Precalculus – Section 3.1

An exponential function is a function of the form

We call b the base of the exponential function.a is a constant multiplier (think stretch/shrink).

Requirements:b greater than zero and not equal to onex is any real number

xbaxf )(

xxf 2)( xxf 21)(

)1,0(

The graph of the function crosses the y-axis at the point (0,a).

)2,1(

The graph also contains the point (1,a⋅b).

)2,1(

)1,0(

To get more points on the graph:

As you increase the value of x by 1, you multiply the previous y-value by the base b.

As you decrease the value of x by 1, you divide the y-value by b.

)4,2(

)8,3(

)2

1,1(

…then the function grows.

…then the function decays.

If the base is greater than 1…

If the base is less than 1…

x

xf

2

3)(

x

xf

3

1)(

Graph the exponential function. xxf 3)(

x y

0 1

1 .3333

2 .1111

-1 3

-2 9

p. 206: 1-4, 7-10, 19-24

Precalculus – Section 3.1

A common choice for the base of an exponential function is e .

e is the called the natural base because it naturally occurs in things such as:

compound interest radioactive decay science applications

The value of e to 15 decimal places…

e = 2.718281828459045…

Think of President Andrew Jackson (the guy on the $20 bill)!

Good enough for precalculus use: 2.718281828

Evaluate each function. Round to 3 decimal places.

1.

2.

1.4 when )( xexf x

-1 when )( 3 xexf x

1.4)1.4( ef 1.4)718281828.2(

340.60

231)1( eef

389.7

Graph the function.xexg 3)(

x f (x)

0 3

1 8.155

2 22.167

-1 1.103

-2 0.406

p. 206: 5,6, 25-30

You may want to scale your graphs to fit them on the paper.

Tomorrow: using exponential functions to problem solve!

Precalculus – Section 3.1

To find interest that is compounded continuously:

P = principle (amount invested)r = interest rate (as a decimal)t = time (in years)

rtPeA

Find the current balance of a $7000 savings fund after 1 year if the interest is compounded continuously at 8%.

rtPeA)1)(08.0()7000( eA

01.7583$A

To find the amount of a radioactive substance that remains after t years:

N = initial quantityt = time (in years)H = half-life of substance (in years)

H

t

NQ

2

1

A certain radioactive substance has a half-life of 825 years. Find the amount of substance that remains after 1000 years if the initial amount is 50 pounds.

H

t

NQ

2

1

825

1000

2

150

Q

pounds 58.21Q

p. 207: 51-53, 55-58