Problem 24 - Calculus Single Variable

Question

Compute \(F(c)\) from the given information. $$ F^{\prime}(x)=1 / x, F(1)=3, c=-e^{2} $$

Step-by-Step Solution

Answer

\( F(-e^2) = 5 \).

Step 1: Recognize the Derivative Function

We are given that the derivative of the function is \( F'(x) = \frac{1}{x} \). This means that the function \( F(x) \) is the antiderivative of \( \frac{1}{x} \), which is \( \ln|x| + C \) where \( C \) is a constant of integration.

Step 2: Use Initial Condition to Find C

We know \( F(1) = 3 \). Substituting in \( F(x) = \ln|x| + C \), we have \( F(1) = \ln|1| + C = 3 \). Since \( \ln(1) = 0 \), it follows that \( C = 3 \). Thus, \( F(x) = \ln|x| + 3 \).

Step 3: Evaluate F at c

Now that we have \( F(x) = \ln|x| + 3 \), we need to evaluate \( F(c) \) where \( c = -e^2 \). Since \( |c| = e^2 \), substitute to get \( F(-e^2) = \ln(e^2) + 3 \).

Step 4: Simplify the Expression

Calculate \( \ln(e^2) = 2 \ln(e) = 2 \times 1 = 2 \). Thus, \( F(-e^2) = 2 + 3 = 5 \).

Key Concepts

Derivative Functions

When you hear the term "derivative," think of it as a tool that tells you how a function changes at any given point. In this exercise, we're dealing with the derivative function ul>

Derivative Function: Given by \( F'(x) = \frac{1}{x} \), this tells us the rate of change of the original function \( F(x) \) with respect to \( x \).

Antiderivative: To find the original function (called the antiderivative), we determine what could have been differentiated to become \( F'(x) \). Here, the antiderivative of \( \frac{1}{x} \) is the natural logarithm function, \( \ln|x| \).

The beauty of derivative functions lies in their ability to reverse-engineer or backtrack, giving us the original function when combined with initial conditions.

Constant of Integration

In calculus, every time we find an antiderivative, we add a constant of integration. This constant is crucial, because derivatives do not reveal it:

• Why Add It?: The derivative of a constant is zero, so any real number added to \( \ln|x| \) would differentiate back to \( \frac{1}{x} \) and thus we add \( C \) to account for any such "invisible" constants.
• Expression: Once we have the antiderivative, we write it as \( F(x) = \ln|x| + C \).

By finding \( C \), we customize the antiderivative to fit particular conditions or data points provided in a problem.

Initial Conditions

To find any unknown constants in the antiderivative, we apply initial conditions from the problem:

• Use the Given Data Point: We were told \( F(1) = 3 \). Using \( F(x) = \ln|x| + C \), we substitute to find the specific value of \( C \).
• Solving for C: Plug \( x = 1 \) into the equation to get \( 3 = \ln(1) + C \). Since \( \ln(1) = 0 \), we determine that \( C = 3 \).

Initial conditions help anchor our function to real situations, allowing us to solve concrete problems accurately.