1. Warm-Up 4/21 D. Rigor: You will learn how to evaluate, analyze, graph and solve exponential functions. Relevance: You will be able to solve population.

1.Warm-Up 4/21

D

36+

36−

26

46

Rigor:You will learn how to evaluate, analyze, graph

and solve exponential functions.

Relevance:You will be able to solve population problems and

solve half-life chemistry problems using exponential functions.

3-1 Exponential Functions

Algebraic Functions are functions are solved using algebraic operations.Transcendental Functions are functions that can not be expressed in terms of algebraic operations. They transcend Algebra.

Exponential and Logarithmic Functions are transcendental functions.

Example 1a: Sketch and analyze the graph of the function. Describe its domain, range, intercepts, asymptotes, end behavior, and where the function is increasing, or decreasing.𝑓 (𝑥 )=3𝑥

x y

– 3 0.04

– 2 0.11

– 1 0.33

0 1

1 3

2 9

3 27

Domain:Range:y-intercept:Asymptotes:End Behavior:

Increasing/Decreasing:

(−∞ ,∞)(0 ,∞)

(0 ,1)𝑦=0

lim𝑥→−∞

𝑓 (𝑥)=0 lim𝑥→∞

𝑓 (𝑥)=∞and

Increasing :(−∞ ,∞)

Example 1b: Sketch and analyze the graph of the function. Describe its domain, range, intercepts, asymptotes, end behavior, and where the function is increasing, or decreasing.𝑓 (𝑥 )=2−𝑥

x y

– 3 8

– 2 4

– 1 2

0 1

1 .5

2 .25

3 .125

Domain:Range:y-intercept:Asymptotes:End Behavior:

Increasing/Decreasing:

(−∞ ,∞)(0 ,∞)

(0 ,1)𝑦=0

lim𝑥→−∞

𝑓 (𝑥)=∞ lim𝑥→∞

𝑓 (𝑥)=0and

Decreasing :(−∞, ∞)

Example 2a: Use the graph of to describe the transformation of the function. Then sketch both functions.

𝑔 (𝑥 )=2𝑥+1

x y

– 4 .125

– 3 .25

– 2 .5

– 1 1

0 2

1 4

2 8

𝑓 (𝑥 )=2𝑥

x y

– 3 .125

– 2 .25

– 1 .5

0 1

1 2

2 4

3 8

𝑔 (𝑥 )= 𝑓 (𝑥+1)The graph of is the graph of translated 1 unit to the left.

Example 2b: Use the graph of to describe the transformation of the function. Then sketch both functions.

h (𝑥 )=2−𝑥

x y

– 3 8

– 2 4

– 1 2

0 1

1 .5

2 .25

3 .125

𝑓 (𝑥 )=2𝑥

x y

– 3 .125

– 2 .25

– 1 .5

0 1

1 2

2 4

3 8

h (𝑥 )= 𝑓 (− 𝑥)The graph of is the graph of reflected in the y-axis.

Example 2c: Use the graph of to describe the transformation of the function. Then sketch both functions.

𝑗 (𝑥 )=−3(2)𝑥

x y

– 3 – .375

– 2 – .75

– 1 – 1.5

0 – 3

1 – 6

2 – 12

3 – 24

𝑓 (𝑥 )=2𝑥

x y

– 3 .125

– 2 .25

– 1 .5

0 1

1 2

2 4

3 8

𝑗 (𝑥 )=−3 𝑓 (𝑥)The graph of is the graph of reflected in the x-axis and expanded vertically by a factor of 3.

Example 3a: Use the graph of to describe the transformation of the function. Then sketch both functions.

𝑎 (𝑥 )=𝑒4𝑥

x y

– 3 6.1x10–6

– 2 3.4x10–4

– 1 .01832

0 1

1 54.598

2 2981

3 162755

𝑓 (𝑥 )=𝑒𝑥

x y

– 4 .01832

– 3 .04979

– 2 .13534

– 1 .36788

0 1

1 2.7183

2 7.3891

3 20.086

4 54.598

𝑎 (𝑥 )= 𝑓 (4 𝑥 )The graph of is the graph of compressed horizontally by a factor of 4.

Example 3b: Use the graph of to describe the transformation of the function. Then sketch both functions.

𝑏 (𝑥 )=𝑒−𝑥+3

x y

– 3 23.086

– 2 10.389

– 1 5.7183

0 4

1 3.3679

2 3.1353

3 3.0498

𝑓 (𝑥 )=𝑒𝑥

x y

– 3 .04979

– 2 .13534

– 1 .36788

0 1

1 2.7183

2 7.3891

3 20.086

𝑏 (𝑥 )= 𝑓 (−𝑥 )+3The graph of is the graph of reflected in the y-axis andtranslated 3 units up.

Example 3c: Use the graph of to describe the transformation of the function. Then sketch both functions.

𝑐 (𝑥 )=12𝑒𝑥

x y

– 3 .02489

– 2 .06767

– 1 .18394

0 .5

1 1.3591

2 3.6945

3 10.043

𝑓 (𝑥 )=𝑒𝑥

x y

– 3 .04979

– 2 .13534

– 1 .36788

0 1

1 2.7183

2 7.3891

3 20.086

𝑐 (𝑥 )=12𝑓 (𝑥 )

The graph of is the graph of compressed vertically by a factor of .

Example 4: Krysti invest $300 in an account with 6% interest rate, making no other deposits or withdrawals. what will Krysti’s account balance be after 20 years if the interest is compounded:

a. semiannually?

b. Monthly?

c. Daily?

𝐴=𝑃 (1+ 𝑟𝑛 )

𝑛𝑡

P = 300, r = 0.06, t = 20

n = 2𝐴=300(1+ 0.06

𝑛 )𝑛(20)

𝐴=300(1+ 0.062 )

2 (20 )

𝐴≈978.61

n = 12𝐴=300(1+ 0.06

12 )12(20)

𝐴≈993.06

n = 365𝐴=300(1+ 0.06

365 )365(20)

𝐴≈995.94

Example 5: Suppose Krysti invest $300 in an account with 6% interest rate, making no other deposits or withdrawals. What will Krysti’s account balance be after 20 years if the interest is compounded continuously?

𝐴=𝑃 𝑒𝑟 𝑡P = 300, r = 0.06, t = 20

𝐴=300𝑒.06 (20)

𝐴≈996.04

a. 1.42% annually

b. 1.42% continuously

Example 6: Mexico has a population of approximately 110 million. If Mexico’s population continues to grow at the described rate, predict the population of Mexico in 10 and 20 years.

𝑁=𝑁 0𝑒𝑘𝑡

N0 = 110,000,000

r = 0.0142

,t = 10 and t = 20

𝑁=𝑁 0 (1+𝑟 )𝑡

𝑁=110,000,000 (1+0.0142 )10 𝑁=110,000,000 (1+0.0142 )20

𝑁 ≈126,656,869 𝑁 ≈145,836,022 k = 0.0142

𝑁=110,000,000𝑒0.0142(10)

𝑁 ≈126,783,431𝑁=110,000,000𝑒0.0142(20)

𝑁 ≈146,127,622

√−1math!

3-1 Assignment: TX p166, 4-32 EOE Test Corrections Due Friday 4/25Chapter 3 test Thursday 5/1