Problem 33 - Essential Calculus Early Transcendentals

Question

Find \(f\) $$f^{\prime \prime}(x)=x^{-2}, \quad x>0, \quad f(1)=0, \quad f(2)=0$$

Step-by-Step Solution

Answer

The function is \( f(x) = (\ln 2)(x - 1) - \ln|x| \).

Step 1: Determine the First Derivative

Given the second derivative of the function: \( f''(x) = x^{-2} \). The first step is to find the first derivative by integrating \( f''(x) \). We integrate: \[ f'(x) = \int x^{-2} \, dx = -x^{-1} + C_1 \], where \( C_1 \) is the constant of integration.

Step 2: Determine the Function

Now, integrate the first derivative \( f'(x) = -x^{-1} + C_1 \) to find \( f(x) \). Perform the integration: \[ f(x) = \int (-x^{-1} + C_1) \, dx = -\ln|x| + C_1 x + C_2 \], where \( C_2 \) is another constant of integration.

Step 3: Apply Initial Condition \(f(1) = 0\)

Use the condition \( f(1) = 0 \) to solve for one of the constants. Substitute \( x = 1 \) into the equation: \( 0 = -\ln|1| + C_1\cdot1 + C_2 \), leading to \( C_2 = -C_1 \), since \( \ln(1) = 0 \).

Step 4: Apply Initial Condition \(f(2) = 0\)

Use the second condition \( f(2) = 0 \) to find \( C_1 \). Substitute \( x = 2 \) into \( f(x) \): \( 0 = -\ln2 + C_1\cdot2 - C_1 \), simplifying gives \( C_1 = \ln2 \).

Step 5: Write the Final Function

With \( C_1 \) determined, substitute back into our function. Thus, the function \( f(x) \) is \[ f(x) = -\ln|x| + (\ln2)x - \ln2 \]. Simplifying, we get \[ f(x) = (\ln2)(x - 1) - \ln|x| \].

Key Concepts

Differential Equations

Differential equations involve functions and their derivatives and play a significant role in calculus and applied mathematics. They can describe various phenomena such as motion, growth, decay, and other dynamic processes.
The problem we are examining deals with finding a function based on its second derivative. More specifically, the differential equation is given as:

• The second derivative, denoted as \( f''(x) = x^{-2} \), provides a rate of change of the slope of the original function.

To progress from the second derivative to the first derivative and eventually to the original function \( f(x) \), we use a process known as integration. This makes differential equations indispensable in predicting how systems evolve over time.

• Integration reverses differentiation, allowing us to backtrack from a rate of change (derivative) to a cumulative sum (original function).

Constant of Integration

In calculus, when integrating a function, a constant of integration, typically denoted as \( C \), is added. This constant accounts for any constant value that may have been "lost" during differentiation.
Let's look at its role in solving the problem:

• When the second derivative \( f''(x) \) is integrated to find \( f'(x) \), it results in \( f'(x) = -x^{-1} + C_1 \). Here, \( C_1 \) represents an unknown constant added because differentiating a constant results in zero; thus, it vanishes when we derive from \( f' (x) \) to \( f''(x) \).

As we integrate \( f'(x) \) to find \( f(x) \), another constant of integration \( C_2 \) appears:

• The expression \( f(x) = -\ln|x| + C_1 x + C_2 \) shows the accumulation of two constants, indicating that these constants tailor the function to fit specific criteria or conditions.

By applying initial conditions, these constants are determined, transforming the general solution into a specific solution that solves the exercise.

Initial Conditions

Initial conditions are specific values given at a particular point to find a unique solution for differential equations. These conditions refine the solution by determining the constants of integration.
In our exercise, we are given:

• \( f(1) = 0 \): Means that when \( x = 1 \), the function \( f(x) \) evaluates to 0.
• \( f(2) = 0 \): Means that when \( x = 2 \), the function \( f(x) \) also evaluates to 0.

These initial conditions help us solve for the values of \( C_1 \) and \( C_2 \):

• From \( f(1) = 0 \), we find that \( C_2 = -C_1 \), aligning with the balance required in our function.
• From \( f(2) = 0 \), substituting and solving lets us find \( C_1 = \ln2 \), thus giving a precise form to \( f(x) \).

These tailored conditions ensure the resulting function satisfies the specific behavior outlined by the exercise's initial conditions.