Mathematics Functions and Relations: Exponential Functions

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Cornelia Reynolds
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1 a place of mind F A C U L T Y O F E D U C A T I O N Department of Curriculum and Pedagog Mathematics Functions and Relations: Exponential Functions Science and Mathematics Education Research Group Supported b UBC Teaching and Learning Enhancement Fund

Exponential Functions Retrieved from http://maths.

2 Exponential Functions Retrieved from

3 Exponential Functions I What is the domain and range for this exponential function? = 2 x A. { x x R},{ 2, Z} B. { x x R},{ 0, Z} C. { x x R},{ 0, R} D. { x x R},{ 0, R} E. { x x R},{ R}

4 Solution Answer: D Justification: For = 2 x, there is no restriction that prohibits what x could be. Therefore, the domain of x is: x R. For our range, when x is a positive number, should also be positive. When x = 0, is 1 as the exponents rule: 0 = 1, 0 We can also see that there is no value of x that will give us = 0. When x is negative value will never be negative as: x n = 1 x n

5 Solution Continued Thus, since our values will alwas be greater than 0 for all x values, we know that all the values for = 2 x will alwas be above the x axis, creating the horizontal asmptote of = 0. As x becomes smaller, values approaches to 0, but will never actuall be equal to zero. This concept introduces the horizontal asmptote along the x-axis (=0) To sum up, our range is 0, R. Thus, our answer is D.

6 Exponential Functions II x Which of the following graphs corresponds to f ( x) 0. 5? A. B. C. D.

7 Solution Answer: C Justification: From the previous problem, we know that there is no x-intercept. Our -intercept is (0,1). Note: 0.5 = 1 and so -intercept 2 0 f (0) 0.5 f x = 0.5 x = 1x = 1 2 x 2x = 2 x 1 When x, and also when x, 0. approaches Now we know that as x increases, then decreases. The onl trend that displas these two facts is C. Thus, our answer is C.

8 Exponential Functions III x Which of the following graphs corresponds to f ( x) 0. 5? A. B. C. D.

9 Solution Answer: D Justification: From the previous problem, we know that there is no x-intercept. Our -intercept is (0,1). Note: 0.5 = 1 and so -intercept 2 0 f (0) 0.5 f x = 0.5 x = 1 x 2 x = 1 2 x 1 f x = 2 x When x, 0 and also when x,. Now we know that as x increases, increases as well. The onl trend that displas these two facts is D. Thus, our answer is D.

10 Exponential Functions IV Which of the following equations corresponds to the graph below? 1 4 E. 1 3 D. 1 3 C. 1 2 B. 1 2 A. x x x x x

11 Solution Answer: C Justification: First, notice that we have applied transformations (vertical translation) to the exponential functions for creating our new functions. In our case, ever exponential function is shifted verticall b +1 unit. As a result of the vertical shift, the horizontal asmptote has moved from = 0 to = 1. (1) Second, our graph represents a function that is increasing. A function is increasing on an interval, if for an x 1 and x 2 in the interval then x 1 < x 2 implies f(x 1 ) < f(x 2 ). Thus, B and D cannot be our answer since these two functions are decreasing functions. A function is decreasing on an interval, if for an x 1 and x 2 in the interval then x 1 < x 2 implies f x 1 > f(x 2 ). (2)

12 Solution Continued Third, the -intercept is (0, 2) and another point on the graph is (2, 10) (this point was chosen because it was eas to read the values off the graph). That is, when x = 2, = = 10. (3) Consequentl, the option that satisfies (1), (2), and (3) is C. Thus, our answer is C. Horizontal Asmptote at = 1 (2, 10) (0, 2)

Exponential Functions V Titanium 44 or Ti-44 is an important radioactive isotope that is produced in significant quantities during the core-collapse of supernovae.

13 Exponential Functions V Titanium 44 or Ti-44 is an important radioactive isotope that is produced in significant quantities during the core-collapse of supernovae. Ti-44 has a half-life of 60 ears and decas b electron capture. If ou begin with a sample of N 0 quantit (measured in grams, moles, etc.) of Ti-44, what exponential function, N(t), can be used to represent the radioactive deca of Ti-44 after some time t? A. N t = 1 N 60 t 2 0 B. N t = 1 2 N 0 (60/t) C. N t = 1 2 N 0 (t/60) D. N t = N 0 ( 1 2 )(60/t) E. N t = N 0 ( 1 2 )(t/60) Retrieved from

14 Solution Answer: E Justification: Ti-44 has a half-life of 60 ears and decas b electron capture. This means that after 60 ears, a sample of Ti-44 will have lost one half of its original radioactivit. In general, exponential deca processes can be described b N t = N 0 e λt or N t = N 0 ( 1 2 )(t/t 1/2), where t is the time, t 1/2 is the half-life of the decaing quantit, N t is the remaining quantit (not et decaed) after time t, N 0 is the initial quantit (when t = 0) of the substance, and λ is a positive number called the deca constant.

15 Solution Answer: E Options A and C: N t = 1 2 N 0 60 t and N t = 1 2 N 0 (t/60). When t, N t growths., which means that A and C describe exponential Option B: N(t) = 1 2 N 0 (60/t). When t, that N 0 (60/t) N 0 (0) 1. That is, N(t) 1 2. Option D: N t = N 0 ( 1 2 )(60/t). When t, that ( 1 2 )(60/t) ( 1 2 )(0) 1. That is, N t N t 0, which means 60 t 0, which means

16 Solution Answer: E Option E: N t = N 0 ( 1 2 )(t/60). When t, that ( 1 2 )(t/60) ( 1 2 )( ) 0. That is, N t 0. t 60, which means Note that in options A, B, C, and D, N t does not approach 0. Remember, in an exponential deca, the remaining quantit, N t, of a substance approaches zero as t approaches infinit. Thus, E is the correct answer. Ti- 44: Half-life:

Exponential Functions VI Customers of the Bank of Montreal (BMO) can open savings account to earn interest on their investments at an annual interest rate of 0.75%, compounded monthl.

17 Exponential Functions VI Customers of the Bank of Montreal (BMO) can open savings account to earn interest on their investments at an annual interest rate of 0.75%, compounded monthl. If our initial investment with BMO is P 0, what exponential function, P(t), can be used to represent the future value of our investment? Let t be the number of ears our investment is left in the bank. A. B. C. D. E. P( t) P 0 P( t) 1 P 12t t 0 P( t) 1 (0.0075P ) P( t) P P( t) P (1.0075) 0 12t (0.0075) 12t P 12t 0

18 Solution Answer: C 0.75% = Justification: There are several was to earn interest on the mone ou deposit in a bank. If the interest is calculated once a ear, then the interest is called a simple interest. If the interest is calculated more than once a ear, then it is called a compound interest. In our case, it will be a 0.75% annual interest rate compounded monthl. That is, the interest will be compounded 12 times per ear. Options A and E: P t = P 0 12t and P t = P 0 12t P 0. These two options are too good to be true. Imagine if ou were to invest $100 with BMO, then b the end of the first ear, ou would have made more than a septillion dollars (more than ).

19 Solution Answer: C Options B and D: P t = 1 + ( P 0 ) 12t and P t = 1 + P 0 (0.0075) 12t. You will lose our investment with these two options. Imagine if ou were to invest $100 with BMO, then b the end of the first ear and beond, ou would have lost $99. In fact, over time (5, 10, or more ears later), our investment would onl be worth $1. Thus, C is the correct answer. Check the table below: Year P 0 A B C D E 1 $100 $10 24 $1.03 $ $1.00 $ $100 $10 48 $1.00 $ $1.00 $ $100 $10 72 $1.00 $ $1.00 $ $100 $10 96 $1.00 $ $1.00 $10 96

Extend Your Learning with Desmos Desmos graphing calculator: https://www.desmos.

20 Extend Your Learning with Desmos Desmos graphing calculator: Desmos is a free and intuitive tool to help students experience functions. We recommend it to ou!

EXPONENTIAL FUNCTION BASICS COMMON CORE ALGEBRA II BASIC EXPONENTIAL FUNCTIONS

Name: Date: EXPONENTIAL FUNCTION BASICS COMMON CORE ALGEBRA II You studied eponential functions etensivel in Common Core Algebra I. Toda's lesson will review man of the basic components of their graphs