Then for $0\le x\le 4$
$f(g(x))= \sqrt{16-(-\sqrt{16-x^2})^2}$
$=\sqrt{16-(16-x^2)}$

$= \sqrt{x^2}$

$=x$
Similarly $0\le f(x) \le 4$, so $f(x)$ is a valid input into $g$ .

Then for $-4\le x \le 0$,
$g(f(x)) = - \sqrt{16-(\sqrt{16-x^2})^2}$
$=-\sqrt{16-(16-x^2)}$
$= -\sqrt{x^2}$
$=x$, noting that for $-4\le x \le 0$, the negative square root is implicit in the value of $x$.
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