1. –47 Assignment, pencil, red pen, highlighter, textbook, GP notebook 2. Find the “undo” equation for f(x). Write each step for f(x) and the “undo” equation g(x). 1. Solve and graph the inequality. 2x – 3 > 112x – 3 < –11 or 2x < –82x > 14 x < –4x > 7 +1 total: +1 f(x) = 5x + 8 1) multiply by 5 2) add 8 g(x) 1) subtract 8 2) divide by 5 +1

2. Definition of a function: For _______ x, there is ______ _____ y. x 1 –5 y 12 17 x 9 4 y –8 x 3 y 2 11 Examples: Determine whether or not the following are functions based on the definition of function. eachonlyone Function! Not a function! 3 has 2 y’s

3. If you pass a _______ line across a graph and it _____ touches the graph at ____ point at a time, then it is a function. If you pass a ________ line across your graph and it touches the graph at _____ than ____ point at a time, then it is _____ a function. vertical only one vertical more onenot

4. Practice: Determine whether or not each graph is a function by applying the vertical line test. Function Not a function

5. There is a name for the “undo” functions: Notation: Given f(x), the _______ of f(x) is ______. ______________ - function that ________ another function. Inverse Function “undoes” inverse f –1 (x)

6. y x 8 8 –8 1. Graph each function. Determine f –1 (x), then graph y = f –1 (x). a) f(x) = 2x + 4 x–int: _____ y–int: _____ x–int: _____ y–int: _____ x–int: 0 = 2x + 4 –4 = 2x 2 2 x = –2 (–2, 0) (0, 4) Multiply by 2 Add 4 Subtract 4 Divide 2 (x – 4) 2 x–int: 0 = (x – 4) 2 0 = x – 4 x = 4 (4, 0) (0, –2) f(x) f –1 (x) f –1 (x) = or

7. x–int: _____ y–int: _____ x–int: _____ y–int: _____ y x 8 8 –8 b) f(x) = x 3 + 1 (–1, 0) (0, 1) Cube x Add 1 Subtract 1 Cube root (1, 0) (0, –1) x–70129 y (0,1) –2–10 1 2 f(x) f –1 (x) f –1 (x) = Cubic What is the name of this function? What is the inflection [locator] point? Compare the points on f(x) and f –1 (x). What do you notice? The x’s and y’s are switched!

8. ►What is the line of reflection between each pair of functions? Conclusions: ►The x– and y–intercepts are ____________ with inverses. Draw the line on each graph. y = x interchanged

9. y x 8 8 –8 a) f(x) = 2x + 4 x–int: _____ y–int: _____ x–int: _____ y–int: _____ (–2, 0) (0, 4) (x – 4) 2 (4, 0) (0, –2) f(x) f –1 (x) f –1 (x) = y x 8 8 –8 f(x) f –1 (x) x–int: _____ y–int: _____ x–int: _____ y–int: _____ (–1, 0) (0, 1) (1, 0) (0, –1) b) f(x) = x 3 + 1 f –1 (x) = y = x Notice that both lines cross the line of reflection at the same point. As a rule of thumb, whenever f(x) crosses the line of reflection, f –1 (x) will always cross at the same point.

10. y x y x 2. Sketch y = x and y = f –1 (x). (2, 3) (0, 1) (–2, 0.5) (3,4) (0,2) (–2, –1) (3, 2) (1, 0) (0.5, –2) (4, 3) (2, 0) (–1, –2) y = x To graph the inverse f –1 (x), switch the x– and y–values, plot the new points, and draw the curve through the new points. Draw y = x, f –1 (x) f(x) Label f –1 (x).

11. y x 5 5 –5 Domain: _________ Range: __________ 3. Sketch y = x 2 and y = x. (–∞, ∞) [0, ∞) No, it fails the vertical line test. y = x Sketch a graph of the inverse. Is the inverse a function? f(x) Be sure to switch the x– and y–values. (–2, 4) (–1, 1) (0, 0) (1, 1) (2, 4) (4, –2) (1, –1) (4, 2)

12. y x 5 5 –5 Domain: _________ Range: __________ Domain: ___________ Range: __________ 4. In order for f –1 (x) to exist, we must ________ the domain of f(x) = x 2. restrict y = x f(x) = x 2 [0, ∞) We are just graphing the positive half of the parabola. f –1 (x) = [0, ∞) f(x) f –1 (x)

14. Old Slides

15. Practice: Determine whether or not each graph is a function by applying the vertical line test. Function Not a function Function Not a function

16. x–int: _____ y–int: _____ x–int: _____ y–int: _____ y x 8 8 –8 b) f(x) = x 3 + 1 x–int: 0 = x 3 + 1 –1 = x 3 (–1, 0) y–int: y = (0) 3 + 1 y = 1 (0, 1) Cube x Add 1 Subtract 1 Cube root x–int: y–int: (1, 0) (0, –1) x–70129 y locator point: (0,1) –2–10 1 2 f(x) f –1 (x) f –1 (x) = –1 = x

17. c) x–int: _____ y–int: _____ x–int: _____ y–int: _____ (–6, 0) (0, 2) 3(x – 2) (2, 0) (0, –6) y x 8 8 –8 f –1 (x) = f(x) f –1 (x) y = x Notice that both lines cross the line of reflection at the same point. As a rule of thumb, whenever f(x) cross the line of reflection, f –1 (x) will always cross at the same point.