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7. Inverse Functions and Logarithms
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United States of America
Grades 9-12

7.04 Exponential functions and the natural base e

Lesson
Practice

Introduction

We were first introduced to exponential function in Math 1 lesson  4.01 Exponential functions  . We looked more specifically at growth and decay functions in the Math 1 lessons  4.02 Exponential growth  and  4.03 Exponential decay  . We will review these functions, their key features, and transformations in this lesson in preparation for finding the inverse function of exponentials in the next lesson.

Exponential functions

Exponential functions can be classified as exponential growth functions or exponential decay functions based on the value of the constant factor.

\displaystyle f\left(x\right)=ab^x
\bm{a}
The y-intercept
\bm{b}
The constant factor
Exponential growth

Increasing by a constant factor, b, where b>1

Exponential decay

Decreasing by a constant factor, b, where 0<b<1

Growth factor

The constant factor of an exponential growth function.

Decay factor

The constant factor of an exponential decay function.

Exponential functions can also be expressed in terms of their constant percent rate of change.

\displaystyle f\left(x\right)=a\left(1\pm r\right)^x
\bm{a}
The initial value
\bm{r}
The growth or decay rate
Growth rate

The fixed percent by which an exponential function increases.

In the form f\left(x\right)=a\left(1+r\right)^x, r is a growth rate, and the constant percent rate of change is positive.

Decay rate

The fixed percent by which an exponential function decreases.

In the form f\left(x\right)=a\left(1-r\right)^x, r is a decay rate, and the constant percent rate of change is negative.

We can determine the key features of an exponential function from its equation or graph:

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  • f\left(x\right)=ab^x, b>1

  • f\left(x\right)=a\left(1+r\right)^x, r>0

  • The graph is increasing

  • y approaches a minimum value of 0

  • The domain is \left(-\infty, \infty\right)

  • The range is \left(0, \infty\right)

  • The y-intercept is at \left(0,\, a\right)

  • The common ratio is b

  • The horizontal asymptote is y=0

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y

  • f\left(x\right)=ab^x, 0<b<1

  • f\left(x\right)=a\left(1-r\right)^x, r>0

  • The graph is decreasing

  • y approaches a minimum value of 0

  • The domain is \left(-\infty, \infty\right)

  • The range is \left( -\infty, 0\right)

  • The y-intercept is at \left(0,\, a\right)

  • The common ratio is b

  • The horizontal asymptote is y=0

The exponential parent function y=b^{x} can be transformed to y=ab^{c\left(x-h\right)}+k

  • If a<0, the graph is reflected across the x-axis

  • The graph is stretched or compressed vertically by a factor of a

  • If c<0, the graph is reflected across the y-axis

  • The graph is stretched or compressed horizontally by a factor of c

  • The graph is translated horizontally by h units

  • The graph is translated vertically by k units

Examples

Example 1

Consider the table of values for function f\left(x\right).

x-101234
f\left(x\right)\dfrac{5}{3}51545135405
a

Determine whether the function represents an exponential function or not.

Worked Solution
Create a strategy

To determine if a set of data represents an exponential function, we want to compare successive y-values, and determine if there is a common ratio.

Apply the idea

We can see that f\left(-1\right)=\dfrac{5}{3} and f\left(0\right)=5, so we want to calculate the common ratio that takes us from \dfrac{5}{3} to 5. \frac{5}{3} \cdot b=5 \longrightarrow b=5 \cdot \frac{3}{5} = 3 The common ratio is 3. We now want to check if this factor works for the remaining y-values:\begin{aligned}5 \cdot 3&=15 \\ 15 \cdot 3 &= 45\\ 45\cdot 3&=135\\ 135\cdot 3&=405\end{aligned}

The function represents an exponential function.

Reflect and check

Instead, we could have divided successive outputs to make sure there is a constant common ratio: \begin{aligned}5 \div \frac{5}{3}&=3 \\ 15 \div 5 &= 3\\ 45\div 15&=3\\ 135\div 45&=3\\405\div 135 &=3\end{aligned}

b

Write an equation to represent the exponential function.

Worked Solution
Create a strategy

The function has a y-intercept when x=0. The y-intercept is at y=5 and the growth factor is 3.

Apply the idea

An equation that represents this exponential function is: f\left(x\right)=5\left(3\right)^x

Reflect and check

If we wanted to know the percent at which the function grows, we can use the fact that b=1+r in the exponential growth functions. \begin{aligned}3&=1+r\\2&=r\end{aligned} Changing this into a percentage, we find the growth rate is 200\%.

Example 2

Consider the following piecewise function:

f\left(x\right) = \begin{cases} \left(\frac{1}{3}\right)^x, & \enspace x \leq 0 \\ -x^{3}, & \enspace x \gt 0 \end{cases}

a

Sketch a graph of the piecewise function.

Worked Solution
Create a strategy

Since the piecewise changes behavior at x=0, the graph will have different rates of change on the left versus the right side of the y-axis. We can use a table of values to find and plot specific points, then connect the points to determine the function.

Apply the idea

When x\leq 0, the graph is an exponential curve represented by f\left(x\right)=\left(\frac{1}{3}\right)^x. We can use the values x=-3,\,-2,\,-1,\,0 in a table.

x-3-2-10
f\left(x\right)27931

Note that the point \left(0,1\right) is included, so a filled in point should be in its place.

When x\gt 0, the graph is a cubic curve represented by f\left(x\right)=-x^3. We can use the values {x=0,\,1,\,2,\,3} in the next part of the table.

x0123
f\left(x\right)0-1-8-27

Note that the point \left(0,0\right) is not included, so an unfilled point should be in its place.

Plotting these points and connecting them, we get the graph of f\left(x\right).

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b

State the domain and range.

Worked Solution
Create a strategy

We can use the graph or the definition of the piecewise function to help us determine the domain and range. Since the domain is included in the definition of a piecewise function, we can use it to find the domain. To find the range, it may be easier to use the graph.

Apply the idea

Looking at the definition of the function, all values less than or equal to 0 are included in the first piece of the domain. All values greater than 0 are included in the second piece. Therefore, the domain of the function is \left(-\infty,\infty\right).

Looking at the graph to determine the range, we can see the end behavior tends toward negative and positive infinity. However, we need to talk a closer look at what happens in the middle of the graph. Changing the scale of the y-axis:

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It is now easy to see that there are values that are not included in the range. The values between y=0 and y=1 are not within the range. We can also see y=0 is not in the range but y=1 is in the range.

The range of the function is \left(-\infty,0\right)\cup\left[1,\infty\right).

Example 3

For each scenario, find the equation of the transformed function.

a

The graph of f\left(x\right) is translated to get the graph of g\left(x\right) show below.

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Worked Solution
Create a strategy

From the graph, we notice the y-intercept of f\left(x\right) has been translated 2 units to the left and 3 units up to obtain the point \left(-2, 4\right) on the graph of g\left(x\right).

A translation of 2 units to the left is equivalent to adding 2 to the input values. A translation of 3 units up is equivalent to adding 3 to output values.

Since the graph has only been translated, the base of both exponential functions will be the same. We can determine the base by finding the ratio of consecutive outputs of the parent function or by finding the ratio of the differences of consecutive outputs of the transformed function.

Apply the idea

The transformed function will be in the form y=b^{x+h}+k where h=2 and k=3.

Looking at the transformed function, we can easily identify the points \left(-2,4\right), \left(-1,6\right), and \left(0,12\right). Analyzing the change in the y-values, we find:

xy\text{Change in }y
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-16\text{Increases by }2
012\text{Increases by }6

The ratio of the differences in these outputs is \frac{6}{2}=3. This means b=3.

Therefore, g\left(x\right)=3^{x+2}+3.

Reflect and check

We could have used consecutive outputs of the parent function to find the base instead. \begin{aligned}3\div 1 &= 3\\9\div 3&=3 \end{aligned}However, dividing consecutive outputs to find the base only works for exponential functions of the form y=ab^x.

b

The graph of y = \left(\dfrac{1}{2}\right)^{x} is reflected across the x-axis and stretched vertically by a factor of 2.

Worked Solution
Create a strategy

We can represent this change graphically:

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  • A reflection across the x-axis is equivalent to multiplying the output values by -1.

  • A vertical stretch by a factor of 2 is equivalent to multiplying the output values by 2.

Apply the idea

The transformed function is y=-2\left(\dfrac{1}{2}\right)^{x}

Reflect and check

To graph this, we could have built a table of values for the parent function of y=\left(\dfrac{1}{2}\right)^x and multiplied the outputs by -2 to find points on the new graph. We will call the parent function f\left(x\right) and the transformed function g\left(x\right).

x-3-2-1012
f\left(x\right)8421\frac{1}{2}\frac{1}{4}
g\left(x\right)-16-8-4-2-1-\frac{1}{2}

Example 4

The graph of the function f\left(x\right) = 5\cdot 2^{x} is translated 10 units down to give a new function g\left(x\right).

Complete the table of values for f\left(x\right) and the transformed function g\left(x\right), then sketch the graphs of both functions.

x-2-101234
f\left(x\right)
g\left(x\right)
Worked Solution
Create a strategy

To find the values for f\left(x\right) we can substitute in the values from the table and evaluate.

As the graph of f\left(x\right) = 5 \cdot 2^{x} is translated 10 units down, g\left(x\right)=f\left(x\right)-10, we can subtract 10 from each value in the table for f\left(x\right) to find the corresponding value for g\left(x\right)

Similarly, the graph of g\left(x\right) will be the graph of f\left(x\right) translated down 10 units.

Apply the idea
x-2-101234
f\left(x\right)\dfrac{5}{4}\dfrac{5}{2}510204080
g\left(x\right)-\dfrac{35}{4}-\dfrac{15}{2}-50103070
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Reflect and check

The graph being translated in this problem was already transformed from the parent function. The parent function is p\left(x\right)=2^x. The parent was stretched vertically by a factor of 5 to obtain the graph of f\left(x\right). Then, we translated f\left(x\right) down 10 units.

If the graph had been translated first, then stretched vertically, this graph would have looked very different.

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If two vertical or two horizontal transformations are applied one after the other, the order in which they are applied will affect the final outcome.

Idea summary

Exponential functions grow or decay at a constant percent rate of change. The function increases when b>1 and decreases when 0<b<1.

The constant rate in decimal form can be found by solving b=1+r for r in growth functions or by solving b=1-r for r in decay functions.

Euler's constant

We are familiar with the famous irrational constant, \pi, but there are other irrational constants that are just as valuable in upper-level mathematics. One of those was discovered by and named after mathematician Leonhard Euler (pronounced "oiler").

Euler's number - e

An irrational number approximately equal to 2.71828.

Euler's constant has many applications in natural growth and compound interest. We can evaluate powers of e using the button on our calculator.

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The graph of e^x appears similar to the graph of 2^x or 3^x since 2<e<3.

Exploration

Recall the formula for compound interest: A=P\left(1+\frac{r}{n}\right)^{nt} where A is the future value, P is in principle or initial amount, r is the interest rate, n is the number of times the interest is compounded, and t is the time in years.

Suppose you borrow money from someone charging 100\% interest for 1 year. This situation would be modeled by: A=P\left(1+\frac{1}{n}\right)^{n} Let's examine what happens as the number of times the interest is compounded during the year increases.

  1. Complete the table of values.

    \text{Compound}\\ \text{period}\text{Annual}\text{Quarter}\text{Month}\text{Biweekly}\text{Week}\text{Day}\text{Hour}
    n141226523658760
    \left(1+\frac{1}{n}\right)^{n}

  2. Create a hypothesis for what will happen to the value of \left(1+\frac{1}{n}\right)^{n} as the number of compounding periods increases.

  3. Confirm or refute your hypothesis using the GeoGebra animation below. Either click the "Start Animation" button or slide the blue slider through the different values of n.

Loading interactive...

Euler's number can be used to model continuous growth and decay:

\displaystyle f\left(x\right)=ae^{kx}
\bm{a}
Initial value
\bm{k}
Growth or decay rate

When k>0, the function represents continuous growth. When k<0, the function has been reflected across the y-axis and represents continuous decay.

Examples

Example 5

Consider the exponential function f\left(x\right)=e^{-x}-2

a

Find the coordinates of the y-intercept.

Worked Solution
Create a strategy

We can see that the parent function y=e^x has been reflected across the y-axis and translated down 2 units.

The y-intercept of y=e^{-x}-2 occurs when x=0, so we can substitute x=0, or we can consider what would happen when we perform the transformations on the y-intercept of y=e^x, which is at \left(0,1\right).

Apply the idea

Substituting x=0 into f\left(x\right)=e^{-x}-2 gives

\displaystyle f\left(0\right)\displaystyle =\displaystyle e^{-0}-2
\displaystyle =\displaystyle -1

The y-intercept of f\left(x\right)=e^{-x}-2 is at \left(0,-1\right).

Reflect and check

Reflecting across the y-axis does not change the y-intercept. Translating \left(0,1\right) down 2 units puts the y-intercept at \left(0,-1\right).

b

State the domain and range.

Worked Solution
Create a strategy

The domain of the parent function is \left(-\infty, \infty\right) and the range is \left(0, \infty\right). Neither the reflection across the y-axis, or the vertical translation will alter the domain. The range will be shifted down 2 units.

Apply the idea

The domain of f\left(x\right)=e^{-x}-2 is \left(-\infty, \infty\right) and the range is \left(-2, \infty\right).

c

Sketch a graph of the function.

Worked Solution
Create a strategy

The graph of the function will be the graph of the parent function, reflected across the y-axis, and translated 2 units downwards. We know it has a y-intercept at \left(0,-1\right), domain of \left(-\infty, \infty\right), and range of \left(-2, \infty\right). The asymptote will be shifted down 2 units also.

Apply the idea

To find the key points on the parent function, y=e^x, we can build a table of values. The values need to be approximated with a calculator.

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y0.1350.36812.7187.389

Reflecting these values across the y-axis:

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y7.3892.71810.3680.135

And shifting them down 2:

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y5.3890.718-1-1.632-1.865

Plotting these points, we get the graph of the transformed function f\left(x\right)=e^{-x}-2.

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Reflect and check

From the graph, we can see that the reflection across the y-axis changes the function to an exponential decay function. Algebraically, we can rewrite f\left(x\right)=e^{-x} as f\left(x\right)=\left(\frac{1}{e}\right)^{x}. 0<\frac{1}{e}<1 Since 0<b<1, this represents exponential decay.

Example 6

A small town is experiencing a population boom and is currently growing continuously at a rate of 5\% every year. The mayor wants to ensure that the housing needs of the growing population are met and passes a bill to support housing for 1000 additional people every year.

The current town population is 5000 residents and the town currently has enough housing to support a population of 8000.

a

Create an algebraic model for the population and for the amount of housing in the town after x years.

Worked Solution
Create a strategy

The population is growing exponentially 'continuously', so we can use the function f\left(x\right)=ae^{kx} to model the population. The housing is growing linearly so we can use g\left(x\right)=mx+b to model the available housing.

Apply the idea

The initial population is 5000 residents, so a=5000 in f\left(x\right). The rate is increasing by 5\%, so k=0.05. The population can be modeled with f\left(x\right)=5000(e)^{0.05x}

There are currently enough houses for 8000 people, so b=8000 in g\left(x\right). The houses will support an additional 1000 people each year, so m=1000. The housing can be modeled with g\left(x\right)=1000x+8000

Reflect and check

We can use properties of exponents to write f\left(x\right) as

\displaystyle f\left(x\right)\displaystyle =\displaystyle 5000\left(e\right)^{0.05x}
\displaystyle =\displaystyle 5000\left(e^{0.05}\right)^x

Then, we can use our calculator to approximate e^{0.05}.

\displaystyle f\left(x\right)\displaystyle =\displaystyle 5000\left(1.0513\right)^x

Because we are talking about people, we will need to round final answers down to the nearest whole number.

b

Create a graph of both functions on the same coordinate plane to determine when the town will run out of available housing. Choose an appropriate scale and label the axes of your graph.

Worked Solution
Create a strategy

We can use a table of values to get an idea of how the population and the housing grows over time:

Years0102030405060
Population (people)5000824313\,59122\,40836\,94560\,912100\,427
Housing (people)800018\,00028\,00038\,00048\,00058\,00068\,000

This table shows us that the population exceeds the available housing between 40 and 50 years and also gives us an idea of the scale needed to graph the relationship.

The scale of the x-axis can increase by intervals of ten, and the domain should be [0,60]. The range of the y-axis should be [0,100\,000] and increase by intervals of ten thousand.

Apply the idea
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\text{Years}
10000
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30000
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50000
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\text{People}

The population exceeds the available housing between 48 and 50 years.

Reflect and check

When using technology to graph the functions, we can determine the exact point where the functions intersect. Your graphing device should tell you the curves intersect at about x=48.454. This means the population will exceed the available housing around 48.5 years later.

Idea summary

Euler's constant, e, is approximately 2.71828.

Euler's constant can be used to model continuous growth and decay:

\displaystyle f\left(x\right)=ae^{kx}
\bm{a}
Initial value
\bm{k}
Growth or decay rate

The function f\left(x\right) represents exponential growth when k>0 and exponential decay when k<0.

Outcomes

A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems.

A.CED.A.2

Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.

F.IF.B.4

For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.

F.IF.C.7.E

Graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude.

F.IF.C.8

Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function.

F.IF.C.8.B

Use the properties of exponents to interpret expressions for exponential functions.

F.BF.B.3

Identify the effect on the graph of replacing f(x) by f(x) + k, kf(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them.

F.LE.A.3

Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function.

F.LE.B.5

Interpret the parameters in a linear or exponential function in terms of a context.

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