Problem 47 - Calculus Single Variable

Question

In each of Exercises \(41-48\), use the given information to find \(F(c)\). $$ F^{\prime}(x)=2 / x^{2}, \quad F(-1)=4, \quad c=4 $$

Step-by-Step Solution

Answer

The value of \( F(4) \) is \( \frac{3}{2} \).

Step 1: Understand the Problem

We need to find the value of the function \( F(c) \) where \( c = 4 \). We are given the derivative of the function, \( F'(x) = 2 / x^2 \), and the condition \( F(-1) = 4 \).

Step 2: Integrate the Derivative

To find \( F(x) \), we integrate \( F'(x) = \frac{2}{x^2} \). Recall that \( \int x^{-n} \, dx = \frac{x^{-n+1}}{-n+1} + C \) for \( n eq 1 \). Here, integrate \( 2x^{-2} \), yielding:\[ F(x) = -\frac{2}{x} + C \]

Step 3: Use the Initial Condition to Solve for C

Substitute \( x = -1 \) into the integrated function and \( F(-1) = 4 \):\[ 4 = -\frac{2}{-1} + C \]\[ 4 = 2 + C \]Solving for \( C \) gives:\[ C = 2 \]

Step 4: Formulate the Function F(x)

Now substitute back the value of \( C \) into the expression for \( F(x) \):\[ F(x) = -\frac{2}{x} + 2 \]

Step 5: Calculate F(c) for c = 4

Substitute \( c = 4 \) into \( F(x) \) to find \( F(4) \):\[ F(4) = -\frac{2}{4} + 2 \]\[ F(4) = -\frac{1}{2} + 2 \]\[ F(4) = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \]

Key Concepts

Derivative

Understanding derivatives is crucial in calculus. They represent the rate of change of a function with respect to a variable. In simpler terms, a derivative tells you how a function is behaving at any given point. It provides the slope of the tangent line to the curve of the function at any specific point.

Key ideas to remember about derivatives include:

• Derivatives can help predict how one variable changes with another.
• The notation for the derivative of a function \( F(x) \) is commonly written as \( F'(x) \) or \( \frac{dF}{dx} \).
• A positive derivative indicates increasing behavior, while a negative derivative suggests decreasing behavior.

In our exercise, the function's derivative was provided as \( F'(x) = \frac{2}{x^2} \). This means the function is decreasing at varying rates depending on the value of \( x \). Understanding the derivative helps us later when we integrate.

Initial Value Problem

An initial value problem requires you to find a particular solution to a differential equation. This is done by using specific conditions provided. In the exercise, we are given an initial condition: \( F(-1) = 4 \).

Steps to handle an initial value problem include:

• Start by finding a general solution to the differential equation.
• Apply the initial condition to solve for the constant in the solution.
• Use the particular solution to meet the conditions provided.

This exercise guides us through finding a specific \( F(x) \) by using the derivative provided and applying \( F(-1) = 4 \), allowing us to solve this initial value problem by determining the constant of integration.

Indefinite Integral

An indefinite integral, also known as the antiderivative, is essentially the reverse process of taking a derivative. When you integrate a function, you find its original function given its derivative. This process introduces a constant because integration results in a family of functions.

Here's what you need to know:

• Notated by \( \int \), an indefinite integral doesn't have upper or lower limits.
• Integrals help recover the function from its rate of change.
• It results in a general solution, \( F(x) + C \), where \( C \) is an unknown constant.

In our problem, we integrate \( F'(x) = \frac{2}{x^2} \) to find \( F(x) \). The integral \( \int \frac{2}{x^2} \, dx \) leads to a function \( F(x) = -\frac{2}{x} + C \). This gives us a general solution before finding \( C \).

Constant of Integration

Upon integration, you encounter the constant of integration, denoted as \( C \). This constant occurs because the process of differentiating a constant results in zero, which means when you integrate, any constant could have been there initially.

Key points include:

• The constant of integration arises because integration reverses differentiation.
• Its purpose is to account for all the vertical shifts of a function.
• The value is determined using initial or boundary conditions.

In our exercise, the initial condition \( F(-1) = 4 \) was used to calculate \( C = 2 \) after integrating. By substituting \( x = -1 \) and using the known value of \( F(-1) = 4 \), the constant can be uniquely determined. It ensures that the integrated function meets the specific requirements of the problem.