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Chapter 6: Problem 1

The value of \(c\) in L.M.V, theorem for the function \(f(x)=x^{2}\) in the interval \([-1,1]\) is (a) 0 (b) \(\frac{1}{2}\) (c) \(-\frac{1}{2}\) (d) non existent

Short Answer

Expert verified
The value for \(c\) is \(0\)

Step by step solution

01

State the Function and Interval

The function \(f(x)\) given is \(x^{2}\), within the interval \([-1,1]\)
02

Apply the Lagrange's Mean Value Theorem

To find the value of \(c\) in the Lagrange's Mean Value Theorem \(f'(c) = \frac{f(b) - f(a)}{b - a}\), we need to determine \(f'(c)\) that is the derivative of \(f(x)\), and \(f(b) - f(a)\)
03

Calculate the Derivative of the Function

The derivative of \(f(x)\) is \(f'(x) = 2x\). This will be \(f'(c) = 2c\) in our Mean Value theorem.
04

Substitute the Interval Values into Function

Substitute the interval values into \(f(x)\) to determine \(f(b) - f(a)\). So, \(f(-1) = (-1)^2 = 1\) and \(f(1) = (1)^2 = 1\). Thus, \(f(b) - f(a) = f(1) - f(-1) = 1 - 1 = 0\)
05

Solve for the value of 'c'

Substitute the found values in the Mean Value theorem: \(2c = 0\). Solving for 'c' gives us \(c = 0\)

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Key Concepts

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

Understanding Differentiation
Differentiation is a core concept in calculus that helps us understand how a function changes at any given point. Differentiation is the process of finding the derivative of a function.
  • It tells us the rate of change or the velocity at which the function's value is changing with respect to its input value, often denoted as \(x\).
  • To perform differentiation, we apply specific rules and formulas to a function to derive its derivative.
  • A derivative is represented as \(f'(x)\) or \(\frac{dy}{dx}\), expressing the change in \(y\) over the change in \(x\).
Differentiation allows us to understand the behavior of curves and helps find tangents at various points of the curve. For example, in our function \(f(x) = x^2\), differentiation gives us the slope of the tangent line at any particular point \(x\). This is essential in solving calculus problems like those involving Lagrange's Mean Value Theorem.
Exploring the Derivative of a Function
A derivative is an important concept describing the relationship between variables in a function. For a function \(f(x) = x^2\), the process of differentiation helps calculate the derivative, \(f'(x) = 2x\).
  • This derivative, \(2x\), gives us the slope of the tangent to the curve at any point on \(f(x)\).
  • The derivative tells us how \(f(x)\) is increasing or decreasing as \(x\) changes.
To find the derivative:
  • We apply the power rule, \(d(x^n)/dx = nx^{n-1}\). For \(x^2\), this becomes \(2x\).
The derivative's role in finding \(c\) in Lagrange's Theorem is crucial, as \(f'(c) = 2c\) refers to our rate of change needed to find such \(c\). By finding \(c\), we essentially locate a specific point within a specified interval where the function's slope equals the average rate of change over that interval.
Solving Calculus Problems with Step-by-Step Methods
Solving calculus problems can be challenging, but breaking them into systematic steps makes it manageable. Applying Lagrange鈥檚 Mean Value Theorem requires such a structured approach:
  • State the Function and Interval: First, identify the function and the interval. Here, \(f(x) = x^2\) over \([-1,1]\).
  • Apply Lagrange鈥檚 Mean Value Theorem: The theorem states that \(f'(c) = \frac{f(b) - f(a)}{b - a}\).
  • Calculate the Derivative: Find \(f'(x)\) for the function. For \(x^2\), it is \(2x\).
  • Substitute the Interval Values: Compute \(f(b) - f(a)\) where \(f(1) = 1^2\) and \(f(-1) = (-1)^2\), yielding zero.
  • Solve for \(c\): Substitute these values into the theorem to solve. This will give \(2c = 0\), resulting in \(c = 0\).
These steps demonstrate how important it is to understand each part of a calculus problem, from understanding derivatives to applying the correct theorems, ensuring comprehensive problem-solving skills.

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Most popular questions from this chapter

Let \(f(x)=\left|\begin{array}{lll}\cos x & \sin x & \cos x \\ \cos 2 x & \sin 2 x & 2 \cos 2 x \\ \cos 3 x & \sin 3 x & 2 \cos 3 x\end{array}\right|\) Then find the value of \(f^{\prime}\left(\frac{\pi}{2}\right)\).

If \(y=A \cos (\log x)+B \sin (\log x)\), then prove that \(x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0\)

Suppose \(f\) is a differentiable function such that \(f(g(x))=x\) and \(f^{\prime}(x)=1+(f(x))^{2}\), then prove that \(g^{\prime}(x)=\frac{1}{1+x^{2}}\)

If \(y=\cos ^{-1}(2 x)+2 \cos ^{-1}\left(\sqrt{1-4 x^{2}}\right)\) \(0

If \(y=x \sin x\), then prove that, \(x^{2} \frac{d^{2} y}{d x^{2}}-2 x \frac{d y}{d x}+\left(x^{2}+2\right) y=0\)
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