Problem 38 - Thomas Calculus

Question

Find the function with the given derivative whose graph passes through the point \(P.\) \begin{equation}g^{\prime}(x)=\frac{1}{x^{2}}+2 x, \quad P(-1,1)\end{equation}

Step-by-Step Solution

Answer

The function is \( g(x) = -\frac{1}{x} + x^2 - 1 \).

Step 1: Understand the Problem

We need to find the function \( g(x) \) whose derivative is given by \( g'(x) = \frac{1}{x^2} + 2x \) and passes through the point \( P(-1, 1) \). This involves integrating the derivative to find the general function and using the given point to find any constants.

Step 2: Integrate the Derivative

Find the indefinite integral of \( g'(x) = \frac{1}{x^2} + 2x \):\[ g(x) = \int \left( \frac{1}{x^2} + 2x \right) \, dx \]The integral of \( \frac{1}{x^2} \) is \( -\frac{1}{x} \) and the integral of \( 2x \) is \( x^2 \). Thus, the general integral is:\[ g(x) = -\frac{1}{x} + x^2 + C \] where \( C \) is the constant of integration.

Step 3: Apply the Initial Condition

Use the point \( P(-1, 1) \) to find the constant \( C \). Substitute \( x = -1 \) and \( g(x) = 1 \) into the equation:\[ 1 = -\frac{1}{-1} + (-1)^2 + C \]\[ 1 = 1 + 1 + C \]Solve for \( C \):\[ C = 1 - 2 = -1 \]

Step 4: Write the Specific Function

Substitute \( C = -1 \) back into the general function to obtain the specific function:\[ g(x) = -\frac{1}{x} + x^2 - 1 \]

Key Concepts

Integration

Integration is a crucial concept in calculus that helps to find a function from its derivative. It is essentially the reverse process of differentiation. When we have a derivative, like in this exercise, and need to find the original function, we use integration. In our problem, the derivative provided is \[ g'(x) = \frac{1}{x^2} + 2x \]To find the original function \( g(x) \), we integrate this expression. Here's how we do it step-by-step:

• Integrate \( \frac{1}{x^2} \). The result is \( -\frac{1}{x} \) because it is the antiderivative of the power \( x^{-2} \).
• Integrate \( 2x \). This is a straightforward operation that yields \( x^2 \).

The general function after integration becomes \[ g(x) = -\frac{1}{x} + x^2 + C \]where \( C \) is the constant of integration. This step is vital to move forward in solving problems involving derivatives.

Derivative

Understanding a derivative is foundational to solving calculus problems. The derivative of a function represents the rate of change, or how a function changes at any given point. In the problem provided, \[ g'(x) = \frac{1}{x^2} + 2x \]indicates how the function \( g(x) \) changes with respect to \( x \). It combines two parts:

• \( \frac{1}{x^2} \) is a component that adds a negative drift to the function due to its negative power when integrated.
• \( 2x \) represents a part that increases the function more rapidly as \( x \) becomes larger.

Differentiation tells us about the slope of the tangent line to the curve at any point on a graph, which is key to understanding changes in functions. When we find a function from its derivative, it's like piecing together a puzzle, stepping backward from changes to identify original characteristics.

Initial condition

The role of the initial condition in integration is to help determine the specific function from a family of functions. In this exercise, the initial condition is the point \( P(-1, 1) \). This tells us that when \( x = -1 \) the function \( g(x) \) should yield 1. Applying this, we set up the equation:
\[ 1 = -\frac{1}{-1} + (-1)^2 + C \]When we solve this equation, we find that \( C = -1 \). The initial condition is crucial as it pinpoints the exact function needed among infinite possibilities that the integration process might suggest. Without this condition, we could only establish a general form of the function which includes the constant \( C \). Therefore, including initial conditions is essential in practical scenarios to obtain a unique and usable answer.

Constant of integration

The constant of integration, denoted as \( C \), emerges whenever we perform indefinite integration. Why does it appear? Because integration is the reverse of differentiation, and when you differentiate any constant, it disappears (since the derivative of any constant is zero).
This means that without knowing additional information, you can't determine what constant was in the original function. Thus, after integrating the derivative in our problem, we write the solution as \[ g(x) = -\frac{1}{x} + x^2 + C \]Leaving \( C \) allows for every possible function that could have resulted in the same derivative. The initial condition, as discussed earlier, is necessary for determining \( C \). In this exercise, solving for \( C \) using the point \( P(-1, 1) \) gave us \( C = -1 \), which completed our solution to find the specific form of \( g(x) \). Understanding the value and purpose of \( C \) is crucial for interpreting results correctly in calculus problems.