Exponential functions
In a linear function, the output grows by the same amount at each step. In an exponential function, it grows by the same factor at each step. For example, it doubles each time ${x}$ goes up by ${1}$. This kind of growth starts slowly, but soon becomes very fast.
Exponential functions
An exponential function with base ${a}$ is ${f(x)=a^x}$, where ${a>0}$ and ${a\ne 1}$.
The base must be positive: for a negative base, some powers are not real numbers, such as ${(-4)^{1/2}=\sqrt{-4}}$. And for ${a=1}$, the function ${1^x=1}$ is just a constant.
Here are some values of ${f(x)=2^x}$. Each step to the right doubles the output:
| ${x}$ | ${-2}$ | ${-1}$ | ${0}$ | ${1}$ | ${2}$ | ${3}$ |
|---|---|---|---|---|---|---|
| ${2^x}$ | ${\frac{1}{4}}$ | ${\frac{1}{2}}$ | ${1}$ | ${2}$ | ${4}$ | ${8}$ |
The exponent can be any real number, not only a fraction. For example, ${2^{\sqrt{3}}\approx 3.322}$. A calculator finds such values.
Example 1: Let ${f(x)=2^x}$ and ${g(x)=4^x}$. Find:
(a) ${f(5)}$
(b) ${g\left(-\dfrac{1}{2}\right)}$
(c) ${f(-3)}$
Solution:
(a) ${f(5)=2^5=32}$
(b) A negative exponent means a reciprocal, and the exponent ${\dfrac{1}{2}}$ means a square root (see exponents and radicals):
$g\left(-\dfrac{1}{2}\right)=4^{-1/2}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}$
(c) ${f(-3)=2^{-3}=\dfrac{1}{2^3}=\dfrac{1}{8}}$
Graphs of exponential functions
The graph of ${y=\left(\frac{1}{2}\right)^x=2^{-x}}$ is the graph of ${y=2^x}$ reflected across the ${y}$-axis. Every exponential graph has the same basic features:
The graph of ${f(x)=a^x}$ passes through ${(0,1)}$, because ${a^0=1}$.
If ${a>1}$, ${f}$ is increasing. If ${0<a<1}$, ${f}$ is decreasing.
The domain is ${(-\infty,\infty)}$, and the range is ${(0,\infty)}$.
The ${x}$-axis, ${y=0}$, is a horizontal asymptote.
The outputs are always positive, because a positive number to any power is positive. So the graph never touches the ${x}$-axis.
Example 2: Sketch the graph of ${y=2^{x+1}-4}$. Give its asymptote and intercepts.
Solution:
Start with ${y=2^x}$. The ${+1}$ in the exponent moves the graph ${1}$ unit to the left. The ${-4}$ moves it ${4}$ units down (see graphs of functions). The asymptote moves down too, to ${y=-4}$.
The ${y}$-intercept is
${y=2^{1}-4=-2}$.
For the ${x}$-intercept, set ${y=0}$:
$\begin{align*}2^{x+1}&=4=2^2\\x+1&=2\\x&=1\end{align*}$
Example 3: Find an exponential function ${f(x)=Ca^x}$ whose graph passes through ${(0,3)}$ and ${(2,12)}$.
Solution:
Put ${x=0}$: ${f(0)=Ca^0=C}$. So ${C=3}$.
Put ${x=2}$: ${3a^2=12}$, so ${a^2=4}$. The base must be positive, so ${a=2}$.
The function is ${f(x)=3\cdot 2^x}$.
Compound interest
Banks pay interest on savings. With compound interest, the interest is added to the account, and later interest is paid on it too. Let
- ${P}$ be the principal (the amount you put in),
- ${r}$ be the yearly interest rate, as a decimal (for example, ${6\%=0.06}$),
- ${n}$ be the number of times per year the interest is added,
- ${t}$ be the time in years.
In each period, the account grows by the factor ${1+\dfrac{r}{n}}$. There are ${nt}$ periods in ${t}$ years. So the amount after ${t}$ years is
${A=P\left(1+\dfrac{r}{n}\right)^{nt}}$
Example 4: You put ${\$1000}$ in an account that pays ${6\%}$ interest per year. How much is in the account after ${5}$ years, if the interest is compounded
(a) once a year?
(b) every month?
Solution:
Here ${P=1000}$, ${r=0.06}$, and ${t=5}$.
(a) With ${n=1}$:
$\begin{align*}&A\\&=1000(1+0.06)^{5}\\&=1000(1.06)^5\approx 1338.23\end{align*}$
(b) With ${n=12}$, each month adds ${\dfrac{0.06}{12}=0.005}$, and there are ${12\cdot 5=60}$ months:
$\begin{align*}&A\\&=1000(1.005)^{60}\\&\approx 1348.85\end{align*}$
So the account holds ${\$1338.23}$ with yearly compounding, and ${\$1348.85}$ with monthly compounding.
The number ${e}$
What if the interest is added more and more often: every day, every hour, every second? Take ${P=1}$, ${r=1}$ (that is, ${100\%}$), and ${t=1}$. Then ${A=\left(1+\dfrac{1}{n}\right)^n}$. Watch what happens as ${n}$ grows:
| ${n}$ | ${\left(1+\frac{1}{n}\right)^n}$ |
|---|---|
| ${1}$ | ${2}$ |
| ${10}$ | ${2.59374}$ |
| ${100}$ | ${2.70481}$ |
| ${1000}$ | ${2.71692}$ |
| ${10{,}000}$ | ${2.71815}$ |
| ${1{,}000{,}000}$ | ${2.71828}$ |
The values do not grow without bound. They approach a number near ${2.71828}$. This number is called ${e}$:
${e\approx 2.71828}$
Like ${\pi}$, the number ${e}$ is irrational: its decimals never end or repeat. The function ${f(x)=e^x}$ is called the natural exponential function. Its base ${e}$ is between ${2}$ and ${3}$, so its graph lies between the graphs of ${2^x}$ and ${3^x}$. The ${e^x}$ key on a calculator gives its values.
When interest is added at every instant, we say it is compounded continuously. Then the amount is
${A=Pe^{rt}}$
Example 5: Find the amount after ${5}$ years when ${\$1000}$ is invested at ${6\%}$ per year, compounded continuously.
Solution:
Here ${rt=0.06\cdot 5=0.3}$:
${A=1000e^{0.3}\approx 1349.86}$
This is only a little more than the ${\$1348.85}$ from monthly compounding in Example 4.
An application
Example 6: A culture of bacteria starts with ${500}$ cells. The number of cells doubles every ${3}$ hours. How many cells are there after ${12}$ hours?
Solution:
After ${3}$ hours there are ${500\cdot 2}$ cells. After ${6}$ hours, ${500\cdot 2^2}$. In ${t}$ hours, the number doubles ${\dfrac{t}{3}}$ times. So
${N(t)=500\cdot 2^{t/3}}$.
After ${12}$ hours, it has doubled ${\dfrac{12}{3}=4}$ times:
${N(12)=500\cdot 2^4=500\cdot 16=8000}$
There are ${8000}$ cells after ${12}$ hours.
Summary
- An exponential function is ${f(x)=a^x}$, with ${a>0}$ and ${a\ne 1}$. Each step of ${1}$ in ${x}$ multiplies the output by ${a}$.
- Its graph passes through ${(0,1)}$ and has the horizontal asymptote ${y=0}$. It rises if ${a>1}$ and falls if ${0<a<1}$.
- Compound interest: ${A=P\left(1+\dfrac{r}{n}\right)^{nt}}$.
- The number ${e\approx 2.71828}$ is the value that ${\left(1+\dfrac{1}{n}\right)^n}$ approaches as ${n}$ grows.
- Continuous compounding: ${A=Pe^{rt}}$.