Problem 30 - University Calculus: Early Transcendentals

Question

Suppose that \(f(0)=5\) and that \(f^{\prime}(x)=2\) for all \(x .\) Must \(f(x)=\) \(2 x+5\) for all \(x ?\) Give reasons for your answer.

Step-by-Step Solution

Answer

Yes, because \( f(x) = 2x + 5 \) satisfies both \( f'(x) = 2 \) and \( f(0) = 5 \).

Step 1: Understanding the Problem

We are given that \( f(0) = 5 \) and \( f'(x) = 2 \) for all \( x \). The task is to determine if \( f(x) \) is equal to \( 2x + 5 \) for all \( x \).

Step 2: Analyze the Derivative of f(x)

Since \( f'(x) = 2 \) is constant for all \( x \), it means that the function \( f(x) \) is linear. Therefore, \( f(x) = 2x + C \), where \( C \) is a constant to be determined.

Step 3: Use the Given Condition f(0) = 5

We know \( f(0) = 5 \). Substitute \( x = 0 \) into the equation \( f(x) = 2x + C \): \[ f(0) = 2(0) + C = C = 5 \]. Thus, \( C = 5 \).

Step 4: Write the Final Function

Substitute \( C = 5 \) back into \( f(x) = 2x + C \). The function becomes \( f(x) = 2x + 5 \).

Step 5: Verify the Conclusion

The derived function \( f(x) = 2x + 5 \) satisfies both \( f'(x) = 2 \) for all \( x \) and \( f(0) = 5 \). Thus, this is the correct function.

Key Concepts

Constant function

A constant function is one of the simplest types of functions in mathematics. It is a function where the output value remains the same regardless of the input. In mathematical terms, a constant function is expressed as \( f(x) = c \) where \( c \) is a constant.
This means that no matter what value of \( x \) you plug into a constant function, the output will always be \( c \). It's like having a straight horizontal line on a graph.

• The derivative of a constant function is always zero, which means its rate of change is zero.
• Constant functions are always parallel to the x-axis and never intersect it.

In the case of integration, the integral of a constant function \( c \) with respect to \( x \) is \( cx + C \), where \( C \) represents the constant of integration.

Linear function

Linear functions are a step up in complexity from constant functions. They have the form \( f(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
This function represents a straight line on a graph. Unlike a constant function's horizontal line, a linear function can have a slope that makes the line rise or fall.

• The slope \( m \) indicates the rate of change; a positive \( m \) means the line ascends, while negative \( m \) means it descends.
• The y-intercept \( b \) indicates where the line crosses the y-axis.

The derivative of a linear function is the constant \( m \), reflecting the constant rate of change. This helps us determine that if \( f'(x) = 2 \), then the function is linear with a slope of 2.

Integration

Integration is the mathematical process of finding the integral of a function. It is essentially the reverse operation of differentiation.
When you integrate a function, you are looking for the function with a derivative that matches the original function. For example, the integral of \( 2 \) with respect to \( x \) is \( 2x + C \), where \( C \) is the integration constant.

• Integration can be thought of as finding the area under the curve of a graphed function.
• Definite integration results in a numerical area value, while indefinite integration includes the \( C \).

In our exercise, integrating \( f'(x) = 2 \) gives us \( f(x) = 2x + C \), which aligns with finding a function whose derivative is a constant.

Initial condition

Initial conditions are critical in determining the specific form of a function when solving differential equations.
These conditions provide a specific value for the function at a particular point, allowing us to find the constant of integration. In our exercise, \( f(0) = 5 \) serves as an initial condition, telling us how the line \( f(x) = 2x + C \) should be positioned on the coordinate plane.

• Substituting the initial condition into the integrated function helps solve for \( C \), ensuring the function matches specified conditions.
• Without an initial condition, there is an infinite number of functions that could satisfy a given differential equation.

Thus, by applying the initial condition \( f(0) = 5 \), we determined that \( C = 5 \), leading to the complete function \( f(x) = 2x + 5 \). This process ensures the solution is unique and satisfies all given parameters.