inverse functions T.T please help !

**iiharthero** · October 30th 2009, 04:57 AM

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 !!!

**Grandad** · October 30th 2009, 09:14 AM

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 $0\le x\le 4$

$f(g(x))= \sqrt{16-(-\sqrt{16-x^2})^2}$