Learning how to do these are the last things I need to learn for today.
Graph each function and its inverse.
1) F(x) = |x - 3| + 2
2) F(x) = (x + 2)2 + 3
(The small 2 represents power)
The first function is an inverse of the second and vice versa. Simply replace the $\displaystyle x$'s in your function with $\displaystyle y$'s to receive the inverse function.
$\displaystyle f(x) = (x + 2)^2 + 3 $
$\displaystyle y = (x + 2)^2 + 3$
$\displaystyle x = (y + 2)^2 + 3$
$\displaystyle -(y+2)^2 = -x + 3$
$\displaystyle (y + 2)^2 = x - 3$
$\displaystyle y + 2 = \sqrt{x - 3}$
$\displaystyle y = \sqrt{x - 3} + 2$
Final Inverse of $\displaystyle f(x) = (x + 2)^2 + 3 $:
$\displaystyle f(x) = \sqrt{x - 3} + 2$
recall that for absolute values you have to consider two cases, when what's inside absolute value signs is negative and when it is positive.
to find the inverse, we switch x and y and solve for y:
$\displaystyle f(x) = |x - 3| + 2 = \left \{ \begin{array}{cc} x - 1, & \mbox { if } x \ge 3 \\ 5 - x, & \mbox { if } x < 3 \end{array} \right.$
work on each piece separately.
For the first graph:
$\displaystyle y = x - 1$ for $\displaystyle x \ge 3$
For inverse, switch x and y:
$\displaystyle x = y - 1$ for $\displaystyle y \ge 3$
$\displaystyle \Rightarrow y = x + 1$ for $\displaystyle y \ge 3$ (which is equivalent to $\displaystyle x \ge 2$)
now for the second graph:
$\displaystyle y = 5 - x$ for $\displaystyle x < 3$
For inverse, switch x and y:
$\displaystyle x = 5 - y$ for $\displaystyle y < 3$
$\displaystyle \Rightarrow y = 5 - x$ for $\displaystyle y < 3$ (which is equivalent to $\displaystyle x > 2$)
thus, putting these together we have:
$\displaystyle f^{-1}(x) = \left \{ \begin{array}{cc} x + 1, & \mbox { if } x \ge 2 \\ 5 - x, & \mbox { if } x > 2 \end{array} \right.$
the graph is below. the green is $\displaystyle f(x)$, the red is $\displaystyle f^{-1}(x)$