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 $\displaystyle y=f(x) = \frac{1}{1+x^2}, x\ge0$

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

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

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

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

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

2) $\displaystyle f(x) = \sqrt{16-x^2}, -4 \le x \le 0$ and $\displaystyle g(x) = -\sqrt{16-x^2}, 0 \le x \le 4$

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

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

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

$\displaystyle = \sqrt{x^2}$

$\displaystyle =x$

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

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

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

$\displaystyle = -\sqrt{x^2}$

$\displaystyle =x$, noting that for $\displaystyle -4\le x \le 0$, the negative square root is implicit in the value of $\displaystyle x$.

Grandad