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Find the average rate of change of the function over the given interval or intervals. \(g(x)=x^{2}-2 x\) a. [1,3] b. [-2,4]

Short Answer

Expert verified
a. 2 b. 0

Step by step solution

01

Understand the Average Rate of Change

The average rate of change of a function over an interval \([a, b]\) is given by the formula \((g(b) - g(a)) / (b - a)\). We will calculate this for each interval provided.
02

Evaluate the Function at the Endpoints for Interval [1,3]

Substitute \(x = 1\) into the function: \(g(1) = 1^2 - 2 \times 1 = 1 - 2 = -1\).Substitute \(x = 3\) into the function: \(g(3) = 3^2 - 2 \times 3 = 9 - 6 = 3\).
03

Calculate the Average Rate of Change for Interval [1,3]

Use the formula \((g(3) - g(1)) / (3 - 1)\). \Calculate: \( (3 - (-1)) / (3 - 1) = 4 / 2 = 2\).
04

Evaluate the Function at the Endpoints for Interval [-2,4]

Substitute \(x = -2\) into the function: \(g(-2) = (-2)^2 - 2 \times (-2) = 4 + 4 = 8\).Substitute \(x = 4\) into the function: \(g(4) = 4^2 - 2 \times 4 = 16 - 8 = 8\).
05

Calculate the Average Rate of Change for Interval [-2,4]

Use the formula \((g(4) - g(-2)) / (4 - (-2))\). Calculate: \( (8 - 8)/(4 + 2) = 0/6 = 0\).

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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 fundamental part of algebra and calculus. They are polynomial functions of degree 2, generally expressed in the form:
  • \(f(x) = ax^2 + bx + c\)
  • Here, \(a\), \(b\), and \(c\) are constants, and \(a eq 0\).
The graph of a quadratic function is a parabola. Depending on the sign of \(a\), the parabola opens upwards (\(a > 0\)) or downwards (\(a < 0\)).

In the example of \(g(x) = x^2 - 2x\), the quadratic is already in standard form where \(a = 1\), \(b = -2\), and \(c = 0\). This function will graph as a parabola peeking up (since \(a > 0\).

Knowing the form of a quadratic function helps when calculating things like the average rate of change, as seen in this exercise. Calculating the rate helps us understand how fast the values of the function are changing over a specific interval.
Interval Notation
Understanding interval notation is essential when dealing with functions, particularly in calculus. It is a way of specifying a segment of the function.
  • A closed interval, written as \([a, b]\), means the function is evaluated at both endpoints, \(a\) and \(b\).
  • An open interval, denoted by \((a, b)\), does not include the endpoints.
  • Mixed intervals like \((a, b]\) or \([a, b)\) only include one endpoint.
In this exercise, intervals \([1,3]\) and \([-2,4]\) are closed, meaning they include the endpoints. This is important as the function needs to be evaluated at \(x = 1, 3, -2,\) and \(4\) to find the average rate of change.

Grasping interval notation helps in setting boundaries for calculations, either differentiating or integrating, ensuring the accuracy of mathematical operations performed on a function.
Calculus Introduction
Calculus delves into how things change; that is essentially what the average rate of change measures. It's a stepping stone to more advanced concepts:
  • The rate of change of a function is a foundational concept in calculus.
  • For functions, you often need to calculate how the output — the \(y\)-value — changes as the \(x\)-value shifts; this is also denoted by \((g(b) - g(a)) / (b - a)\).
Exploring average rates of change prepares you for understanding derivatives, which are central to calculus.

When you compute the average rate of change, as done in the provided step-by-step solution, you're finding out how the function behaves overall between two points. This is particularly useful in predicting trends or simplifying complex changes into understandable pieces. Embracing these initial calculus principles sets a strong foundation for tackling more complex calculus problems later on.

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Most popular questions from this chapter

Gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon \geq 0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. $$\begin{aligned}&f(x)=m x+b, \quad m>0, \quad L=m+b, \quad c=1\\\&\epsilon=0.05\end{aligned}$$

Are the functions continuous at the point being approached? \(\lim _{x \rightarrow 1} \cos ^{-1}(\ln \sqrt{x})\)

Show that the equation \(x^{3}-15 x+1=0\) has three solutions in the interval [-4,4]

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form $$\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ occur frequently in calculus. Evaluate this limit for the given value of \(x\) and function \(f.\) $$f(x)=x^{2}, \quad x=-2$$

Use a CAS to perform the following steps: a. Plot the function \(y=f(x)\) near the point \(c\) being approached. b. Guess the value of the limit \(L\) and then evaluate the limit symbolically to see if you guessed correctly. c. Using the value \(\epsilon=0.2,\) graph the banding lines \(y_{1}=L-\epsilon\) and \(y_{2}=L+\epsilon\) together with the function \(f\) near \(c\) d. From your graph in part (c), estimate a \(\delta>0\) such that for all \(x\) \(0<|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon\) Test your estimate by plotting \(f, y_{1},\) and \(y_{2}\) over the interval \(0<|x-c|<\delta .\) For your viewing window use \(c-2 \delta \leq\) \(x \leq c+2 \delta\) and \(L-2 \epsilon \leq y \leq L+2 \epsilon .\) If any function values lie outside the interval \([L-\epsilon, L+\epsilon],\) your choice of \(\delta\) was too large. Try again with a smaller estimate. e. Repeat parts (c) and (d) successively for \(\epsilon=0.1,0.05,\) and 0.001 $$f(x)=\frac{x(1-\cos x)}{x-\sin x}, \quad c=0$$

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