# Thread: Graph each function and its inverse

1. ## Graph each function and its inverse

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)

2. The first function is an inverse of the second and vice versa. Simply replace the $x$'s in your function with $y$'s to receive the inverse function.

$f(x) = (x + 2)^2 + 3$
$y = (x + 2)^2 + 3$
$x = (y + 2)^2 + 3$
$-(y+2)^2 = -x + 3$
$(y + 2)^2 = x - 3$
$y + 2 = \sqrt{x - 3}$
$y = \sqrt{x - 3} + 2$

Final Inverse of $f(x) = (x + 2)^2 + 3$:

$f(x) = \sqrt{x - 3} + 2$

3. Originally Posted by Failbait
$y + 2 = \sqrt{x - 3}$
$y = \sqrt{x - 3} + 2$
Wouldn't the 2 change to a -2 when it switches from the side with the Y to the side with the square root of X - 3?

4. Ah yes, sorry about that.

5. How would I solve the first problem?

Graph each function and its inverse.

1) F(x) = |x - 3| + 2

6. Originally Posted by Brazuca
How would I solve the first problem?

Graph each function and its inverse.

1) F(x) = |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:

$f(x) = |x - 3| + 2 = \left \{ \begin{array}{cc} x - 1, & \mbox { if } x \ge 3 \\ 5 - x, & \mbox { if } x < 3 \end{array} \right.$

work on each piece separately.

For the first graph:

$y = x - 1$ for $x \ge 3$

For inverse, switch x and y:

$x = y - 1$ for $y \ge 3$

$\Rightarrow y = x + 1$ for $y \ge 3$ (which is equivalent to $x \ge 2$)

now for the second graph:

$y = 5 - x$ for $x < 3$

For inverse, switch x and y:

$x = 5 - y$ for $y < 3$

$\Rightarrow y = 5 - x$ for $y < 3$ (which is equivalent to $x > 2$)

thus, putting these together we have:

$f^{-1}(x) = \left \{ \begin{array}{cc} x + 1, & \mbox { if } x \ge 2 \\ 5 - x, & \mbox { if } x > 2 \end{array} \right.$

the graph is below. the green is $f(x)$, the red is $f^{-1}(x)$