Graph Of E X 2

Exponential Functions

At this signal in our study of algebra we begin to wait at transcendental functions or functions that seem to "transcend" algebra. Nosotros take studied functions with variable bases and constant exponents such as $x^2$ or $y^{-\mathrm{three}}.$ In this section we explore functions with a abiding base and variable exponents. Given a real number $b>0$ where $b\ne 1$ an exponential partWhatever function with a definition of the class $f\left(x\right)=b^x$ where $b>0$ and $b\ne 1.$ has the form,

$$\begin{array}{c}f\left(\mathrm{ten}\right)=b^{\mathrm{ten}}Expon\mathrm{eastward}ntialFunctio\mathrm{due\; north}\end{array}$$

For example, if the base b is equal to two, and then we have the exponential function divers by $f\left(x\right)=2^x.$ Here nosotros tin see the exponent is the variable. Up to this point, rational exponents accept been divers but irrational exponents have non. Consider ${\mathrm{ii}}^{\sqrt{\mathrm{seven}}}$ , where the exponent is an irrational number in the range,

$$2.64<\sqrt{\mathrm{vii}}<2.65$$

We can use these bounds to estimate $2^{\sqrt{7}}$ ,

$$\begin{array}{c}{\mathrm{ii}}^{\mathrm{ii.64}}<2^{\sqrt{7}}<2^{2.65}\\ 6.23<2^{\sqrt{7}}<6.28\end{array}$$

Using rational exponents in this manner, an approximation of ${\mathrm{two}}^{\sqrt{7}}$ can be obtained to any level of accuracy. On a calculator,

$$2^\sqrt{\mathrm{vii}}\approx 6.26$$

Therefore the domain of any exponential function consists of all real numbers $\left(-\infty ,\infty \right).$ Choose some values for 10 and so determine the corresponding y-values.

$$\begin{array}{llllll}x\hfill & y\hfill & f\left(x\right)=2^{\mathrm{10}}\hfill & \hfill & \hfill & Solutio\mathrm{due\; north}s\hfill \\ -\text{2}\hfill & \frac{\mathrm{one}}{\mathrm{iv}}\hfill & y={\mathrm{two}}^{-2}=\frac{\mathrm{i}}{{\mathrm{ii}}^2}=\frac{1}{\mathrm{four}}\hfill & \hfill & \hfill & \left(-2,\frac{\mathrm{ane}}{4}\right)\hfill \\ -1\hfill & \frac{1}{2}\hfill & y={\mathrm{two}}^{-1}=\frac{1}{2^1}=\frac{1}{2}\hfill & \hfill & \hfill & \left(-1,\frac{1}{2}\right)\hfill \\ 0\hfill & \mathrm{ane}\hfill & y=2^0=1\hfill & \hfill & \hfill & \left(0,1\right)\hfill \\ 1\hfill & \mathrm{ii}\hfill & y=2^1=2\hfill & \hfill & \hfill & \left(\text{1},\text{2}\right)\hfill \\ 2\hfill & 4\hfill & y={\mathrm{two}}^2=\mathrm{four}\hfill & \hfill & \hfill & \left(\mathrm{ii},4\right)\hfill \\ \sqrt{7}\hfill & 6.26\hfill & y=2^{\sqrt{\mathrm{seven}}}\approx \mathrm{half\; dozen.26}\hfill & \hfill & \hfill & \left(2.65,\mathrm{six.26}\right)\hfill \end{array}$$

Because exponents are divers for any real number we can sketch the graph using a continuous curve through these given points:

It is important to point out that as x approaches negative infinity, the results become very small but never really attain zero. For example,

$$\begin{array}{ccc}\hfill f\left(-5\right)& =& 2^{-5}=\frac{1}{2^5}\approx 0.03125\hfill \\ \hfill f\left(-\mathrm{ten}\right)& =& 2^{-10}=\frac{1}{2^{10}}\approx 0.0009766\hfill \\ \hfill f\left(-15\right)& =& 2^{-15}=\frac{\mathrm{i}}{2^{-\mathrm{xv}}}\approx .00003052\hfill \end{array}$$

This describes a horizontal asymptote at $y=0$ , the 10-axis, and defines a lower spring for the range of the role: $\left(0,\infty \right).$

The base of operations b of an exponential part affects the rate at which it grows. Below we have graphed $y={\mathrm{two}}^x$ , $y={\mathrm{iii}}^x$ , and $y={\mathrm{x}}^x$ on the aforementioned set of axes.

Notation that all of these exponential functions have the same y-intercept, namely $\left(0,1\right).$ This is because $f\left(0\right)=b^0=\mathrm{i}$ for any function defined using the form $f\left(x\right)=b^x.$ As the functions are read from left to correct, they are interpreted every bit increasing or growing exponentially. Furthermore, any exponential function of this form will have a domain that consists of all real numbers $\left(-\infty ,\infty \right)$ and a range that consists of positive values $\left(0,\infty \right)$ bounded by a horizontal asymptote at $y=0.$

Example 1

Sketch the graph and determine the domain and range: $f\left(x\right)={10}^x+5.$

Solution:

The base 10 is used ofttimes, near notably with scientific notation. Hence, 10 is chosen the common base. In fact, the exponential function $y={\mathrm{ten}}^x$ is then of import that you volition find a button $\begin{array}{}{10}^{\mathrm{10}}\end{array}$ dedicated to information technology on most modern scientific calculators. In this example, we will sketch the basic graph $y={\mathrm{ten}}^x$ and then shift information technology up v units.

Note that the horizontal asymptote of the basic graph $y={10}^x$ was shifted up 5 units to $y=5$ (shown dashed). Accept a infinitesimal to evaluate a few values of x with your computer and convince yourself that the upshot will never be less than 5.

Answer:

Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(\mathrm{five},\infty \right)$

Next consider exponential functions with fractional bases $0<b<1.$ For instance, $f\left(\mathrm{ten}\right)={\left(\frac{1}{\mathrm{ii}}\right)}^{\mathrm{ten}}$ is an exponential function with base $b=\frac{\mathrm{i}}{2}.$

$$\begin{array}{llllll}\mathrm{10}\hfill & y\hfill & f\left(x\right)={\left(\frac{1}{\mathrm{two}}\right)}^x\hfill & \hfill & \hfill & Solutions\hfill \\ -2\hfill & \mathrm{four}\hfill & f\left(\frac{\mathrm{i}}{2}\right)={\left(\frac{1}{2}\right)}^{-\mathrm{ii}}=\frac{{\mathrm{ane}}^{-2}}{2^{-2}}=\frac{2^2}{1^2}=4\hfill & \hfill & \hfill & (-\mathrm{ii},4)\hfill \\ -1\hfill & 2\hfill & f\left(\frac{\mathrm{ane}}{2}\right)={\left(\frac{1}{2}\right)}^{-\mathrm{one}}=\frac{1^{-\mathrm{ane}}}{2^{-1}}=\frac{{\mathrm{ii}}^{\mathrm{one}}}{{\mathrm{one}}^{\mathrm{ane}}}=2\hfill & \hfill & \hfill & (-\mathrm{one},2)\hfill \\ 0\hfill & 1\hfill & f\left(\frac{\mathrm{i}}{2}\right)={\left(\frac{1}{\mathrm{ii}}\right)}^0=1\hfill & \hfill & \hfill & (0,1)\hfill \\ 1\hfill & \frac{\mathrm{one}}{2}\hfill & f\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^1=\frac{1}{2}\hfill & \hfill & \hfill & (1,\frac{\mathrm{ane}}{2})\hfill \\ 2\hfill & \frac{1}{4}\hfill & f\left(\frac{1}{2}\right)={\left(\frac{\mathrm{ane}}{2}\right)}^2=\frac{1}{4}\hfill & \hfill & \hfill & (2,\frac{1}{4})\hfill \end{array}$$

Plotting points we have,

Reading the graph from left to right, it is interpreted as decreasing exponentially. The base affects the charge per unit at which the exponential role decreases or decays. Beneath we take graphed $y={\left(\frac{1}{2}\right)}^x$ , $y={\left(\frac{1}{3}\right)}^x$ , and $y={\left(\frac{\mathrm{i}}{10}\right)}^x$ on the same set of axes.

Remember that $x^{-1}=\frac{1}{x}$ and so nosotros can express exponential functions with fractional bases using negative exponents. For instance,

$$\mathrm{grand}\left(x\right)={\left(\frac{1}{2}\right)}^x=\frac{1^{\mathrm{ten}}}{{\mathrm{ii}}^x}=\frac{1}{2^x}=2^{-x}.$$

Furthermore, given that $f\left(\mathrm{10}\right)=2^{\mathrm{ten}}$ we can see $g\left(x\right)=f\left(-x\right)=2^{-\mathrm{ten}}$ and can consider m to be a reflection of f nigh the y-centrality.

In summary, given $b>0$

And for both cases,

$$\begin{array}{cc}\hfill Domai\mathrm{northward}:& \left(-\infty ,\infty \right)\hfill \\ \hfill Ra\mathrm{north}ge:\text{}& \left(0,\infty \right)\hfill \\ \hfill y-intercept:& \left(0,\mathrm{ane}\right)\hfill \\ \hfill Asymptot\mathrm{east}:\text{}\text{}\text{}& y=0\hfill \end{array}$$

Furthermore, notation that the graphs pass the horizontal line test and thus exponential functions are one-to-ane. We use these basic graphs, along with the transformations, to sketch the graphs of exponential functions.

Example 2

Sketch the graph and make up one's mind the domain and range: $f\left(x\right)=5^{-x}-\mathrm{x}.$

Solution:

Brainstorm with the basic graph $y=5^{-x}$ and shift it down x units.

The y-intercept is $\left(0,-9\right)$ and the horizontal asymptote is $y=-10.$

Answer:

Domain: $\left(-\text{∞},\infty \right)$ ; Range: $\left(-\mathrm{ten},\infty \right)$

Note: Finding the ten-intercept of the graph in the previous example is left for a later section in this chapter. For now, nosotros are more concerned with the general shape of exponential functions.

Case 3

Sketch the graph and determine the domain and range: $g\left(x\right)=-2^{x-\mathrm{three}}.$

Solution:

Begin with the bones graph $y=2^{\mathrm{ten}}$ and identify the transformations.

$$\begin{array}{cccc}y& =& {\mathrm{two}}^x\hfill & Basicgraph\hfill \\ y& =& -2^x\hfill & R\mathrm{eastward}flectio\mathrm{north}aboutth\mathrm{due\; east}x-a\mathrm{ten}is\hfill \\ y& =& -2^{\mathrm{10}-3}\hfill & \mathrm{Southward}hiftright3u\mathrm{north}it\mathrm{due\; south}\hfill \end{array}$$

Note that the horizontal asymptote remains the same for all of the transformations. To cease we ordinarily desire to include the y-intercept. Remember that to notice the y-intercept we set $\mathrm{ten}=0.$

$$\begin{array}{ccc}\hfill g\left(0\right)& =& -2^{0-3}\hfill \\ & =& -2^{-\mathrm{iii}}\hfill \\ & =& -\frac{1}{2^3}\hfill \\ & =& -\frac{\mathrm{one}}{8}\hfill \end{array}$$

Therefore the y-intercept is $\left(0,-\frac{\mathrm{one}}{8}\right)$ .

Answer:

Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(-\infty ,0\right)$

Try this! Sketch the graph and decide the domain and range: $f\left(x\right)={\mathrm{two}}^{\mathrm{ten}-\mathrm{ane}}+\mathrm{iii}.$

Answer:

Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(3,\infty \right)$

Natural Base of operations e

Some numbers occur often in common applications. One such familiar number is pi (π), which nosotros know occurs when working with circles. This irrational number has a defended button on nearly calculators $\begin{array}{}\mathit{\pi }\end{array}$ and approximated to five decimal places, $\mathit{\pi }\approx 3.14159.$ Another of import number eastward occurs when working with exponential growth and decay models. It is an irrational number and approximated to 5 decimal places, $e\approx \mathrm{ii.71828}.$ This constant occurs naturally in many existent-globe applications and thus is called the natural base of operations. Sometimes e is called Euler'south abiding in honor of Leonhard Euler (pronounced "Oiler").

Effigy 7.1

Leonhard Euler (1707–1783)

In fact, the natural exponential function, $$f\left(x\right)=e^x$$ is then important that yous volition find a button $\begin{array}{}{e}^{\mathrm{10}}\end{array}$ dedicated to information technology on whatsoever modern scientific computer. In this section, nosotros are interested in evaluating the natural exponential function for given real numbers and sketching its graph. To evaluate the natural exponential function, defined by $f\left(x\right)={\mathrm{east}}^{\mathrm{10}}$ where $x=-2$ using a calculator, you may need to employ the shift push. On many scientific calculators the caret will display every bit follows,

$$f\left(-2\right)=e^\left(-2\right)\approx 0.13534$$

Afterwards learning how to apply your particular calculator, you can now sketch the graph by plotting points. (Round off to the nearest hundredth.)

$$\begin{array}{llllll}x\hfill & y\hfill & f\left(x\right)=e^x\hfill & \hfill & \hfill & \mathrm{Due\; south}o\mathrm{fifty}ution\mathrm{south}\hfill \\ -\mathrm{ii}\hfill & 0.14\hfill & f\left(-2\right)=e^{-2}=0.14\hfill & \hfill & \hfill & (-2,0.14)\hfill \\ -\mathrm{i}\hfill & 0.37\hfill & f\left(-1\right)={\mathrm{due\; east}}^{-1}=0.37\hfill & \hfill & \hfill & (-1,0.37)\hfill \\ 0\hfill & 1\hfill & f\left(0\right)=e^0=1\hfill & \hfill & \hfill & (0,\mathrm{ane})\hfill \\ \mathrm{i}\hfill & \mathrm{ii.72}\hfill & f\left(1\right)={\mathrm{eastward}}^{\mathrm{one}}=2.72\hfill & \hfill & \hfill & (\mathrm{i},\mathrm{two.72})\hfill \\ \mathrm{two}\hfill & 7.39\hfill & f\left(2\right)={\mathrm{east}}^2=7.39\hfill & \hfill & \hfill & (\mathrm{two},7.39)\hfill \end{array}$$

Plot the points and sketch the graph.

Note that the function is like to the graph of $y=3^x.$ The domain consists of all real numbers and the range consists of all positive existent numbers. There is an asymptote at $y=0$ and a y-intercept at $\left(0,\mathrm{i}\right).$ We can apply the transformations to sketch the graph of more complicated exponential functions.

Example 4

Sketch the graph and determine the domain and range: $\mathrm{one\; thousand}\left(x\right)={\mathrm{east}}^{\mathrm{10}+2}-\mathrm{iii}.$

Solution:

Identify the basic transformations.

$$\begin{array}{cccc}\hfill y& =& {\mathrm{eastward}}^x\hfill & Basickraph\hfill \\ \hfill y& =& {\mathrm{due\; east}}^{x+2}\hfill & Shiftleft2u\mathrm{north}its\hfill \\ \hfill y& =& e^{x+2}-\mathrm{iii}\hfill & Shiftdow\mathrm{north}3unit\mathrm{due\; south}\hfill \end{array}$$

To determine the y-intercept set $x=0.$

$$\begin{array}{ccc}\hfill \mathrm{grand}\left(0\right)& =& e^{0+\mathrm{two}}-\mathrm{iii}\hfill \\ & =& {\mathrm{eastward}}^2-3\hfill \\ & \approx & \mathrm{four.39}\hfill \end{array}$$

Therefore the y-intercept is $\left(0,e^{\mathrm{ii}}-3\right).$

Answer:

Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(-\mathrm{three},\infty \right)$

Try this! Sketch the graph and decide the domain and range: $f\left(x\right)=e^{-x}+2.$

Respond:

Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(2,\infty \right)$

Chemical compound Involvement Formulas

Exponential functions appear in formulas used to calculate involvement earned in most regular savings accounts. Compound interest occurs when interest accumulated for one menstruation is added to the primary investment before calculating interest for the next period. The amount accrued in this manner over time is modeled past the compound interest formulaA formula that gives the amount accumulated past earning involvement on chief and interest over time: $A\left(t\right)=P{\left(1+\frac{r}{\mathrm{northward}}\right)}^{nt}.$ :

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

Here the amount A depends on the time t in years the principal P is accumulating compound interest at an almanac involvement rate r. The value n represents the number of times the interest is compounded in a year.

Example 5

An investment of $500 is made in a 6-twelvemonth CD that earns $4\frac{1}{2}$ % annual interest that is compounded monthly. How much will the CD be worth at the end of the 6-twelvemonth term?

Solution:

Here the main $P=\$500$ , the involvement charge per unit $r=4\frac{1}{2}\%=0.045$ , and because the interest is compounded monthly, $n=12.$ The investment is modeled by the post-obit,

$$A\left(t\right)=500{\left(1+\frac{0.045}{12}\right)}^{12t}$$

To determine the corporeality in the account after 6 years evaluate $A\left(\mathrm{vi}\right)$ and circular off to the nearest cent.

$$\begin{array}{ccc}\hfill A\left(6\right)& =& 500{\left(1+\frac{0.045}{12}\right)}^{12\left(\mathrm{six}\right)}\hfill \\ & =& 500{\left(1.00375\right)}^{72}\hfill \\ & =& 654.65\hfill \end{array}$$

Answer: The CD volition be worth $654.65 at the end of the 6-year term.

Next we explore the furnishings of increasing n in the formula. For the sake of clarity we allow P and r equal 1 and calculate accordingly.

Annual compounding	${\left(\mathrm{ane}+\frac{\mathrm{i}}{n}\right)}^n$
Yearly ( $\mathrm{north}=1$ )	${\left(\mathrm{i}+\frac{1}{\mathrm{one}}\right)}^{\mathrm{i}}=\mathrm{two}$
Semi-annually ( $n=2$ )	${\left(1+\frac{1}{2}\right)}^2=\mathrm{two.25}$
Quarterly ( $n=4$ )	${\left(1+\frac{1}{4}\right)}^4\approx 2.44140$
Monthly ( $n=12$ )	${\left(1+\frac{1}{12}\right)}^{12}\approx \mathrm{two.61304}$
Weekly ( $n=52$ )	${\left(1+\frac{1}{52}\right)}^{52}\approx \mathrm{two.69260}$
Daily ( $n=365$ )	${\left(1+\frac{1}{365}\right)}^{365}\approx 2.71457$
Hourly ( $n=8760$ )	${\left(1+\frac{\mathrm{i}}{8760}\right)}^{8760}\approx \mathrm{two.71813}$

Standing this pattern, as n increases to say compounding every infinitesimal or even every 2d, we can see that the result tends toward the natural base $\mathrm{east}\approx 2.71828.$ Compounding interest every instant leads to the continuously compounding interest formulaA formula that gives the amount accumulated by earning continuously compounded interest: $A\left(t\right)=P{e}^{rt}.$ ,

$$A\left(t\right)=P{e}^{rt}$$

Hither P represents the initial chief corporeality invested, r represents the annual interest charge per unit, and t represents the fourth dimension in years the investment is allowed to accrue continuously compounded interest.

Case 6

An investment of $500 is fabricated in a 6-year CD that earns $\mathrm{four}\frac{\mathrm{ane}}{2}$ % annual interest that is compounded continuously. How much will the CD be worth at the end of the 6-year term?

Solution:

Here the principal $P=\$500$ , and the interest charge per unit $r=4\frac{\mathrm{i}}{2}\%=0.045.$ Since the interest is compounded continuously we will apply the formula $A\left(t\right)=P{e}^{rt}.$ The investment is modeled by the following,

$$A\left(t\right)=500{\mathrm{east}}^{0.045t}$$

To decide the amount in the business relationship after 6 years, evaluate $A\left(6\right)$ and round off to the nearest cent.

$$\begin{array}{ccc}\hfill A\left(6\right)& =& 500{\mathrm{due\; east}}^{0.045\left(6\right)}\hfill \\ & =& 500e^{0.27}\hfill \\ & =& 654.98\hfill \end{array}$$

Reply: The CD will be worth $654.98 at the end of the 6-twelvemonth term.

Compare the previous two examples and note that compounding continuously may not be as benign as it sounds. While information technology is better to compound interest more than oftentimes, the difference is non that profound. Certainly, the interest rate is a much greater factor in the stop upshot.

Effort this! How much will a $ane,200 CD, earning 5.ii% almanac involvement compounded continuously, be worth at the end of a 10-yr term?

Respond: $two,018.43

Key Takeaways

• Exponential functions have definitions of the form $f\left(x\right)=b^{\mathrm{ten}}$ where $b>0$ and $b\ne \mathrm{ane}.$ The domain consists of all real numbers $\left(-\infty ,\infty \right)$ and the range consists of positive numbers $\left(0,\infty \right).$ Also, all exponential functions of this form have a y-intercept of $\left(0,\mathrm{ane}\right)$ and are asymptotic to the 10-axis.
• If the base of an exponential office is greater than ane ( $b>1$ ), then its graph increases or grows as it is read from left to right.
• If the base of operations of an exponential function is a proper fraction ( $0<b<1$ ), and so its graph decreases or decays as it is read from left to right.
• The number 10 is called the common base and the number e is called the natural base.
• The natural exponential function divers by $f\left(\mathrm{10}\right)=e^x$ has a graph that is very similar to the graph of $k\left(x\right)=3^{\mathrm{ten}}.$
• Exponential functions are one-to-one.

Topic Exercises

Function A: Exponential Functions

Evaluate.

1. $f\left(\mathrm{10}\right)=3^x$ where $f\left(-\mathrm{ii}\right)$ , $f\left(0\right)$ , and $f\left(2\right).$

2. $f\left(x\right)={10}^x$ where $f\left(-1\right)$ , $f\left(0\right)$ , and $f\left(1\right).$

3. $k\left(x\right)={\left(\frac{1}{3}\right)}^x$ where $g\left(-\mathrm{one}\right)$ , $g\left(0\right)$ , and $g\left(\mathrm{iii}\right).$

4. $\mathrm{chiliad}\left(\mathrm{ten}\right)={\left(\frac{\mathrm{iii}}{4}\right)}^x$ where $g\left(-2\right)$ , $g\left(-1\right)$ , and $\mathrm{one\; thousand}\left(0\right).$

5. $h\left(x\right)=9^{-\mathrm{ten}}$ where $h\left(-\mathrm{one}\right)$ , $h\left(0\right)$ , and $h\left(\frac{\mathrm{ane}}{2}\right).$

6. $h\left(x\right)=4^{-x}$ where $h\left(-1\right)$ , $h\left(-\frac{1}{2}\right)$ , and $h\left(0\right).$

7. $f\left(x\right)=-2^{\mathrm{ten}}+1$ where $f\left(-\mathrm{i}\right)$ , $f\left(0\right)$ , and $f\left(\mathrm{iii}\right).$

8. $f\left(\mathrm{10}\right)=2-3^x$ where $f\left(-1\right)$ , $f\left(0\right)$ , and $f\left(2\right).$

9. $k\left(x\right)={10}^{-x}+20$ where $g\left(-2\right)$ , $\mathrm{chiliad}\left(-\mathrm{i}\right)$ , and $\mathrm{thousand}\left(0\right).$

10. $g\left(\mathrm{ten}\right)=1-{\mathrm{ii}}^{-\mathrm{10}}$ where $\mathrm{thousand}\left(-1\right)$ , $\mathrm{thou}\left(0\right)$ , and $\mathrm{thousand}\left(1\right).$

Use a calculator to approximate the following to the nearest hundredth.

11. $f\left(x\right)={\mathrm{ii}}^x+5$ where $f\left(2.5\right).$

12. $f\left(x\right)=3^{\mathrm{10}}-\mathrm{ten}$ where $f\left(3.2\right).$

13. $\mathrm{one\; thousand}\left(\mathrm{ten}\right)={\mathrm{four}}^{\mathrm{ten}}$ where $g\left(\sqrt{2}\right).$

14. $g\left(x\right)=5^x-1$ where $\mathrm{yard}\left(\sqrt{3}\right).$

15. $h\left(\mathrm{10}\right)={\mathrm{x}}^x$ where $h\left(\mathit{\pi }\right).$

16. $h\left(x\right)={10}^x+\mathrm{ane}$ where $h\left(\frac{\mathit{\pi }}{3}\right)$

17. $f\left(x\right)={\mathrm{x}}^{-x}-\mathrm{ii}$ where $f\left(1.5\right).$

18. $f\left(\mathrm{ten}\right)=5^{-\mathrm{10}}+\mathrm{three}$ where $f\left(1.3\right).$

19. $f\left(x\right)={\left(\frac{\mathrm{two}}{3}\right)}^x+\mathrm{i}$ where $f\left(-\mathrm{2.vii}\right).$

20. $f\left(x\right)={\left(\frac{\mathrm{iii}}{5}\right)}^{-x}-\mathrm{i}$ where $f\left(1.4\right).$

Sketch the function and decide the domain and range. Draw the horizontal asymptote with a dashed line.

21. $f\left(\mathrm{ten}\right)=4^x$

22. $g\left(\mathrm{10}\right)=3^{\mathrm{ten}}$

23. $f\left(x\right)=4^{\mathrm{ten}}+\mathrm{ii}$

24. $f\left(x\right)=3^x-6$

25. $f\left(x\right)=2^{\mathrm{10}-2}$

26. $f\left(\mathrm{10}\right)={\mathrm{four}}^{\mathrm{10}+2}$

27. $f\left(x\right)=3^{\mathrm{10}+1}-4$

28. $f\left(x\right)={10}^{x-\mathrm{four}}+2$

29. $h\left(\mathrm{ten}\right)={\mathrm{two}}^{\mathrm{ten}-3}-2$

30. $h\left(x\right)=3^{x+2}+4$

31. $$f\left(x\right)={\left(\frac{1}{4}\right)}^x$$

32. $$h\left(x\right)={\left(\frac{1}{\mathrm{iii}}\right)}^{\mathrm{ten}}$$

33. $$f\left(x\right)={\left(\frac{1}{4}\right)}^x-2$$

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

35. $g\left(x\right)={\mathrm{two}}^{-x}-3$

36. $\mathrm{one\; thousand}\left(x\right)={\mathrm{iii}}^{-x}+1$

37. $f\left(x\right)=6-{10}^{-x}$

38. $g\left(\mathrm{10}\right)=5-{\mathrm{four}}^{-\mathrm{ten}}$

39. $f\left(x\right)=\mathrm{v}-{\mathrm{two}}^x$

40. $f\left(x\right)=3-3^x$

Part B: Natural Base e

Find $f\left(-1\right)$ , $f\left(0\right)$ , and $f\left(\frac{\mathrm{iii}}{\mathrm{two}}\right)$ for the given office. Use a calculator where advisable to approximate to the nearest hundredth.

41. $f\left(x\right)={\mathrm{east}}^x+\mathrm{two}$

42. $f\left(\mathrm{ten}\right)={\mathrm{east}}^x-\mathrm{four}$

43. $f\left(x\right)=\mathrm{five}-3e^x$

44. $f\left(x\right)=e^{-x}+3$

45. $f\left(x\right)=1+e^{-x}$

46. $f\left(x\right)=3-\mathrm{ii}{e}^{-x}$

47. $f\left(x\right)=e^{-\mathrm{two}x}+\mathrm{ii}$

48. $f\left(x\right)=e^{-{\mathrm{ten}}^2}-1$

Sketch the role and determine the domain and range. Draw the horizontal asymptote with a dashed line.

49. $f\left(x\right)={\mathrm{due\; east}}^x-\mathrm{three}$

50. $f\left(x\right)=e^x+2$

51. $f\left(x\right)={\mathrm{east}}^{x+1}$

52. $f\left(x\right)={\mathrm{east}}^{x-3}$

53. $f\left(x\right)=e^{\mathrm{ten}-2}+\mathrm{i}$

54. $f\left(\mathrm{10}\right)={\mathrm{due\; east}}^{\mathrm{ten}+2}-1$

55. $g\left(\mathrm{10}\right)=-e^{\mathrm{10}}$

56. $g\left(x\right)={\mathrm{east}}^{-x}$

57. $h\left(\mathrm{ten}\right)=-e^{x+\mathrm{ane}}$

58. $h\left(\mathrm{10}\right)=-{\mathrm{east}}^{\mathrm{10}}+\mathrm{iii}$

Function C: Compound Involvement Formulas

59. Jim invested $750 in a 3-year CD that earns 4.2% annual involvement that is compounded monthly. How much will the CD be worth at the terminate of the 3-year term?

60. Jose invested $2,450 in a 4-year CD that earns 3.6% almanac involvement that is compounded semi-annually. How much will the CD be worth at the stop of the iv-twelvemonth term?

61. Jane has her $v,350 savings in an business relationship earning $3\frac{5}{8}$ % annual interest that is compounded quarterly. How much will exist in the account at the end of 5 years?

62. Bill has $12,400 in a regular savings account earning $4\frac{\mathrm{two}}{\mathrm{three}}$ % annual interest that is compounded monthly. How much volition exist in the business relationship at the terminate of 3 years?

63. If $85,200 is invested in an account earning 5.viii% annual involvement compounded quarterly, and so how much interest is accrued in the first 3 years?

64. If $124,000 is invested in an account earning 4.6% annual interest compounded monthly, then how much interest is accrued in the first 2 years?

65. Bill invested $ane,400 in a 3-twelvemonth CD that earns 4.2% annual involvement that is compounded continuously. How much volition the CD exist worth at the end of the iii-yr term?

66. Brooklyn invested $2,850 in a 5-year CD that earns 5.3% annual interest that is compounded continuously. How much volition the CD be worth at the cease of the 5-twelvemonth term?

67. Omar has his $4,200 savings in an business relationship earning $\mathrm{four}\frac{\mathrm{iii}}{8}$ % almanac involvement that is compounded continuously. How much will be in the business relationship at the cease of $2\frac{1}{\mathrm{two}}$ years?

68. Nancy has her $8,325 savings in an account earning $\mathrm{v}\frac{7}{8}$ % almanac interest that is compounded continuously. How much volition exist in the business relationship at the end of $\mathrm{v}\frac{\mathrm{i}}{2}$ years?

69. If $12,500 is invested in an account earning 3.8% annual interest compounded continuously, then how much interest is accrued in the showtime 10 years?

70. If $220,000 is invested in an account earning four.5% almanac interest compounded continuously, then how much interest is accrued in the get-go 2 years?

71. The population of a sure small-scale town is growing co-ordinate to the function $P\left(t\right)=\mathrm{12,500}{\left(\mathrm{i.02}\right)}^t$ where t represents fourth dimension in years since the terminal census. Use the function to determine the population on the day of the census (when t = 0) and approximate the population in 6 years from that time.

72. The population of a certain small town is decreasing co-ordinate to the role $P\left(t\right)=\mathrm{22,300}{\left(0.982\right)}^t$ where t represents time in years since the last census. Utilize the role to determine the population on the day of the census (when t = 0) and estimate the population in 6 years from that time.

73. The decreasing value, in dollars, of a new car is modeled by the formula $V\left(t\right)=\mathrm{28,000}{\left(0.84\right)}^t$ where t represents the number of years afterward the car was purchased. Use the formula to determine the value of the car when it was new (t = 0) and the value after 4 years.

74. The number of unique visitors to the college website tin can exist approximated by the formula $N\left(t\right)=410{\left(1.32\right)}^t$ where t represents the number of years later 1997 when the website was created. Approximate the number of unique visitors to the college website in the twelvemonth 2020.

75. If left unchecked, a new strain of influenza virus tin can spread from a single person to others very quickly. The number of people affected can exist modeled using the formula $P\left(t\right)=e^{0.22t}$ where t represents the number of days the virus is allowed to spread unchecked. Guess the number of people infected with the virus after 30 days and after 60 days.

76. If left unchecked, a population of 24 wild English rabbits tin grow according to the formula $P\left(t\right)=24{\mathrm{eastward}}^{0.19t}$ where the time t is measured in months. How many rabbits would be present $3\frac{\mathrm{one}}{\mathrm{ii}}$ years later?

77. The population of a certain urban center in 1975 was 65,000 people and was growing exponentially at an annual rate of 1.7%. At the time, the population growth was modeled by the formula $P\left(t\right)=\mathrm{65,000}{\mathrm{east}}^{0.017t}$ where t represented the number of years since 1975. In the year 2000, the census adamant that the actual population was 104,250 people. What population did the model predict for the year 2000 and what was the actual error?

78. Because of radioactive decay, the amount of a 10 milligram sample of Iodine-131 decreases co-ordinate to the formula $A\left(t\right)=10e^{-0.087t}$ where t represents time measured in days. How much of the sample remains later 10 days?

79. The number of cells in a bacteria sample is approximated past the logistic growth model $\mathrm{North}\left(t\right)=\frac{1.2\times {\mathrm{ten}}^5}{\mathrm{ane}+9{\mathrm{due\; east}}^{-0.32t}}$ where t represents time in hours. Determine the initial number of cells and then decide the number of cells half-dozen hours after.

80. The market place share of a product, as a percentage, is approximated past the formula $P\left(t\right)=\frac{100}{\mathrm{two}+{\mathrm{east}}^{-0.44t}}$ where t represents the number of months subsequently an aggressive advertising campaign is launched. By how much can nosotros expect the market share to increase later on the start 3 months of ad?

Function D: Discussion Board

81. Why is b = i excluded as a base in the definition of exponential functions? Explain.

82. Explain why an exponential part of the form $y=b^{\mathrm{10}}$ can never exist negative.

83. Research and talk over the derivation of the compound interest formula.

84. Research and discuss the logistic growth model. Provide a link to more data on this topic.

85. Research and discuss the life and contributions of Leonhard Euler.

Answers

1. $f\left(-2\right)=\frac{1}{\mathrm{ix}}$ , $f\left(0\right)=1$ , $f\left(2\right)=9$

3. $\mathrm{grand}\left(-1\right)=\mathrm{iii}$ , $\mathrm{thou}\left(0\right)=\mathrm{one}$ , $\mathrm{thou}\left(3\right)=\frac{1}{27}$

5. $h\left(-1\right)=9$ , $h\left(0\right)=1$ , $h\left(\frac{\mathrm{ane}}{2}\right)=\frac{1}{3}$

7. $f\left(-1\right)=\frac{1}{\mathrm{ii}}$ , $f\left(0\right)=0$ , $f\left(\mathrm{iii}\right)=-7$

9. $g\left(-\mathrm{two}\right)=120$ , $\mathrm{yard}\left(-\mathrm{i}\right)=\mathrm{xxx}$ , $\mathrm{chiliad}\left(0\right)=21$

11. ten.66

13. 7.10

15. 1385.46

17. −1.97

19. 3.99

21. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(0,\infty \right)$

23. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(\mathrm{two},\infty \right)$

25. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(0,\infty \right)$

27. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(-\mathrm{iv},\infty \right)$

29. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(-2,\infty \right)$

31. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(0,\infty \right)$

33. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(-2,\infty \right)$

35. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(-\mathrm{iii},\infty \right)$

37. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(-\infty ,6\right)$

39. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(-\infty ,5\right)$

41. $f\left(-1\right)\approx 2.37$ , $f\left(0\right)=3$ , $f\left(\frac{3}{2}\right)\approx 6.48$

43. $f\left(-\mathrm{one}\right)\approx \mathrm{three.90}$ , $f\left(0\right)=2$ , $f\left(\frac{\mathrm{iii}}{2}\right)\approx -\mathrm{eight.45}$

45. $f\left(-1\right)\approx 3.72$ , $f\left(0\right)=\mathrm{ii}$ , $f\left(\frac{3}{2}\right)\approx 1.22$

47. $f\left(-\mathrm{i}\right)\approx 9.39$ , $f\left(0\right)=\mathrm{iii}$ , $f\left(\frac{3}{2}\right)\approx 2.05$

49. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(-\mathrm{three},\infty \right)$

51. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(0,\infty \right)$

53. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(1,\infty \right)$

55. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(-\infty ,0\right)$

57. 

    Domain: $\left(-\infty ,\infty \right)$ ; Range: $\left(-\infty ,0\right)$

59. $850.52

61. $half-dozen,407.89

63. $sixteen,066.xiii

65. $1,588.00

67. $iv,685.44

69. $5,778.56

71. Initial population: 12,500; Population half dozen years later on: 14,077

73. New: $28,000; In 4 years: $13,940.40

75. Later 30 days: 735 people; Subsequently lx days: 540,365 people

77. Model: 99,423 people; fault: four,827 people

79. Initially there are 12,000 cells and 6 hours after there are 51,736 cells.

81. Answer may vary

83. Respond may vary

85. Reply may vary

Graph Of E X 2,

Source: https://saylordotorg.github.io/text_intermediate-algebra/s10-02-exponential-functions-and-thei.html

Posted by: jonesdescuseence.blogspot.com

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