/*! 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

Use the model \(A=P\left(1+\frac{r}{n}\right)^{n t} .\) The variable A represents the future value of P dollars invested at an interest rate \(r\) compounded \(n\) times per year for \(t\) years. If 4000 is put aside in a money market account with interest reinvested monthly at 2.2%, find the time required for the account to earn 1000. Round to the nearest month.

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A farmer depreciates a \(\$ 120,000\) tractor. He estimates that the resale value \(V(t)\) (in \(\$ 1000\) ) of the tractor \(t\) years after purchase is \(80 \%\) of its value from the previous year. Therefore, the resale value can be approximated by \(V(t)=120(0.8)^{t}\). a. Find the resale value 5 yr after purchase. Round to the nearest \(\$ 1000\). b. The farmer estimates that the cost to run the tractor is \(\$ 18 / \mathrm{hr}\) in labor, \(\$ 36 / \mathrm{hr}\) in fuel, and \(\$ 22 / \mathrm{hr}\) in overhead costs (for maintenance and repair). Estimate the farmer's cost to run the tractor for the first year if he runs the tractor for a total of \(800 \mathrm{hr}\). Include hourly costs and depreciation.

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