/*! 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 45 Find the average rate of change ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the average rate of change of each ficnetion on the given interval. $$f(x)=3 x^{2}-1 ; \text { interval: }[2,3]$$

Short Answer

Expert verified
The average rate of change of the function \(f(x) = 3x^2 - 1\) over the interval [2, 3] is 15.

Step by step solution

01

Compute \(f(3)\) and \(f(2)\)

The first thing to do is to substitute \(x = 3\) and \(x = 2\) into the given function \(f(x) = 3x^2 - 1\). \nThis gives us \(f(3) = 3*3^2 - 1 = 26\) and \(f(2) = 3*2^2 - 1 = 11\).
02

Apply the formula for the average rate of change

Then, we use the formula for the average rate of change, which is \(\frac{f(b) - f(a)}{b - a}\), substituting \(a = 2\), \(b = 3\), \(f(2) = 11\), and \(f(3) = 26\) as per our previous step. This gives us: \(\frac{f(3) - f(2)}{3 - 2} = \frac{26 - 11}{3 - 2} = 15\).
03

Interpret the result

The result, 15, is the average rate of change of the function \(f(x) = 3x^2 - 1\) over the interval [2, 3]. This value describes how much our function changes, on average, for each unit increase in \(x\) over our interval.

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Key Concepts

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

Understanding Quadratic Functions
Quadratic functions are a fundamental part of algebra and mathematics education. These functions are represented by the general form \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. In the exercise, the quadratic function given is \(f(x) = 3x^2 - 1\).

Quadratic functions create a parabolic curve on a graph. This means that when you plot them, they have a distinct 'U' shape, either opening upwards or downwards depending on the coefficient \(a\). If \(a\) is positive, like in our example \(3x^2 - 1\), the parabola opens upward. Quadratic functions are important because they model many real-world phenomena, from the trajectory of a thrown ball to the business world where profit maximization is considered.
  • The vertex of this parabola represents the maximum or minimum point.
  • The axis of symmetry is the vertical line that passes through the vertex.
  • Quadratic functions are applicable in areas like physics, engineering, and economics.
Breaking Down Intervals
An interval in mathematics denotes a specific segment of the number line. In this context, the interval \([2, 3]\) means we are examining how the quadratic function changes between \(x = 2\) and \(x = 3\).

Intervals are crucial when we're trying to understand how functions behave over specific ranges of input values. By focusing on an interval, like \([2, 3]\), we can determine the rate at which a function's output changes. This is akin to checking the speed of a car over a specific distance rather than its entire journey.
  • Closed intervals, denoted by brackets \([ ]\), include their endpoints.
  • Open intervals, denoted by parentheses \(( )\), do not include endpoints.
  • Intervals allow us to calculate average rates and other statistical measures for that segment.
Function Evaluation Made Easy
Function evaluation is a straightforward process that involves substituting specific values into a function to ascertain the output. It's a bit like testing a recipe by adding certain ingredients to see what flavor you get.

For the function \(f(x)=3x^2-1\), the task was to evaluate it at \(x = 2\) and \(x = 3\). This means replacing \(x\) with each of these values in the formula, yielding \(f(2) = 3(2)^2 - 1 = 11\) and \(f(3) = 3(3)^2 - 1 = 26\). By calculating these values, we find the outputs or the 'flavor' of the function at specific input 'ingredients'.
  • Always follow order of operations: parenthesis, exponents, multiplication, division, addition, and subtraction.
  • Check your work to avoid common human errors, like incorrect arithmetic.
  • Function evaluation prepares the ground for more complex topics like calculus.
Simplified Mathematics Education Approach
Mathematics education aims to build a coherent understanding of fundamental concepts that help solve real-world problems. Instructors and educational content aim to break down complex ideas into digestible formats so that learners can efficiently process, apply, and retain knowledge.

When learning concepts like quadratic functions and interval calculations, visualization tools and step-by-step guidance are beneficial. These educational strategies enhance student engagement and understanding by making abstract concepts more concrete.
  • Develop practical skills through exercises and real-life applications.
  • Use resources like step-by-step solutions, video tutorials, and quizzes to reinforce learning.
  • Foster critical thinking by encouraging problem-solving from multiple angles.
Using diverse teaching methods and focusing on different learning styles ensures all students have the opportunity to grasp essential mathematical concepts.

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Most popular questions from this chapter

Name at least two features of a quadratic function that differ from those of a linear function.

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=x-2 ; g(x)=2 x^{2}-x+3$$

In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=(3 x-7)^{10}+5(3 x-7)^{2}$$

When designing buildings, engineers must pay careful attention to how different factors affect the load a structure can bear. The following table gives the load in terms of the weight of concrete that can be borne when threaded rod anchors of various diameters are used to form joints. $$\begin{array}{cc} \text { Diameter (in.) } & \text { Load (1b) } \\ \hline 0.3750 & 2105 \\ 0.5000 & 3750 \\ 0.6250 & 5875 \\ 0.7500 & 8460 \\ 0.8750 & 11,500 \end{array}$$ (a) Examine the table and explain why the relationship between the diameter and the load is not linear. (b) The function $$f(x)=14,926 x^{2}+148 x-51$$ gives the load (in pounds of concrete) that can be borne when rod anchors of diameter \(x\) (in inches) are employed. Use this function to determine the load for an anchor with a diameter of 0.8 inch. (c) since the rods are drilled into the concrete, the manufacturer's specifications sheet gives the load in terms of the diameter of the drill bit. This diameter is always 0.125 inch larger than the diameter of the anchor. Write the function in part (b) in terms of the diameter of the drill bit. The loads for the drill bits will be the same as the loads for the corresponding anchors. (Hint: Examine the table of values and see if you can present the table in terms of the diameter of the drill bit.)

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=\frac{x^{2}+1}{x^{2}-1} ; g(x)=|x|$$

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