Problem 47 - Calculus and its applications

Question

Find f such that: $$ f^{\prime}(x)=x-3, \quad f(2)=9 $$

Step-by-Step Solution

Answer

The function is \( f(x) = \frac{x^2}{2} - 3x + 13 \).

Step 1: Understand the Problem

We need to find the function \( f(x) \) given its derivative \( f'(x) = x - 3 \) and a specific point \( f(2) = 9 \). To find \( f(x) \), we'll integrate the derivative.

Step 2: Integrate the Derivative

The antiderivative of \( f'(x) = x - 3 \) is found by integrating it with respect to \( x \). The general form will be: \[ f(x) = \int (x - 3) \, dx = \frac{x^2}{2} - 3x + C \] where \( C \) is the constant of integration.

Step 3: Solve for Constant C

Use the given condition \( f(2) = 9 \) to find \( C \). Substitute 2 for \( x \) and 9 for \( f(x) \): \[ \frac{2^2}{2} - 3(2) + C = 9 \] \[ 2 - 6 + C = 9 \] \[ C = 9 + 4 = 13 \].

Step 4: Write the Solution

Now that we know \( C = 13 \), we can write the function: \[ f(x) = \frac{x^2}{2} - 3x + 13 \]. This is the function whose derivative is \( x - 3 \) and that satisfies the condition \( f(2) = 9 \).

Key Concepts

Understanding Antiderivatives

When given the derivative of a function, finding the original function requires us to use the concept of the antiderivative. The antiderivative is essentially the reverse process of differentiation. While differentiation gives us the rate at which a function changes, antiderivation tells us about the original function before it was differentiated.

For example, if you have a derivative function like \( f'(x) = x - 3 \), the task is to find its antiderivative. This means we need to determine a function \( f(x) \) such that its derivative is \( x - 3 \).

The process involves integrating the derivative function. Integration will result in a general form of the function, often denoted as \( f(x) = \int (x - 3) \, dx = \frac{x^2}{2} - 3x + C \). Notice the addition of the \( C \), which we will explain in the next section.

The Role of the Constant of Integration

When integrating a derivative to find its antiderivative, you will always encounter a term called the constant of integration, often represented by the letter \( C \).

The reason for this constant is simple: differentiation removes constant terms. If \( f(x) \) was differentiated to become \( f'(x) = x - 3 \), there could have been any constant value added to \( f(x) \) like \( +1 \) or \( -7 \), etc.

When reversing the differentiation process through integration, we cannot pinpoint this original constant value just by examining the derivative. Thus, we denote it simply as \( C \).

Adding this constant of integration accounts for every possible vertical shift of the antiderivative function graph.

Solving for the Initial Condition

After finding the general antiderivative including the constant of integration, we can use specific conditions to find its exact form. This involves finding a missing piece of the puzzle through initial or boundary conditions.

An initial condition specifies a particular point through which the function passes, like \( f(2) = 9 \). By plugging these values into the general antiderivative \( f(x) = \frac{x^2}{2} - 3x + C \), we can solve for \( C \).

For example, substitute \( x = 2 \) and \( f(2) = 9 \):

• \( \frac{2^2}{2} - 3(2) + C = 9 \)
• Simplify to find \( C = 13 \)

With \( C \) determined, our specific function becomes \( f(x) = \frac{x^2}{2} - 3x + 13 \), ensuring it fits the original condition \( f(2) = 9 \).