U6D5 Have out: Bellwork: Complete Example #1 on the worksheet. total: Assignment, red pen, highlighter, textbook, GP notebook U6D5 Have out: Bellwork: Complete Example #1 on the worksheet. Given and g(x) = x2 + 1, find: a) f(g(10)) b) g(f(10)) c) f(g(x)) d) g(f(x)) +1 +1 +1 g(10) = (10)2 + 1 f(10) +1 = f(x2 + 1) = g( ) +1 = 100 + 1 +1 = 101 +1 = 3 +1 +1 +1 +1 +1 = x – 1 + 1 f(101) g(3) = (3)2 + 1 = x +2 = x +2 +1 +1 = 9 + 1 +1 = 10 +1 f(g(x)) = x g(f(x)) = x = 10 f(g(10)) = 10 g(f(10)) = 10 total:

Notice that the outputs equal the inputs for each example. Given and g(x) = x2 + 1, find: a) f(g(10)) b) g(f(10)) c) f(g(x)) d) g(f(x)) = 10 = 10 = x = x Notice that the outputs equal the inputs for each example. Why does this work? Let’s graph both functions to see how they relate to each other.

Graph y = f(x) and y = g(x) for the domain {x | 0 ≤x < ∞} x f(x) 1 2 5 10 g(x) y = x x y 4 6 8 10 2 –2 –4 –6 –8 –10 1 g(x) = x2 + 1 2 Graph y = x. 3 f(x) Since f(x) and g(x) are _________ to each other with respect to the line y = x, then g = ____ . symmetric f –1 For any function f(x) and g(x), if f(g(x)) = g(f(x)) = x, then g = ____ . f –1 In other words, The composition of a function and its inverse results in an “undoing” of each other. Each output will equal the original input.

Example #2: Prove f(x) and g(x) are inverses given: f((g(x)) g(f(x)) Since f(g(x)) = g(f(x)) = x, then f(x) and g(x) are inverses.

a) Given f(x) = (x – 2)2 with domain {x | 2 ≤ x < ∞}. Example #3: a) Given f(x) = (x – 2)2 with domain {x | 2 ≤ x < ∞}. Determine the following information: f(x) y x –5 11 (5, 9) Domain: Range: x–int: y–int: (4, 4) (2, 0) None (3, 1) (2, 0)

Graph y = f(x), y = f–1(x), and y = x. Domain: Range: x–int: y–int: f(x) y x –5 11 (5, 9) (5, 9) (2, 0) f–1(x) None (9, 5) (4, 4) (4, 4) (1, 3) (0, 2) (3, 1) (3, 1) (2, 0) (2, 0) Domain: Range: x–int: y–int: None (0, 2)

Calculate the inverse: When we write the inverse, we need to take into account that the domain is restricted. Look at the previous graph. Did we graph the “positive” half or the “negative” half of the parabola? +2 +2 Positive… Thus, we are only going to consider the “positive” half our the inverse.

Quiz time!! It’s… When you finish, continue working on: Clear your desk except for a pencil, ruler, and highlighter. No Calculator!!! When you finish, continue working on: Day 5 WS and CC 47 – 50, 52 – 54

Example #3: b) Given f(x) = ½(x + 4)2 + 1 with domain {x |–4 ≤ x < ∞}. Determine the following information: f(x) y x –5 11 (0, 9) Domain: Range: x–int: y–int: (–1, 5.5) (–2, 3) None (–3, 1.5) (0, 9) (–4, 1) locator point: (–4, 1) stretch factor:

Graph y = f(x), y = f–1(x), and y = x. Domain: Range: x–int: y–int: f(x) y = x y x –5 11 (0, 9) None (0, 9) (–1, 5.5) (–2, 3) (–3, 1.5) f–1(x) (–4, 1) (5.5, –1) Domain: Range: x–int: y–int: (0, 9) (3, –2) (1.5, –3) (0, 9) (1, –4) None

Calculate the inverse: Switch x and y, and solve for y. Which half did we graph of the original parabola: the “positive” half or the “negative” half? –1 –1 Positive, so we are only going to consider the “positive” half our the inverse. 2( ) 2( ) –4 –4

Mixed Practice 1) Write the inverse for the following function. c) Remember how to write a root as an exponent? -3 -3

2) Given f(x) = 2x – 5 and g(x) = x2, determine: a) f(g(2)) c) f(g(x))

Example #2: Prove f(x) and g(x) are inverses given: f((g(x)) g(f(x)) Since f(g(x)) = g(f(x)) = x, then f(x) and g(x) are inverses.