Problem 29 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

a. 4, b. 3, c. 3

Step 1: Understand the problem

We are given the derivative of a function \(f(x)\) as \(f'(x) = 2x\). We need to find the function \(f(x)\) by integrating \(f'(x)\), and then solve for specific initial conditions to find \(f(2)\).

Step 2: Integrate the derivative

To find \(f(x)\), integrate \(f'(x) = 2x\):\[\int 2x\,dx = x^2 + C\]where \(C\) is the constant of integration.

Step 3: Solve for \(C\) given \(f(0) = 0\)

Use the condition \(f(0) = 0\) to find \(C\):\[0^2 + C = 0\]Thus, \(C = 0\), and \(f(x) = x^2\).

Step 4: Find \(f(2)\) when \(f(0)=0\)

Substitute \(x = 2\) in \(f(x) = x^2\):\[f(2) = 2^2 = 4\]

Step 5: Solve for \(C\) given \(f(1) = 0\)

Use the condition \(f(1) = 0\) to find \(C\):\[1^2 + C = 0 \Rightarrow C = -1\]Thus, \(f(x) = x^2 - 1\).

Step 6: Find \(f(2)\) when \(f(1)=0\)

Substitute \(x = 2\) in \(f(x) = x^2 - 1\):\[f(2) = 2^2 - 1 = 3\]

Step 7: Solve for \(C\) given \(f(-2) = 3\)

Use the condition \(f(-2) = 3\) to find \(C\):\[(-2)^2 + C = 3 \Rightarrow 4 + C = 3 \Rightarrow C = -1\]Thus, \(f(x) = x^2 - 1\).

Step 8: Find \(f(2)\) when \(f(-2)=3\)

Substitute \(x = 2\) in \(f(x) = x^2 - 1\):\[f(2) = 2^2 - 1 = 3\]

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus used to find the original function from its derivative. When given a derivative like \(f'(x) = 2x\), the process of integrating involves finding the function \(f(x)\) whose derivative is \(2x\). The technique to achieve this is called indefinite integration. The result of integrating \(2x\) is \(x^2 + C\), where \(C\) is a constant of integration.

Why is there a constant \(C\)? This constant arises because the process of differentiation removes any constant term from the function. When we reverse this process using integration, we add the constant back, which could be any real number.

• Perform indefinite integration by finding the antiderivative.
• Remember to add the constant \(C\) to represent any constant lost during differentiation.

This constant is crucial when applying initial conditions to find a particular solution.

Initial Conditions

Initial conditions are specific values that allow us to determine the constant \(C\) in an integrated function. They provide specific information about the function at certain points. In the original exercise, initial conditions like \(f(0)=0\), \(f(1)=0\), and \(f(-2)=3\) were provided to find the exact form of \(f(x)\).

Here's how to use initial conditions:

• Substitute the initial condition into the integrated function \(x^2 + C\).
• Solve for \(C\) to find a unique function for each condition.

This process allows us to tailor the general integral to fit specific scenarios.

For example, if \(f(0) = 0\), substituting gives \(0^2 + C = 0\), resulting in \(C = 0\) and therefore \(f(x) = x^2\). Similar steps are repeated for other given conditions to find different functions.

Function Evaluation

Once the function is found using integration and initial conditions, the final step is evaluating the function at a given point. This is necessary to find the function's value at that particular argument, like \(f(2)\), in our problem. Here's how you can accomplish this:

• Insert the desired value of \(x\) into your specific function formula (e.g., \(f(x) = x^2\) or \(f(x) = x^2 - 1\)).
• Perform the calculations to find the output value.

This step provides the solution to questions asking for the function's value at certain points. For instance, evaluating \(f(2)\) when \(f(0) = 0\) gives \(4\), and when \(f(1) = 0\) or \(f(-2) = 3\), using the specific function \(f(x) = x^2 - 1\), it results in \(3\).

In essence, function evaluation is about finding out what the function's output is for given input once you've defined the function completely.