Problem 25 - Thomas Calculus

Question

Suppose that \(f^{\prime}(x)=2 x\) for all \(x .\) Find \(f(2)\) if a. \(f(0)=0 \quad\) b. \(f(1)=0 \quad\) c. \(f(-2)=3\)

Step-by-Step Solution

Answer

a. 4 b. 3 c. 3

Step 1: Understand the Problem

We are given a derivative of a function, \(f^{\prime}(x)=2x\), and initial condition values: \(f(0)=0\), \(f(1)=0\), \(f(-2)=3\). These initial conditions will help us find particular solutions of the function \(f\). Our task is to find \(f(2)\) for each case.

Step 2: Integrate the Derivative

To find \(f(x)\), we need to integrate the given derivative \(f^{\prime}(x)=2x\). The integral of \(2x\) is \(x^2 + C\), where \(C\) is the constant of integration. Therefore, \(f(x) = x^2 + C\).

Step 3: Apply Initial Condition a

For case (a), we use the initial condition \(f(0) = 0\). Substituting into the equation, \(f(0) = 0^2 + C = 0\), we find \(C = 0\). Thus, \(f(x) = x^2\). Now, find \(f(2)\):\[f(2) = 2^2 = 4\].

Step 4: Apply Initial Condition b

For case (b), use the initial condition \(f(1) = 0\). Substituting into the equation, \(f(1) = 1^2 + C = 0\) gives \(1 + C = 0\) or \(C = -1\). Therefore, \(f(x) = x^2 - 1\). Find \(f(2)\):\[f(2) = 2^2 - 1 = 3\].

Step 5: Apply Initial Condition c

For case (c), use the initial condition \(f(-2) = 3\). Substituting into the equation, \(f(-2) = (-2)^2 + C = 3\), results in \(4 + C = 3\), hence \(C = -1\). Therefore, \(f(x) = x^2 - 1\). Find \(f(2)\):\[f(2) = 2^2 - 1 = 3\].

Key Concepts

Integration

Integration is a fundamental technique in calculus. It is often described as the process of finding antiderivatives. It's like working backwards from derivatives to determine the original function. In the given exercise, we have a derivative expressed as a simple algebraic expression: \(f'(x) = 2x\).
To find the original function \(f(x)\), we need to integrate \(2x\). When integrating powers of \(x\), we use the basic rule: the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\).
Applying this to \(2x\), we get:

• The integral of \(2x\) is \(x^2 + C\).

Here, \(C\) represents the constant of integration, an indefinite number that appears because integration is an inverse operation to differentiation. Understanding how to perform integration is crucial for solving calculus problems like finding the original function from its derivative.

Initial Conditions

Initial conditions play a critical role in determining the specific solution to a differential equation. After integrating a derivative, we often end up with a general form of the antiderivative, which includes a constant \(C\).
To find a particular solution, we need to use given conditions at specific points, known as initial conditions.
Let's consider how initial conditions are used in the example:

• For \(f(0) = 0\), substituting gives the equation \(0 = 0^2 + C\), leading to \(C = 0\).
• For \(f(1) = 0\), follow the equation \(0 = 1^2 + C\), simplifying to \(C = -1\).
• Finally, for \(f(-2) = 3\), substitute into \(f(x) = x^2 + C\) to find \(4 + C = 3\), thus \(C = -1\).

Each initial condition offers a unique value for \(C\) that best fits the scenario and provides the specific solution we need. This way, the initial conditions complete the process of finding the function accurately.

Function Evaluation

Function evaluation involves substituting specific values into the function to discover further properties or verify results. In our exercise, after determining the function \(f(x)\) with the proper \(C\), we substitute \(x = 2\) to find \(f(2)\).
Let's see how this works in the context of our exercise:

• When \(C = 0\), we find \(f(x) = x^2\), leading to \(f(2) = 2^2 = 4\).
• With \(C = -1\), the function becomes \(f(x) = x^2 - 1\), and \(f(2) = 2^2 - 1 = 3\).

By substituting \(x = 2\) into these expressions, we calculate the specific value of the function, which represents a real-world quantity or satisfies the condition of a problem. Function evaluation is particularly useful in practical applications like physics or engineering, wherein one needs to assess specific points within a system. This practice enables deeper insights and predictions based on calculated scenarios.