Problem 27
Question
\(17-28\) A function is given. Determine the average rate of change of the function between the given values of the variable. $$ f(t)=\frac{2}{t} ; \quad t=a, t=a+h $$
Step-by-Step Solution
Verified Answer
The average rate of change is \(-\frac{2}{a(a+h)}\).
1Step 1: Identify the formula for the average rate of change
The average rate of change of a function over an interval is given by the formula \( \frac{f(b) - f(a)}{b - a} \), where \(a\) and \(b\) are the endpoints of the interval.
2Step 2: Substitute specific endpoints into the formula
Here, the endpoints are \(t = a\) and \(t = a + h\). Therefore, the average rate of change is \( \frac{f(a+h) - f(a)}{(a+h) - a} = \frac{f(a+h) - f(a)}{h} \).
3Step 3: Calculate \(f(a)\) and \(f(a+h)\)
Using the function \(f(t) = \frac{2}{t}\), calculate:- \(f(a) = \frac{2}{a}\)- \(f(a+h) = \frac{2}{a+h}\).
4Step 4: Substitute \(f(a)\) and \(f(a+h)\) back into the rate of change formula
Plug \(f(a)\) and \(f(a+h)\) into the average rate of change formula: \[\text{Average rate of change} = \frac{\frac{2}{a+h} - \frac{2}{a}}{h}\].
5Step 5: Simplify the expression
Simplify the numerator: \[\frac{2}{a+h} - \frac{2}{a} = \frac{2a - 2(a+h)}{a(a+h)} = \frac{2a - 2a - 2h}{a(a+h)} = \frac{-2h}{a(a+h)}\]. Thus, the average rate of change becomes: \[\frac{-2h}{a(a+h)h} = \frac{-2}{a(a+h)}\].
6Step 6: Final verification
Make sure that all calculations are correct by re-evaluating each simplification step and ensuring the formula used was correct throughout.
Key Concepts
FunctionsInterval EndpointsSimplifying ExpressionsRational Functions
Functions
Understanding functions is essential in mathematics. Functions are relationships between sets, where each element in the first set (domain) maps to exactly one element in the second set (range).
The notation used is typically \( f(t) \), where \(t\) is the input from the domain, and \( f(t) \) is the output in the range.
- Imagine a function as a machine: you input a number and it gives you a result according to a defined rule.- Example: In the function \( f(t) = \frac{2}{t} \), if you input \( t = 4 \), the output will be \( f(4) = \frac{2}{4} = \frac{1}{2} \).
Understanding how the function behaves as the input changes will help you grasp more complex topics like average rate of change.
The notation used is typically \( f(t) \), where \(t\) is the input from the domain, and \( f(t) \) is the output in the range.
- Imagine a function as a machine: you input a number and it gives you a result according to a defined rule.- Example: In the function \( f(t) = \frac{2}{t} \), if you input \( t = 4 \), the output will be \( f(4) = \frac{2}{4} = \frac{1}{2} \).
Understanding how the function behaves as the input changes will help you grasp more complex topics like average rate of change.
Interval Endpoints
When dealing with functions, it's important to know about intervals. Intervals specify a part of the function's domain.
Interval endpoints mark the beginning and end of these intervals.
- For example, if we have an interval from \( t = a \) to \( t = a + h \), \( a \) and \( a + h \) are the interval endpoints.- In problems related to average rate of change, you'll often see questions framed with such intervals to focus on a specific range of values.
Understanding interval endpoints is crucial to setting up your calculations correctly, especially when applying formulas like the average rate of change.
Interval endpoints mark the beginning and end of these intervals.
- For example, if we have an interval from \( t = a \) to \( t = a + h \), \( a \) and \( a + h \) are the interval endpoints.- In problems related to average rate of change, you'll often see questions framed with such intervals to focus on a specific range of values.
Understanding interval endpoints is crucial to setting up your calculations correctly, especially when applying formulas like the average rate of change.
Simplifying Expressions
Simplifying expressions is a foundational skill in mathematics that makes equations easier to handle.
It involves reducing an expression to its simplest form.
- For instance, to simplify \( \frac{2}{a+h} - \frac{2}{a} \), you'll manipulate it into a common format that can be easily understood or further calculated.- The expression is transformed into \( \frac{2a - 2(a+h)}{a(a+h)} \), which simplifies to \( \frac{-2h}{a(a+h)} \).
Simplification helps in reducing errors and making the resulting expression easier to interpret and work with, particularly in solving problems like finding the average rate of change.
It involves reducing an expression to its simplest form.
- For instance, to simplify \( \frac{2}{a+h} - \frac{2}{a} \), you'll manipulate it into a common format that can be easily understood or further calculated.- The expression is transformed into \( \frac{2a - 2(a+h)}{a(a+h)} \), which simplifies to \( \frac{-2h}{a(a+h)} \).
Simplification helps in reducing errors and making the resulting expression easier to interpret and work with, particularly in solving problems like finding the average rate of change.
Rational Functions
Rational functions are fractions that contain polynomials in both the numerator and the denominator.
The function \( f(t) = \frac{2}{t} \) is an example of a rational function.
- These types of functions are everywhere in calculus and algebra because they describe various real-world phenomena, like speed and growth rates.- Analyzing rational functions can involve finding limits, asymptotes, and average rates of change.
When working with rational functions, you often need to be careful about where the function is undefined, typically where the denominator equals zero. Understanding these principles makes analyzing and interpreting functions much easier.
The function \( f(t) = \frac{2}{t} \) is an example of a rational function.
- These types of functions are everywhere in calculus and algebra because they describe various real-world phenomena, like speed and growth rates.- Analyzing rational functions can involve finding limits, asymptotes, and average rates of change.
When working with rational functions, you often need to be careful about where the function is undefined, typically where the denominator equals zero. Understanding these principles makes analyzing and interpreting functions much easier.
Other exercises in this chapter
Problem 27
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