2 - 1 © 2012 Pearson Education, Inc.. All rights reserved. Chapter 2 Nonlinear Functions
2 - 2 © 2012 Pearson Education, Inc.. All rights reserved. Figure 1
2 - 3 © 2012 Pearson Education, Inc.. All rights reserved. Section 2.1 Properties of Functions
2 - 4 © 2012 Pearson Education, Inc.. All rights reserved.
2 - 5 © 2012 Pearson Education, Inc.. All rights reserved. Figure 2
2 - 6 © 2012 Pearson Education, Inc.. All rights reserved. Figure 3
2 - 7 © 2012 Pearson Education, Inc.. All rights reserved. Figure 4
2 - 8 © 2012 Pearson Education, Inc.. All rights reserved. Figure 5a - 5b
2 - 9 © 2012 Pearson Education, Inc.. All rights reserved. Figure 5c - 5d
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Figure 6
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 Find the domain and range for the function Solution: The domain includes only those values of x satisfying since the denominator cannot be zero. Using the methods for solving a quadratic inequality produces the domain Because the numerator can never be zero, the denominator can take on any positive real number except for 0, allowing y to take on any positive value except for 0, so the range is
© 2012 Pearson Education, Inc.. All rights reserved. Figure 7
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Given the function find each of the following. (a) (b) All values of x such that (a) Solution: Replace x with the expression x + h and simplify. (b) Solution: Set f (x) equal to − 5 and then add 5 to both sides to make one side equal to 0. Continued
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 continued This equation does factor as Set each factor equal to 0 and solve for x.
© 2012 Pearson Education, Inc.. All rights reserved. Figure 8
© 2012 Pearson Education, Inc.. All rights reserved. Figure 9
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Figure 10
© 2012 Pearson Education, Inc.. All rights reserved. Figure 11
© 2012 Pearson Education, Inc.. All rights reserved. Figure 12
© 2012 Pearson Education, Inc.. All rights reserved. Figure 13
© 2012 Pearson Education, Inc.. All rights reserved. Section 2.2 Quadratic Functions; Translation and Reflection
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Figure 14
© 2012 Pearson Education, Inc.. All rights reserved. Figure 15
© 2012 Pearson Education, Inc.. All rights reserved. Figure 16
© 2012 Pearson Education, Inc.. All rights reserved. Figure 17
© 2012 Pearson Education, Inc.. All rights reserved. Figure 18
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 For the function (a) complete the square, (b) find the y-intercept, (c) find the x intercepts, (d) find the vertex, and (e) sketch the graph. Solution (a): To begin, factor 2 from the x-terms so the coefficient of x 2 is 1: Next, we make the expression inside the parentheses a perfect square by adding the square of one-half of the coefficient of x, Continued
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 continued Solution (b):The y-intercept (where x = 0) is − 1. Solution (c): To find the x-intercepts, solve Use the quadratic formula to verify that the x-intercepts are at Solution (d): The function is now in the form Continued
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 continued
© 2012 Pearson Education, Inc.. All rights reserved. Figure 19
© 2012 Pearson Education, Inc.. All rights reserved. Figure 20
© 2012 Pearson Education, Inc.. All rights reserved. Figure 21-22
© 2012 Pearson Education, Inc.. All rights reserved. Figure 23-25
© 2012 Pearson Education, Inc.. All rights reserved. Figure
© 2012 Pearson Education, Inc.. All rights reserved. Figure
© 2012 Pearson Education, Inc.. All rights reserved. Figure 31
© 2012 Pearson Education, Inc.. All rights reserved. Figure 32
© 2012 Pearson Education, Inc.. All rights reserved. Section 2.3 Polynomial and Rational Functions
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Figure 33
© 2012 Pearson Education, Inc.. All rights reserved. Figure 34
© 2012 Pearson Education, Inc.. All rights reserved. Figure 35
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 Graph Solution: Using the principles of translation and reflection, we recognize that this is similar to the graph of but reflected vertically (because of the negative in front of x 6 ) and 64 units up.
© 2012 Pearson Education, Inc.. All rights reserved. Figure
© 2012 Pearson Education, Inc.. All rights reserved. Figure 38
© 2012 Pearson Education, Inc.. All rights reserved. Figure 39
© 2012 Pearson Education, Inc.. All rights reserved. Figure 40
© 2012 Pearson Education, Inc.. All rights reserved. Figure 41
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Figure 42
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Figure 43
© 2012 Pearson Education, Inc.. All rights reserved. Figure 44
© 2012 Pearson Education, Inc.. All rights reserved. Figure 45
© 2012 Pearson Education, Inc.. All rights reserved. Section 2.4 Exponential Functions
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Figure 46
© 2012 Pearson Education, Inc.. All rights reserved. Figure 47
© 2012 Pearson Education, Inc.. All rights reserved. Figure 48
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 Solve Solution: Since the bases must be the same, write 25 as 5 2 and 125 as 5 3.
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Find the interest earned on $4400 at 3.25% interest compounded quarterly for 5 years. Solution: Use the formula for compound interest with P = 4400, r = , m = 4, and t = 5. The investment plus the interest is $ The interest amounts to $ − $4400 = $
© 2012 Pearson Education, Inc.. All rights reserved. Figure 49
© 2012 Pearson Education, Inc.. All rights reserved. Figure 50
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 3 Find the amount after 4 years if $800 is invested in an account earning 3.15% compounded continuously. Solution: In the formula for continuous compounding, let P = 800, t = 4 and r = to get or $
© 2012 Pearson Education, Inc.. All rights reserved. Figure
© 2012 Pearson Education, Inc.. All rights reserved. Figure 53
© 2012 Pearson Education, Inc.. All rights reserved. Section 2.5 Logarithmic Functions
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Example 1
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Figure 54
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Example 3 If all the following variable expressions represent positive numbers, then for a > 0, a ≠ 1, the statements in (a)–(c) are true.
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 3 Write the expression as a sum, difference, or product of simpler logarithms. Solution: Using the properties of logarithms,
© 2012 Pearson Education, Inc.. All rights reserved. Figure 55
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 4 Evaluate Solution: Using the change-of-base theorem for logarithms with x = 50 and a = 3 gives
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 5 Solve for x: Solution: This leads to two solutions: x = − 4 and x = 2. But notice that x = − 4 is not a valid value for x in the original equation, since the logarithm of a negative number is undefined. The only solution is, therefore, x = 2.
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 6 Solve for x: Solution: Taking natural logarithms on both sides gives
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Figure 56
© 2012 Pearson Education, Inc.. All rights reserved. Figure 57
© 2012 Pearson Education, Inc.. All rights reserved. Section 2.6 Applications: Growth and Decay; Mathematics of Finance
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 Yeast in a sugar solution is growing at a rate such that 5 g grows exponentially to 18 g after 16 hours. Find the growth function, assuming exponential growth. Solution: The values of y 0 and k in the exponential growth function y = y 0 e kt must be found. Since y 0 is the amount present at time t = 0, y 0 = 5. To find k, substitute y = 18, t = 16, and y 0 = 5 into the equation y = y 0 e kt. Now take natural logarithms on both sides and use the power rule for logarithms and the fact that Continued
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 continued The exponential growth function is where y is the number of grams of yeast present after t hours.
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Estimate the age of a sample with 1/10 the amount of carbon- 14 as a live sample. Solution: Continued
© 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Continued
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved. Figure 58
© 2012 Pearson Education, Inc.. All rights reserved. Graphs of Basic Functions
© 2012 Pearson Education, Inc.. All rights reserved. Chapter 2 Extended Application
© 2012 Pearson Education, Inc.. All rights reserved. Figure 59
© 2012 Pearson Education, Inc.. All rights reserved. Figure 60
© 2012 Pearson Education, Inc.. All rights reserved. Figure 61