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Chapter 5: Problem 1

Let \(f(x)=x^{2} .\) Find a value \(c \in(-1,2)\) so that \(f^{\prime}(c)\) equals the slope between the endpoints of \(f(x)\) on [-1,2]

Short Answer

Expert verified
The value of \( c \) is \( \frac{1}{2} \).

Step by step solution

01

Find the slope between the endpoints

First, identify the endpoints of the function on the interval [-1, 2]. Calculate the slope using the formula for the slope of a line, which is \( m = \frac{f(b) - f(a)}{b - a} \), where \( f(x) = x^2 \), \( a = -1 \), and \( b = 2 \). This gives:\[f(-1) = (-1)^2 = 1, \quad f(2) = 2^2 = 4\]Thus, the slope \( m \) is:\[m = \frac{4 - 1}{2 + 1} = \frac{3}{3} = 1\]
02

Compute the derivative of the function

Find the derivative of the function \( f(x) = x^2 \). Use the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \). For \( x^2 \), the derivative is:\[f'(x) = 2x\]
03

Set derivative equal to slope and solve for c

Now, equate the derivative \( f'(x) = 2x \) to the slope found in Step 1, which is 1:\[2c = 1\]Solve for \( c \):\[c = \frac{1}{2}\]
04

Verify that c is within the given interval

Ensure that the value of \( c \) we found is within the interval (-1, 2). Since \( c = \frac{1}{2} \), verify:\[-1 < \frac{1}{2} < 2\]Thus, \( c \) is indeed within the given interval.

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Key Concepts

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

Slope of a Line
Understanding the slope of a line is a fundamental concept in mathematics, particularly in calculus and algebra. It represents the rate at which one variable changes in relation to another. For a straight line, the slope is constant and can be quickly calculated using two points on the line.
  • The slope is determined by the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \), where \((x_1, y_1)\) and \((x_2, y_2)\) are two distinct points on the line.
  • This formula gives the vertical change (rise) per unit of horizontal change (run).
  • In our exercise, the points are derived from the function values at the endpoints of the interval, which were \((-1,1)\) and \((2,4)\) on the curve \(f(x) = x^2\).
Calculating the slope between these curve points helps us find places where the derivative of the function matches this rate of change. This is crucial for applying the Mean Value Theorem in continuous functions.
Derivative
In calculus, the derivative is a measure of how a function changes as its input changes. Think of it as a tool that tells you the slope of the tangent line to the function at any point, providing a snapshot of the function's behavior.
  • For a function \( f(x) \), the derivative \( f'(x) \) quantifies the instantaneous rate of change of \( f \) with respect to \( x \).
  • The process of finding a derivative is called differentiation.
  • Derivatives can vary across the domain of a function. For instance, \( f(x) = x^2 \) has a derivative \( f'(x) = 2x \), meaning the slope of the function's tangent line changes linearly with \( x \).
The derivative is essential for understanding how a function behaves and is instrumental in solving problems involving rates of change, optimization, and motion.
Power Rule
The power rule is a quick and efficient technique for finding the derivative of a function that involves a variable raised to a power. It is one of the building blocks of calculus, simplifying many differentiation tasks.
  • The power rule states that for any real number \( n \), the derivative of \( x^n \) is \( nx^{n-1} \).
  • This rule is particularly useful as it eliminates the need for using the limit-based definition of a derivative for each term.
  • In the given exercise, the power rule was applied to derive \( f'(x) = 2x \) from \( f(x) = x^2 \), demonstrating how easy and efficient it is to analyze the rate of change for polynomials.
Ultimately, the power rule not only provides us with speed but also clarity when working through calculations involving polynomials.

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