Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
3. Section 9.3
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Exercise 113 Page 523

Practice makes perfect
a To determine which function will be ahead when x=10, we have to calculate their values when x is 10. We already know two of the function's rules, which means we can calculate their value directly by substituting 10 for x and evaluating.
g( 10)& = 10( 10)^2=1000 h( 10)& = 1.01^(10)≈ 1.1046 The function g(x) wins over h(x) when x= 1000. Examining the table showing m(x), we notice that it doubles when x increase by 1. With this information, we can continue the pattern to find the value of m(x) when x is 10.

From the table, we see that m(10)=10.24. This means g(x) is ahead when x=10.

b To calculate how fast each function was going during the first two seconds, we have to subtract the function's value when x=0 from the function's value when x=2 and divide by the difference in time. This gives us the function's average rate of change. By using the table we can identify these data points for m(x).

m(x)=m(2)-m(0)/2-0
m(x)=0.04- 0.01/2-0
â–¼
Evaluate right-hand side
m(x)=0.03/2
m(x)=0.015

The average rate of change for m(x) is 0.015. To calculate the remaining average rates of change, we have to determine their values when x=0 and x=2.

g( 0)& = 10( 0)^2=0 g( 2)& = 10( 2)^2=40 [1em] h( 0)& = 1.01^0= 1 h( 2)& = 1.01^2= 1.0201 Let's start by determining the average rate of change for g(x).

g(x)=g(2)-g(0)/2-0
g(x)=40- 0/2-0
â–¼
Evaluate right-hand side
g(x)=40/2
g(x)=20

Next we will determine the average rate of change for h(x).

h(x)=h(2)-h(0)/2-0
h(x)=1.0201- 1/2-0
â–¼
Evaluate right-hand side
h(x)=0.0201/2
h(x)=0.01005

c As explained in Part A, m(x) doubles when the input increases by 1. This means this is an exponential function with a multiplier of b= 2.

m(x)=a( 2)^x

From the table, we also know that m(x)=0.01 when x=0, which means the initial value is a=0.01. Now we can complete the equation.
m(x)=0.01(2)^x
d Let's label what kind of functions g(x), h(x), and m(x) are.

ll g(x)=10x^2 & (Power function) [0.2em] h(x)=1.01^x & (Exponential function) [0.2em] m(x)=0.01(2)^x & (Exponential function) For very large values of x, an exponential function always overtakes a power function which means we are left with two functions to decide between. l g(x)=10x^2 (Power function) [0.2em] h(x)=1.01^x (Exponential function) [0.2em] m(x)=0.01(2)^x (Exponential function) Of these functions, the exponential function with the greater base will end up overtaking the other. Since 2>1.01, we know that m(x) will win the race. l g(x)=10x^2 (Power function) [0.2em] h(x)=1.01^x (Exponential function) [0.2em] m(x)=0.01(2)^x (Exponential function) To determine when m(x) takes the lead we have to investigate when it is greater than both functions. Therefore, let's set m(x) equal to the other functions and solve for x. We will need a graphing calculator for this.

Graphing Calculator

To plot the functions on our calculator, we first push the Y= button and type two of them on two of the rows. Having written the functions, we can push GRAPH to draw them.

We want to know when the exponential function overtakes the power function. This will happen outside of the calculator's standard window. Therefore, let's resize the window by pushing WINDOW and changing the settings.

To find the point of intersection, push 2nd and CALC and choose the fifth option, intersect. Choose the first and second curve, then pick a best guess for the point of intersection.

The functions intersect when x is approximately 18.51. Let's also find when m(x) overtakes h(x) by following the same procedure as above.

The exponential functions intersect at x≈ 6.74. This means when x≈ 18.51, the exponential function m(x) has taken the lead over both functions.