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Evaluate the compositions if: f(x) = x + 2g(x) = 3h(x) = x 2 + 3 1. f(g(x))2. h(f(x))3. h(f(g(x))) f(3) 3 + 2 5 h(x + 2) (x + 2) 2 + 3 x 2 + 4x + 4 + 3.

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Presentation on theme: "Evaluate the compositions if: f(x) = x + 2g(x) = 3h(x) = x 2 + 3 1. f(g(x))2. h(f(x))3. h(f(g(x))) f(3) 3 + 2 5 h(x + 2) (x + 2) 2 + 3 x 2 + 4x + 4 + 3."— Presentation transcript:

1 Evaluate the compositions if: f(x) = x + 2g(x) = 3h(x) = x 2 + 3 1. f(g(x))2. h(f(x))3. h(f(g(x))) f(3) 3 + 2 5 h(x + 2) (x + 2) 2 + 3 x 2 + 4x + 4 + 3 x 2 + 4x + 7 h(f(3)) h(3 + 2) h(5) 5 2 + 3 25 + 3 = 28 f(x) = x + 2h(x) = x 2 + 3

2 Objective – Students will define one-to-one functions and use the horizontal line test. Students will be able to solve problems involving inverse functions. To be able to evaluate logarithmic functions.

3 Definition: A function f is called one-to-one if it never takes on the same value twice; that is... whenever Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more than once

4 STEP 1Switch the “y” and the “x” values. Solving for the Inverse STEP 2Solve for “y”.

5 Example 1: Find the inverse of 10y +2x = 4 Answer: y -1 = -5x + 2 10x + 2y = 4 2y = -10x + 4 y = -5x + 2 2 2

6 Example 2: Find the inverse of y = -3x + 6 Answer: y -1 = (-1/3)x + 2 x = -3y + 6 x – 6 = -3y –6 –3 y = (-1/3)x + 2

7 f -1 (x)= 5 x Example 3: Find the inverse of the function: f(x) = x 5 x = y 5 5 x = y y = x 5

8 Find the Inverse Equation: 1)Y = 2x + 3 2) y = x 2 + 1 3) y = x 3 – 1 4) f(x) = 5) f(x) = 1 x 3 2x + 1 x + 3

9 Pg. 74-75: (5-7, 17, 23-28) all

10 http://www.youtube.com/watch?v=4Y-62Ti5_6s https://www.youtube.com/watch?v=VSgB1IWr6O4 The Richter Scale Magnitude +1 0 1 2 3 4 5 6 7 8 9 E E(30) E(30) 2 E(30) 3 E(30) 4 E(30) 5 E(30) 6 E(30) 7 E(30) 8 E(30) 9 energy released: x 30 Logarithmic Functions

11 DEFINITION OF LOGARITHM WITH BASE b log b y = x if and only if b x = y The expression log b y is read as “log base b of y.”

12 Example 4 Rewrite the logarithmic equation in exponential form. a)log 3 9 = 2 b) log 8 1 = 0 c) log 5 ( 1 / 25 ) = -2 3 2 = 9 8 0 = 15 (-2) = 1 / 25 Example 5 Evaluate the expression. a)log 4 64 b) log 3 27 c) log 6 ( 1 / 36 ) 4 x = 64 x = 3 3 x = 276 x = 1 / 36 x = 3 x = -2

13 Objective – To be able to use properties of logarithms State Standard – 11.0 Students will understand and use simple laws of logarithms. Properties of Logarithms Product Propertylog b uv = log b u + log b v Quotient Propertylog b (u/v) = log b u - log b v Power Propertylog b u n = n log b u

14 log 5 2 + log 5 x 6 Example 2 Expand log 5 2x 6 log 5 2 + 6 log 5 x

15 Example 3 Condense: 2 log 3 7 – 5 log 3 x = log 3 7 2 – log 3 x 5 = log 3 7 2 x 5 = log 3 49 x 5

16 Pg. 76 35, 36, 39, and 40


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