Unit 3: Exponential and Logarithmic Functions Section 3.1 Graphs of Exponential Functions including transformations and base ‘e’ Notes: Exponential Function- function containing a variable exponent f(x) = 𝑏 𝑥 where b is a positive constant other than 1 (b > 0 and b ≠ 1) think about why b ≠ 1 Using a half sheet of graph paper, sketch the graph of f(x) = 2 𝑥 Then find the following: lim 𝑥→−∞ 2 𝑥 lim 𝑥→∞ 2 𝑥 =0 = ∞

Sketch the graphs of the following functions on the same set of axes as you graphed f(x) = 2 𝑥 . g(x) = 3 𝑥 h(x) = 4 𝑥 j(x) = 1 2 𝑥 k(x) = 1 3 𝑥 l(x) = 1 4 𝑥 Then come up with as many properties of exponential functions that you can (there are a total of 7) Domain is (-∞, ∞) Range is (0, ∞) Horizontal asymptote at y = 0 y-intercept = 1 If b > 1, graph is increasing If 0 < b < 1, graph is decreasing When b > 1, as b increases, the graph gets steeper When 0 < b < 1, as b decreases, the graph gets steeper It is a 1:1 function

Using the transformation rules we learned in Chapter 1, describe the transformations that are occurring in the examples below. Be sure to list them in the correct order. f(x) = −2 (−𝑥+4) g(x) = 3· 2 (𝑥−3) − 6 h(x) = 1 2 ( −2 (𝑥+1) ) + 5 left 4, reflect over the x and y axis right 3, vertically stretched by a factor of 3, down 6 left 1, vertically shrunk by a factor of ½, reflect over the x-axis, up 5

Natural Base ‘e’ The natural base ‘e’ is an irrational number symbolized by the letter e. It was discovered by a Swiss mathematician Leonard Euler who proved that: lim 𝑛→∞ 1+ 1 𝑛 𝑛 e where e ≈ 2.71828… e is called the natural base What does the graph of f(x) = 𝑒 𝑥 look like? Sketch the graph on the same grid as all the others.

Example In 1969, the world population was 3.6 billion, with a growth rate of 2% per year. The function f(x) = 3.6 𝑒 0.02𝑥 describes world population, f(x), in billions, x years after 1969. Find the world population for 2020. In 2000, the world population was approximately 6 billion but the growth rate had slowed to 1.3%. Find the world population in 2050. f(x) = 6 𝑒 0.013𝑥 𝑤ℎ𝑒𝑟𝑒 𝑥=# 𝑜𝑓 𝑦𝑒𝑎𝑟𝑠 𝑠𝑖𝑛𝑐𝑒 2000 9.98 billion 11.49 billion

Compound Interest Interest computed on original investment as well as any already accumulated interest. 𝐴= 𝑃(1+ 𝑟 𝑛 ) 𝑛𝑡 Where: A = final balance P = principal (original investment) r = annual percentage rate in decimal form n = number of compounding periods per year t = number of years Be sure to know the following: Annually: n = 1 Semi-annually: n = 2 Quarterly: n = 4

Some banks use continuous compounding where the number of compounding periods increases infinitely. as n →∞, (1+ 1 𝑛 ) 𝑛 →𝑒 Therefore: A = Pert is the formula for continuous compounding Example: Suppose $10000 is invested at an annual rate of 8%. Find the balance in the account after 5 years when interest is compounded: a) Quarterly b) continuously 𝐴= 10000(1+ .08 4 ) 4·5 A = 10000e.08·5 A = $14,918.25 A = $14,859.47

Homework: Pg.396-97/19-24 all, 29, 39, 51 Day 2: Pg. 397/ 53,55,57,61,63,67,73