Precalculus by Richard Wright

Come to me, all you who are weary and burdened, and I will give you rest. Matthew 11:28 NIV

Summary: In this section, you will:

- Evaluate exponential functions with base
*b*. - Graph exponential functions with base
*b*. - Evaluate and graph exponential functions with base
*e*.

SDA NAD Content Standards (2018): PC.4.1, PC.5.3

Credit card companies charge interest every day. Banks pay interest every month. How much money would be owed after several years? Or how much money is in a bank account after 10 years? These questions can be answered using exponential functions.

Repeated addition is multiplication. For example, 4 + 4 + 4 + 4 + 4 = 5(4). Repeated multiplication is exponents. For example, 4 ⋅ 4 ⋅ 4 ⋅ 4 ⋅ 4 = 4^{5}. An exponential function is a function that has the variable in the exponent.

An exponential function is in the form of

*f*(*x*) = *ab*^{x}

where *b* > 0, *b* ≠ 1, and *x* is a real number.

Use a calculator to evaluate the following functions.

*f*(*x*) = 2^{x};*x*= 3*g*(*x*) = 5(3)^{x + 1};*x*= 3*h*(*x*) = 6^{−x};*x*= −2

*f*(3) = 2^{3}In a calculator type 2 ^ 3 ENTER. The display reads 8.*g*(3) = 5(3)^{3 + 1}In a calculator type 5 ( 3 ) ^ ( 3 + 1 ) ENTER. The display reads 405.*h*(−2) = 6^{−(−2)}In a calculator type 6 ^ (-) ( (-) 2 ) ENTER. The display reads 36.

Evaluate *f*(*x*) = 2$\left(\frac{1}{2}\right)$^{x − 1} when *x* = 5

1⁄8 = 0.125

The graphs of exponential functions have a characteristic ever-changing curve as in **figure 2**.

Notice the domain is all real numbers because any real number can be an exponent. The range is *y* > 0 if *a* > 0; and *y* < 0 if *a* < 0. The *y*-intercept is (0, *a*), and horizontal asymptote is *y* = 0.

For *f*(*x*) = *ab*^{x}, then

- Domain is all real numbers.
- Range is
*y*> 0 if*a*> 0. - Range is
*y*< 0 if*a*< 0. *y*-intercept is (0,*a*).- Horizontal asymptote is
*y*= 0. - If
*b*> 1, then the function is exponential growth. - If
*b*< 1, then the function is exponential decay.

- Create a table of values by choosing inputs and calculating outputs.
- Plot the points.
- Draw a curve through the points.
- Verify that the domain, range,
*y*-intercepts, and horizontal asymptote of the graph match the function.

Graph

*f*(*x*) = 3^{x}*g*(*x*) = 5^{x}

Are these exponential growth or decay?

Since *b* > 0 in both cases, this is exponential growth. Notice the graph increases.

Graph

- $f\left(x\right)={\left(\frac{1}{3}\right)}^{x}$
- $g\left(x\right)={\left(\frac{1}{5}\right)}^{x}$

Are these exponential growth or decay?

Since *b* < 0 in both cases, this is exponential decay. Notice the graph decreases.

Graph $h\left(x\right)=\left(\frac{1}{2}\right){3}^{x}$

Notice the graphs of exponential growth functions like those in Example 2 always increase. The graphs of exponential decay functions like those in Example 3 always decrease. Because the graphs never switch from increasing to decreasing or vise versa, each input corresponds to a unique output. This type of function is called one-to-one. Each *x* gives a unique *y*-value that is not paired with another *x*.

Exponential functions follow the rules of transformations as given in Lesson 1-07.

Describe the transformation and graph the following functions.

*g*(*x*) = 3(2)^{x − 1}- $h\left(x\right)=-{\left(\frac{2}{3}\right)}^{x+2}+1$

- Vertical shift by factor of 3, shift right 1
- Reflected over the
*x*-axis, shifted left 2 and up 1

Describe the transformation and graph the following function: *f*(*x*) = −(2)^{x − 2} − 1.

Reflected over the *x*-axis, shifted right 2 and down 1.

Many uses of exponential functions are simplest using the number *e* ≈ 2.718281828…. This number is called the natural base and is found by letting *n* get very large and approach ∞ in the expression ${\left(1+\frac{1}{n}\right)}^{n}$. The natural base, *e*, can be useful in calculus because the slope of the function *f*(*x*) = *e*^{x} is also *e*^{x}. For example, the slope of the graph of *f*(*x*) = *e*^{x} at *x* = 2 is *e*^{2}.

Use a calculator to evaluate the following functions.

*f*(*x*) =*e*^{x};*x*= 3*g*(*x*) = 5(*e*)^{x + 1};*x*= 3*h*(*x*) =*e*^{−x};*x*= −2

*f*(3) =*e*^{3}In a calculator type e^{x}3 ENTER. The display reads 20.086.*g*(3) = 5(*e*)^{3 + 1}In a calculator type 5 e^{x}( 3 + 1 ) ENTER. The display reads 272.991.*h*(−2) =*e*^{−(−2)}In a calculator type e^{x}(-) ( (-) 2 ) ENTER. The display reads 7.389.

Evaluate *j*(*x*) = 2*e*^{x + 1} when *x* = −3.

0.736

Graph $f\left(x\right)=\frac{1}{2}{e}^{x}-1$.

When a credit card company or bank calculates, or compounds, interest, they quote a yearly interest rate. But the banks then calculate interest monthly or daily. To derive a formula for calculating compound interest, look at the pattern. In the following table *P* is the principle and *r* is the yearly interest rate (APR). Each year's interest is calculated by *Pr*. The amount in the account will be *P* + *Pr* which is the principle plus the interest. This factors to *P*(1 + *r*).

Year | Balance |
---|---|

0 | P_{0} = P |

1 | P_{1} = P_{0}(1 + r) = P(1 + r) |

2 | P_{2} = P_{1}(1 + r) = P(1 + r)(1 + r) = P(1 + r)^{2} |

3 | P_{3} = P_{2}(1 + r) = P(1 + r)^{2}(1 + r) = P(1 + r)^{3} |

t |
P_{t} = P(1 + r)^{t} |

If the interest is compounded *n* times per year, the formula needs to be modified to $A=P{\left(1+\frac{r}{n}\right)}^{nt}$.

Now think about what happens as the number of compoundings per year gets very large. The most often interest can be compounded is continuously where the interest is calculated every instant when *n* = ∞. To derive the formula for continuously compounded interest, let *m* = ^{n}⁄_{r}, so *n* = *mr*.

$$\begin{array}{c}A=P{\left(1+\frac{r}{mr}\right)}^{mrt}\\ A=P{\left(1+\frac{1}{m}\right)}^{mrt}\\ A=P{\left({\left(1+\frac{1}{m}\right)}^{m}\right)}^{rt}\end{array}$$

For continuous compounding, *m* becomes large and approaches ∞. The expression in parentheses is the definition of *e* when *m* approaches ∞.

*A* = *Pe*^{rt}

$$A=P{\left(1+\frac{r}{n}\right)}^{nt}$$

where *A* is the current amount, *P* is the principle or initial amount, *r* is the interest rate as a decimal, *n* is the number of compoundings per year, and *t* is the time in years.

**Continuously Compounded Interest**

*A* = *Pe*^{rt}

A high school student deposits $1000 in a mutual fund that averages 8% APR interest. How much is the mutual fund worth after 10 years if the interest is compounded

- monthly?
- daily?
- continuously?

Which scenario gives the most money?

*P* = $1000 because that is the initial amount of money. *r* = 0.08 or 8%, and *t* = 10 yrs.

$$A=1000{\left(1+\frac{0.08}{n}\right)}^{n\left(10\right)}$$

- There are 12 months in a year, so
*n*= 12. $A=1000{\left(1+\frac{0.08}{12}\right)}^{12\left(10\right)}=\mathrm{\$2219.64}$ - There are 365.25 days in a year, so
*n*= 365.25. $A=1000{\left(1+\frac{0.08}{365.25}\right)}^{365.25\left(10\right)}=\mathrm{\$2225.35}$ - Compounding continuously uses the other function.
*A*= 1000*e*^{0.08(10)}= $2225.54

Compounding continuously produces the most money. The higher the value of *n*, the more money there is in the account. That is why banks pay interest monthly, but charge interest on credit cards daily.

A college student goes on a weekend shopping spree and charges $500 on a credit card that charges 25% APR interest compounded daily. If they don't pay anything, how much will they owe after 1 year?

$641.96 plus late payment fees

An exponential function is in the form of

*f*(*x*) = *ab*^{x}

where *b* > 0, *b* ≠ 1, and *x* is a real number.

For *f*(*x*) = *ab*^{x}, then

- Domain is all real numbers.
- Range is
*y*> 0 if*a*> 0. - Range is
*y*< 0 if*a*< 0. *y*-intercept is (0,*a*).- Horizontal asymptote is
*y*= 0. - If
*b*> 1, then the function is exponential growth. - If
*b*< 1, then the function is exponential decay.

- Create a table of values by choosing inputs and calculating outputs.
- Plot the points.
- Draw a curve through the points.
- Verify that the domain, range,
*y*-intercepts, and horizontal asymptote of the graph match the function.

\(A = P\left(1+\frac{r}{n}\right)^{nt}\)

where *A* is the current amount, *P* is the principle or initial amount, *r* is the interest rate as a decimal, *n* is the number of compoundings per year, and *t* is the time in years.

**Continuously Compounded Interest**

*A* = *Pe*^{rt}

Helpful videos about this lesson.

- Which fish population is growing at a faster rate?
- What type of fish was initially introduced in greater quantity?
- Assuming the models are still accurate, which type of fish will have the bigger population after 10 years?
*f*(*x*) = 2⋅3^{x}

*f*(0)*f*(2)*f*(−1)*f*(1⁄2)

*f*(*x*) = −*e*^{x + 1}

*f*(0)*f*(2)*f*(−1)*f*(1⁄2)

- What is an asymptote?
- The graph of
*f*(*x*) = 2^{x}is reflected over the*x*-axis and stretched vertically by a factor of 5. (a) Write the new function*g*(*x*), and (b) State its*y*-intercept, domain, and range. - Shift
*f*(*x*) 3 units right and 2 units down - Reflect
*f*(*x*) over the*y*-axis and shift 2 units up *f*(*x*) = 3^{x − 2}- $g\left(x\right)={\left(\frac{1}{2}\right)}^{x}-3$
*h*(*x*) = 2^{1 − x}+ 1*j*(*x*) = −*e*^{x}+ 2*k*(*x*) = 2*e*^{x}- Sally invests $1500 in an account that pays 7.5% interest. How much is the account worth after 20 years if the interest is compounded

- annually?
- semiannually?
- monthly?
- daily?
- continuously?

- (2-09) Solve $\frac{x+1}{x}\le 0$.
- (2-08) Find the asymptotes and graph $f\left(x\right)=\frac{x+1}{x}$.
- (2-05) Find the rational zeros of
*x*^{4}− 2*x*^{3}− 11*x*^{2}+ 6*x*+ 24. - (2-06) Find the irrational zeros of
*x*^{4}− 2*x*^{3}− 11*x*^{2}+ 6*x*+ 24. - (1-09) Verify that
*f*(*x*) = 2*x*+ 3 and $g\left(x\right)=\frac{x-3}{2}$ are inverses by finding their composition.

Two types of fish have been introduced into a pond. The population of fish A can be modeled by *A*(*t*) = 20(1.05)^{t} and fish B by *B*(*t*) = 40(1.025)^{t} where *t* is in years. (Round your answers to the nearest whole number.)

Evaluate the function for the given values.

Write a function that represents the transformation of *f*(*x*) = 3^{x}.

The graph is a transformation of *y* = 2^{x}. Write an equation describing the transformation.

Graph the function. Then state the domain, range, asymptote, and whether it is exponential growth, decay, or neither.

Mixed Review

- Fish A has a larger base (1.05).
- Fish B had 20 more than fish A.
- Fish A will have 33 fish and B will have 51 fish.
- 2; 18; 2⁄3; $2\sqrt{3}$
- −2.718; −20.086; −1; −4.482
- An asymptote is a line that the graph of a function approaches, as
*x*either increases or decreases without bound or approaches from the left or right. The horizontal asymptote of an exponential function tells us the limit of the function’s values as the independent variable gets either extremely large or extremely small. *g*(*x*) = −5(2)^{x};*y*-int: (0, −5); domain: all real numbers; range: (−∞, 0)*g*(*x*) = 3^{x − 3}− 2*g*(*x*) = 3^{−x}+ 2*y*= −2^{x}+ 2- ; domain: all real numbers; range: (0, ∞);
*y*= 0; growth - ; domain: all real numbers; range: (−3, ∞);
*y*= −3; decay - ; domain: all real numbers; range: (1, ∞);
*y*= 1; decay - ; domain: all real numbers; range: (−∞, 2);
*y*= 2; neither - ; domain: all real numbers; range: (0, ∞);
*y*= 0; growth - $6371.78; $6540.57; $6691.23; $6721.50; $6722.53
- [−1, 0)
- VA:
*x*= 0; HA:*y*= 1; - −2, 4
- $\pm \sqrt{3}$
*f*(*g*(*x*)) =*x*