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Compute the average rate of change of the function on the given interval. $$g(x)=2 x^{2}-4 x \text { on }[-1,3]$$

Short Answer

Expert verified
The average rate of change is 0.

Step by step solution

01

Identify the Function and Interval

The function given is \( g(x) = 2x^2 - 4x \). We need to find the average rate of change of this function over the interval \([-1, 3]\).
02

Calculate Function Values at Interval Endpoints

Evaluate \( g(x) \) at the endpoints of the interval. First, compute \( g(-1) \). \[g(-1) = 2(-1)^2 - 4(-1) = 2(1) + 4 = 6.\]Next, compute \( g(3) \).\[g(3) = 2(3)^2 - 4(3) = 2(9) - 12 = 18 - 12 = 6.\]
03

Apply Average Rate of Change Formula

The formula for the average rate of change of a function \( g(x) \) on the interval \([a, b]\) is:\[\text{Average Rate of Change} = \frac{g(b) - g(a)}{b - a}.\]Substitute \( a = -1 \), \( b = 3 \), \( g(-1) = 6 \), and \( g(3) = 6 \) into the formula:\[\frac{6 - 6}{3 - (-1)} = \frac{0}{4} = 0.\]
04

Conclusion

The average rate of change of \( g(x) = 2x^2 - 4x \) on the interval \([-1, 3]\) is 0. The function's value does not change on this interval.

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Key Concepts

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

Quadratic Functions
Quadratic functions are a type of polynomial function with the form \( ax^2 + bx + c \). The graph of a quadratic function is a parabola, which can open upwards or downwards depending on the sign of \( a \). These functions are fundamental in mathematics because they model various physical phenomena and have applications in diverse fields, including physics and finance.
The function \( g(x) = 2x^2 - 4x \) is a standard quadratic function. Here, the coefficient \( a \) is 2, which means the parabola opens upwards and is relatively narrow compared to other parabolas where \( a \) might be smaller. The function's vertices and intercepts are calculated by evaluating or solving for \( x \) when \( g(x) = 0 \).
Since quadratic functions are smooth and continuous, they are perfect candidates to analyze using calculus principles such as finding the average rate of change. Understanding these functions provides a foundation to solve more complex problems.
Intervals in Calculus
In calculus, intervals help us define the specific section of a function that we are interested in examining. An interval can be closed, open, or half-open based on whether or not their endpoints are included. In our example, the interval is \([-1, 3]\), a closed interval that includes both endpoints -1 and 3.
Intervals are crucial when evaluating functions, like determining the average rate of change. On the interval \([-1, 3]\), we assess the change by calculating the difference in the function's values at these endpoints. Using a defined interval helps us focus our analysis on relevant segments of the graph, capturing the function's behavior in that space.
Being able to work with intervals not only improves understanding of single-variable functions but also lays the groundwork for studying more advanced topics like integrals and differential equations in calculus.
Function Evaluation
Function evaluation is the process of finding the function's output for specific inputs. It involves substituting a value for \( x \) in the equation and performing the operations to get the result. For the function \( g(x) = 2x^2 - 4x \), evaluating at a particular point means replacing \( x \) with that point's value and solving.
In our exercise, we evaluate \( g \) at the endpoints of the interval \([-1, 3]\). First, we substitute \( x = -1 \) into the function to compute \( g(-1) \), which results in 6. Similarly, evaluating at \( x = 3 \) yields \( g(3) = 6 \). These evaluations give us the function values needed to apply formulas, like those for calculating average rate of change.
Understanding how to evaluate functions is vital as it serves as the first step in many calculus problems and aids in graphing functions to predict their behavior accurately.

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Most popular questions from this chapter

Indicate how iteration is used in finding roots of numbers and roots of equations. (The functions that are given in each exercise were determined using Newton's method, a process studied in calculus.) Let \(f(x)=\frac{2 x^{3}+7}{3 x^{2}}\). (a) Compute the first ten iterates of \(x_{0}=1\) under the function \(f .\) What do you observe? (b) Evaluate the expression \(\sqrt[3]{7}\) and compare the answer to your results in part (a). What do you observe? (c) It can be shown that for any positive number \(x_{0}\), the iterates of \(x_{0}\) under the function \(f(x)=\frac{2 x^{3}+7}{3 x^{2}}\) always approach the number \(\sqrt[3]{7} .\) Looking at your results in parts (a) and (b), which is the first iterate that agrees with \(\sqrt[3]{7}\) through the first three decimal places? Through the first eight decimal places?

Let \(a(x)=x^{2}, b(x)=|x|,\) and \(c(x)=3 x-1 .\) Express each of the following functions as a composition of two of the given functions. (a) \(f(x)=(3 x-1)^{2}\) (c) \(h(x)=3 x^{2}-1\) (b) \(g(x)=|3 x-1|\)

Express each function as a composition of two functions. (a) \(F(x)=\sqrt[3]{3 x+4}\) (c) \(H(x)=(a x+b)^{5}\) (b) \(G(x)=|2 x-3|\) (d) \(T(x)=1 / \sqrt{x}\)

Consider the following two rules, \(F\) and \(G,\) where \(F\) is the rule that assigns to each person his or her birth-mother and \(G\) is the rule that assigns to each person his or her aunt. Explain why \(F\) is a function but \(G\) is not.

In this exercise you'll investigate the inverse of a composite function. In parts (b) and (c), which involve graphing, be sure to use the same size unit and scale on both axes so that symmetry about the line \(y=x\) can be checked visually. (a) Let \(f(x)=2 x+1\) and \(g(x)=\frac{1}{4} x-3 .\) Compute each of the following: (i) \(f(g(x))\) \(g^{-1}(x)\) (ii) \(g(f(x))\) (v) \(f^{-1}\left(g^{-1}(x)\right)\) (iii) \(f^{-1}(x)\) (vi) \(g^{-1}\left(f^{-1}(x)\right)\) (b) On the same set of axes, graph the two answers that you obtained in (i) and (v) of part (a). Note that the graphs are not symmetric about \(y=x .\) The conclusion here is that the inverse function for \(f(g(x))\) is not \(f^{-1}\left(g^{-1}(x)\right)\) (c) On the same set of axes, graph the two answers that you obtained in (i) and (vi) of part (a); also put the line\(y=x\) into the picture. Note that the two graphs are symmetric about the line \(y=x .\) The conclusion here is that the inverse function for \(f(g(x))\) is \(g^{-1}\left(f^{-1}(x)\right)\) In fact, it can be shown that this result is true in general. For reference, then, we summarize this fact about the inverse of a composite function in the box that follows.

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