Problem 10
Question
Find the average rate of change of \(f(x)=x^{2}\) over each of the following intervals. (a) \([0,3]\) (b) \([1,4]\) (c) \([2,5]\) (d) \([a, a+3]\) (e) \([a, a+h]\)
Step-by-Step Solution
Verified Answer
The average rates of change are: (a) 3, (b) 5, (c) 7, (d) \(2a+3\), and (e) \(2a+h\).
1Step 1: Average rate of change in the interval [0,3]
For the interval [0,3], we find the rate of change using the formula so it is \[ \frac{f(3)-f(0)}{3-0} = \frac{3^{2}-0^{2}}{3-0} = \frac{9-0}{3} = 3. \]
2Step 2: Average rate of change in the interval [1,4]
For the interval [1,4], we find the rate of change using the formula so it is \[ \frac{f(4)-f(1)}{4-1} = \frac{4^{2}-1^{2}}{4-1} = \frac{16-1}{3} = 5. \]
3Step 3: Average rate of change in the interval [2,5]
For the interval [2,5], we find the rate of change using the formula so it is \[ \frac{f(5)-f(2)}{5-2} = \frac{5^{2}-2^{2}}{5-2} = \frac{25-4}{3} = 7. \]
4Step 4: Average rate of change in the interval [a, a+3]
For the interval [a, a+3], we find the rate of change using the formula so it is \[ \frac{f(a+3)-f(a)}{(a+3)-a} = \frac{(a+3)^{2}-a^{2}}{3} = \frac{9+6a}{3} = 2a+3. \]
5Step 5: Average rate of change in the interval [a, a+h]
For the interval [a, a+h], we find the rate of change using the formula so it is \[ \frac{f(a+h)-f(a)}{(a+h)-a} = \frac{(a+h)^{2}-a^{2}}{h} = \frac{h^{2}+2ah}{h} = 2a+h. \]
Key Concepts
Rate of Change FormulaFunction Intervals in CalculusPolynomial Functions
Rate of Change Formula
Understanding the rate of change is fundamental in calculus as it measures how one quantity changes in relation to another. In practical terms, it is often used to determine how a function behaves between any two points. The average rate of change formula is represented as \( \frac{\Delta y}{\Delta x} \), where \( \Delta y \) is the change in the function's value (\( f(x) \)) and \( \Delta x \) is the change in the independent variable (\( x \) values).
For example, if a function \( f(x) \) is defined as \( x^2 \) and we want to find the average rate of change between \( x = 1 \) and \( x = 4 \) we substitute these values into the positions of \( \Delta y \) and \( \Delta x \). The formula \( \frac{f(4)-f(1)}{4-1} \) would result in the rate of change.
The significance of understanding this formula lies in its ability to depict function behavior on specific intervals, which is an essential aspect of analyzing functions in calculus.
For example, if a function \( f(x) \) is defined as \( x^2 \) and we want to find the average rate of change between \( x = 1 \) and \( x = 4 \) we substitute these values into the positions of \( \Delta y \) and \( \Delta x \). The formula \( \frac{f(4)-f(1)}{4-1} \) would result in the rate of change.
The significance of understanding this formula lies in its ability to depict function behavior on specific intervals, which is an essential aspect of analyzing functions in calculus.
Function Intervals in Calculus
In calculus, the concept of function intervals pertains to the portions of the domain over which we are interested in examining a function's behavior. It allows us to determine the function's characteristics, such as increasing or decreasing behavior and concavity, over specific sections of its domain.
When evaluating the average rate of change, the selection of these intervals is crucial. For instance, given a function \( f(x) = x^2 \) and the task to find its average rate of change over the interval \( [a, a+h] \), we carefully consider the endpoints of this interval, as the value of \( h \) will determine the stretch of the function we're examining.
Determining function behavior over these intervals using the rate of change formula allows us to derive information about the function's growth or decline, which in turn can inform us about real-world phenomena if the function models a particular situation.
When evaluating the average rate of change, the selection of these intervals is crucial. For instance, given a function \( f(x) = x^2 \) and the task to find its average rate of change over the interval \( [a, a+h] \), we carefully consider the endpoints of this interval, as the value of \( h \) will determine the stretch of the function we're examining.
Determining function behavior over these intervals using the rate of change formula allows us to derive information about the function's growth or decline, which in turn can inform us about real-world phenomena if the function models a particular situation.
Polynomial Functions
Polynomial functions are expressions consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. An example of a polynomial function is \( f(x) = x^2 \) which is a simple quadratic equation, a specific type of polynomial with a degree of 2.
Understanding polynomial functions is important when discussing rates of change because their smooth and continuous nature allows for straightforward computation of change across intervals. In our previous examples, finding the average rate of change involves executing basic algebraic manipulations on a polynomial where the intervals can be simple constants like \( [0,3] \) or variables such as \( [a, a+h] \).
This ties back to our exercise, as examining a polynomial function like \( f(x) = x^2 \) over various intervals helps students grasp the concept of how functions change and also lays the foundation for more complex calculus topics, such as differentiation and integration, which heavily rely on the properties of polynomials.
Understanding polynomial functions is important when discussing rates of change because their smooth and continuous nature allows for straightforward computation of change across intervals. In our previous examples, finding the average rate of change involves executing basic algebraic manipulations on a polynomial where the intervals can be simple constants like \( [0,3] \) or variables such as \( [a, a+h] \).
This ties back to our exercise, as examining a polynomial function like \( f(x) = x^2 \) over various intervals helps students grasp the concept of how functions change and also lays the foundation for more complex calculus topics, such as differentiation and integration, which heavily rely on the properties of polynomials.
Other exercises in this chapter
Problem 9
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