# functions

• Jun 19th 2008, 12:35 AM
Tangera
functions
Question:

a) f(x) = | (3-x)/(x-2) |, x > 2

Define the inverse funtion of f corresponding to the domain of f.

b) g(x) = ln (x+2), x > -2
h(x) = 2 + (e ^ -x) , x > b and b > 0

The composite function gh is well defined and the range of gh is given as (ln 4, ln 6]. Find the exact value of b.

Thank you for helping me!
• Jun 19th 2008, 01:15 AM
red_dog
a)
I think f has not an inverse, because f is not one-to-one function:
$\displaystyle f\left(\frac{8}{3}\right)=f(4)=\frac{1}{2}$
• Jun 19th 2008, 04:09 AM
earboth
Quote:

Originally Posted by Tangera
Question:

a) f(x) = | (3-x)/(x-2) |, x > 2

Define the inverse funtion of f corresponding to the domain of f.

...

1. Write the equation of f as a piecewise defined function: (completely drawn in black)

$f(x)=\left\{\begin{array}{lcl}\dfrac{x-3}{2-x}~,& \text{for} & 2 \leq x < 3 \\ \\
\dfrac{x-3}{x-2}~,& \text{for} & x \geq 3 \end{array}\right.$

2. Since f(3) = 0 this is the minimum of f. The behavior of the function changes at x = 3 from monotically decreasing to monotically increasing.

3. Therefore the inverse function must be defined piecewise too:

$f^{-1}(x)=\left\{\begin{array}{lcl}\dfrac{2x+3}{x+1}~, & \text{for} & x > 3 \\ \\
\dfrac{2x-3}{x-1}~,& \text{for} & 0 \leq x \leq1 \end{array}\right.$

EDIT: As red_dog pointed out there doesn't exist one inverse function. Therefore you have to split my last answer into:

$f_1^{-1}(x)= \frac{2x+3}{x+1}~, \text{for } x > 0$ and

$f_2^{-1}(x)= \frac{2x-3}{x-1}~, \text{for } x > 0$