/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 An exponential function \(y=b^{x... [FREE SOLUTION] | 91Ó°ÊÓ

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An exponential function \(y=b^{x}\) with base \(b>1\) increases over its domain. An exponential function with base \(0

Short Answer

Expert verified
a. 0.4444, b. 4, c. 36, d. 324

Step by step solution

01

Identify the function

The given exponential function is \(f(x) = 3^x\).
02

Recall the formula for the average rate of change

The average rate of change on an interval \([a, b]\) for a function \(y = f(x)\) is calculated as \(\frac{f(b) - f(a)}{b - a}\).
03

Calculate for interval [-2,0]

For \([-2, 0]\), calculate \(f(-2)\) and \(f(0)\): \[ f(-2) = 3^{-2} = \frac{1}{9} \] \[ f(0) = 3^0 = 1 \] Substitute these values into the formula: \[ \text{Average rate of change} = \frac{1 - \frac{1}{9}}{0 - (-2)} = \frac{\frac{8}{9}}{2} = \frac{4}{9} \approx 0.4444 \]
04

Calculate for interval [0,2]

For \([0, 2]\), calculate \(f(0)\) and \(f(2)\): \[ f(0) = 3^0 = 1 \] \[ f(2) = 3^2 = 9 \] Substitute these values into the formula: \[ \text{Average rate of change} = \frac{9 - 1}{2 - 0} = \frac{8}{2} = 4 \]
05

Calculate for interval [2,4]

For \([2, 4]\), calculate \(f(2)\) and \(f(4)\): \[ f(2) = 3^2 = 9 \] \[ f(4) = 3^4 = 81 \] Substitute these values into the formula: \[ \text{Average rate of change} = \frac{81 - 9}{4 - 2} = \frac{72}{2} = 36 \]
06

Calculate for interval [4,6]

For \([4, 6]\), calculate \(f(4)\) and \(f(6)\): \[ f(4) = 3^4 = 81 \] \[ f(6) = 3^6 = 729 \] Substitute these values into the formula: \[ \text{Average rate of change} = \frac{729 - 81}{6 - 4} = \frac{648}{2} = 324 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

average rate of change
The concept of average rate of change helps understand how a function behaves over a specific interval. Think of it as a way to calculate the 'speed' at which the function's value changes. For instance, with an exponential function like \(f(x) = 3^x\), we can find its average rate of change between two points, say \(a\) and \(b\). It's calculated by dividing the difference in function values, \( f(b) - f(a) \), by the difference in \(x\) values, \( b - a \). This formula, \( \frac{f(b)-f(a)}{b-a} \), gives a single number that represents the overall change over that interval.
Let's apply this to the exponential function laid out in our problem. For the interval \([-2,0]\), we have: \[ f(-2) = 3^{-2} = \frac{1}{9} \] and \[ f(0) = 3^0 = 1. \] The average rate of change is then: \[ \frac{1 - \frac{1}{9}}{0 - (-2)} = \frac{\frac{8}{9}}{2} = \frac{4}{9} \approx 0.4444. \] It's clear that the function increases, albeit not too rapidly over this interval.
Now, the average rate of change helps illustrate how quickly or slowly a function ascends or descends over different intervals. It's a useful tool because it simplifies the computation to a single value, making comparisons straightforward.
interval calculations
When dealing with functions, intervals represent the span over which we examine the function's behavior. Calculating the average rate of change requires careful choice and evaluation over different intervals.
Consider the interval \([0,2]\) for our function \(f(x) = 3^x\). Evaluate it at the ends: \[ f(0) = 1 \] and \[ f(2) = 9. \] The average rate of change over this interval is: \[ \frac{9 - 1}{2 - 0} = \frac{8}{2} = 4. \] Notice, the rate is higher than the previously calculated interval \([-2,0]\), indicating that 3^x starts to grow faster as \(x\) increases.
Let's go further and examine a longer interval like \([2,4]\). We have: \[ f(2) = 9 \] and \[ f(4) = 81. \] Now, the average rate of change is: \[ \frac{81 - 9}{4 - 2} = \frac{72}{2} = 36. \] Clearly, the growth is much steeper here. The length of the interval plays a role in how significant the rate of change appears, especially in exponential functions.
mathematical functions
Understanding mathematical functions like exponentials is crucial in various fields. A function like \(f(x) = 3^x\) shows how a base number raised to a power \(x\) grows.
Exponential functions have unique properties. With a base greater than 1, they steeply increase as \(x\) becomes larger. Conversely, with a base between 0 and 1, they decrease.
This can be seen in our previous examples: For \([4,6]\), calculate: \[ f(4) = 81 \] and \[ f(6) = 729. \] The average rate of change: \[ \frac{729 - 81}{6 - 4} = \frac{648}{2} = 324. \] The large rate indicates rapid increase, a hallmark of exponential growth. Understanding this helps decipher real-world phenomena like population growth and radioactive decay, where exponential functions commonly apply.
In summary, grasping these concepts not only aids academically but also offers practical insights into various disciplines, making the mastery of mathematical functions indispensable.

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Most popular questions from this chapter

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(10^{5+8 x}+4200=84,000\)

a. The populations of two countries are given for January 1,2000 , and for January 1,2010 . Write a function of the form \(P(t)=P_{0} e^{k t}\) to model each population \(P(t)\) (in millions) \(t\) years after January 1, 2000.$$ \begin{array}{|l|c|c|c|} \hline & \begin{array}{c} \text { Population } \\ \text { in 2000 } \\ \text { (millions) } \end{array} & \begin{array}{c} \text { Population } \\ \text { in 2010 } \\ \text { (millions) } \end{array} & \boldsymbol{P}(t)=\boldsymbol{P}_{0} e^{k t} \\ \hline \text { Switzerland } & 7.3 & 7.8 & \\ \hline \text { Israel } & 6.7 & 7.7 & \\ \hline \end{array}$$ b. Use the models from part (a) to predict the population on January \(1,2020,\) for each country. Round to the nearest hundred thousand. c. Israel had fewer people than Switzerland in the year 2000 , yet from the result of part (b), Israel will have more people in the year \(2020 ?\) Why? d. Use the models from part (a) to predict the year during which each population will reach 10 million if this trend continues.

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(80=320 e^{-0.5 t}\)

Given \(y=\log x,\) the base is understood to be _____ . Given \(y=\ln x,\) the base is understood to be _____ .

In \(1989,\) the Loma Prieta earthquake damaged the city of San Francisco with an intensity of approximately \(10^{6.9} I_{0} .\) Film footage of the 1989 earthquake was captured on a number of video cameras including a broadcast of Game 3 of the World Series played at Candlestick Park. (See Example \(\mathbf{1 0}\) ) a. Determine the magnitude of the Loma Prieta earthquake. b. Smaller earthquakes occur daily in the San Francisco area and most are not detectable without a seismograph. Determine the magnitude of an earthquake with an intensity of \(10^{3.2} I_{0}\). c. How many times more intense was the Loma Prieta earthquake than an earthquake with a magnitude of 3.2 on the Richter scale? Round to the nearest whole unit.

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