Problem 10 - Calculus: Early Transcendentals

Question

Find \(f(x)\) if \(f^{\prime}(x)=e^{-x}\) and \(f(0)=2\).

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem Statement

We need to find the function \(f(x)\) given its derivative \(f'(x) = e^{-x}\) and an initial condition \(f(0) = 2\). This is an integration and constant determination problem.

Step 2: Integrate the Derivative

Integrate the derivative function \(f'(x) = e^{-x}\) with respect to \(x\) to find \(f(x)\). The integration of \(e^{-x}\) is \(-e^{-x}\). Thus, the function \(f(x)\) is \(-e^{-x} + C\), where \(C\) is the integration constant.

Step 3: Utilize the Initial Condition

Use the initial condition \(f(0) = 2\) to solve for the constant \(C\). Substitute \(x = 0\) into \(-e^{-x} + C\) to get \(-e^{0} + C = 2\). Simplifying gives \(-1 + C = 2\), hence \(C = 3\).

Step 4: Write the Final Expression for \(f(x)\)

The function \(f(x)\) is found by substituting \(C = 3\) back into the general solution. Thus, \(f(x) = -e^{-x} + 3\).

Key Concepts

Derivative

The concept of a derivative is all about understanding how a function changes at any given point. Think of it like the slope of a mountain at different spots.
You might see derivatives expressed as \( f'(x) \) or \( \frac{df}{dx} \). This notation shows how the function \( f(x) \) changes as \( x \) changes.
To solve problems like the one given, it's helpful to know that differentiation and integration are opposite processes.
While differentiation tells us the rate of change (how fast something is moving), integration helps us find the original function from its rate.

• This problem starts with \( f'(x) = e^{-x} \), which is our rate of change.
• We want to find \( f(x) \), the original function.

Exponential Function

Exponential functions are characterized by variables in the exponent, like \( e^{-x} \). They have unique and useful properties.
The base \( e \) is a special number, approximately equal to 2.718, known for making the process of differentiation and integration straightforward.
When you differentiate \( e^x \), you get \( e^x \) back.

• The differentiation rule for \( e^{-x} \) results in \( -e^{-x} \).

Understanding these concepts allows you to integrate \( e^{-x} \) smoothly.
In this problem, integrating \( e^{-x} \) gives \( -e^{-x} + C \), because integrating reverses differentiation's effects, bringing us back to the original function.

Initial Condition

Initial conditions offer specific values that help us determine the unknown constants when integrating. They are crucial for pinning down one specific solution out of infinite possibilities.
In integration, you often get a general solution first, which includes a constant term (like \( C \)).
The initial condition tells us that when \( x = 0 \), \( f(x) = 2 \).

• We substitute this into \( f(x) = -e^{-x} + C \) to solve for \( C \).
• When \( x = 0 \), \( -e^{0} + C = 2 \).
• This simplifies to \( -1 + C = 2 \), so \( C = 3 \).

By applying the initial condition, we pinpoint the exact form of \( f(x) \). This allows us to move beyond the general form and find a specific answer.

Constant of Integration

The constant of integration, symbolized as \( C \), arises when we integrate because the process adds an unknown value that wasn't initially there during differentiation.
Without this constant, solutions to differential equations can be incomplete, as they might miss different possible vertical shifts of the function.
Following the integration of \( e^{-x} \), we get \( -e^{-x} + C \). This \( C \) represents any constant value that, when added to the function, still fits the derived rate of change.

• \( C \) is then determined using the initial condition.
• This ensures our function matches the initial condition, giving us \( f(x) = -e^{-x} + 3 \) as the final, complete answer.

So, the constant \( C \) reflects missing heights that weren't visible in the derivative alone, but the initial condition helps us find its true value.