Chapter 4: Problem 1
Find the value or values of \(c\) that satisfy the equation $$\frac{f(b)-f(a)}{b-a}=f^{\prime}(c)$$ in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises \(1-6 .\) $$f(x)=x^{2}+2 x-1, \quad[0,1]$$
Short Answer
Expert verified
The value of \(c\) is \(\frac{1}{2}\).
Step by step solution
01
Understand the Mean Value Theorem (MVT)
The Mean Value Theorem (MVT) states that for a function \(f(x)\) continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists at least one \(c\) in \((a, b)\) such that \(\frac{f(b)-f(a)}{b-a} = f^{\prime}(c)\). In this exercise, we apply it to \(f(x) = x^2 + 2x - 1\) over the interval \([0, 1]\).
02
Compute \(f(a)\) and \(f(b)\)
For \(a = 0\) and \(b = 1\), calculate \(f(0)\) and \(f(1)\): - \(f(0) = (0)^2 + 2(0) - 1 = -1\) - \(f(1) = (1)^2 + 2(1) - 1 = 2\).
03
Calculate the Difference Quotient
The difference quotient is \(\frac{f(b) - f(a)}{b-a}\):\[\frac{2 - (-1)}{1 - 0} = \frac{3}{1} = 3.\]
04
Differentiate the Function
Find the derivative \(f^{\prime}(x)\) of the function \(f(x) = x^2 + 2x - 1\):\[f^{\prime}(x) = 2x + 2.\]
05
Solve for \(c\)
Set the derived expression equal to the difference quotient and solve for \(c\):\[2c + 2 = 3.\] Solve for \(c\): \(2c + 2 = 3 \Rightarrow 2c = 1 \Rightarrow c = \frac{1}{2}.\)
06
Verify \(c\) is in the Interval
Ensure that the value of \(c = \frac{1}{2}\) is within the open interval \((0, 1)\). Since \(\frac{1}{2}\) is between 0 and 1, \(c\) satisfies the conditions of the Mean Value Theorem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiable Function
A differentiable function is one that has a derivative at every point within its domain. This means the function is smooth and without any sharp corners or breaks. For a function to be considered differentiable over an interval, it must be possible to find the rate of change at any point within that interval.
Key characteristics of differentiable functions include:
Key characteristics of differentiable functions include:
- They are continuous, meaning they have no sudden jumps or gaps.
- They possess a tangent line at every point in their domain.
- The existence of a derivative at each point indicates the function's smoothness.
Closed Interval
A closed interval is indicated by symbols like \([a, b]\), suggesting that it includes its end points. Namely, any x-value for this interval can be equal to or fall between a and b.
When applying the Mean Value Theorem, the function must be continuous over the closed interval \([a, b]\). This ensures the behavior of the function is predictable, especially at the endpoints. Closed intervals are key concepts in calculus because:
When applying the Mean Value Theorem, the function must be continuous over the closed interval \([a, b]\). This ensures the behavior of the function is predictable, especially at the endpoints. Closed intervals are key concepts in calculus because:
- They provide limits or boundaries within which the function needs to operate smoothly.
- They define a range within which continuity is essential for applications of the MVT.
- By including the endpoints, the evaluation of the function at these points becomes integral to determining changes over the interval.
Derivative
The derivative of a function represents the concept of instantaneous rate of change or the slope of the tangent line at any point of a curve. It is a fundamental tool in calculus used to understand how a function is changing at any given point.
Here are some insights into derivatives:
Here are some insights into derivatives:
- They allow us to determine how a function behaves locally, determining if it is increasing or decreasing at given points.
- The process of deriving a function, known as differentiation, helps find this rate of change.
- Derivatives provide vital information for solving optimization problems and analyzing how parameters influence functions.
- It helps find the specific point "c" where the instantaneous rate of change equals the average rate of change over the interval.
- Performing differentiation allows us to apply the conditions set forth by the theorem accurately.