Question

If \(\mathrm{f}^{\prime}(\mathrm{x})=-\mathrm{f}(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x})\) and \(\mathrm{F}(\mathrm{x})=|[\mathrm{f}(\mathrm{x} / 2)]|^{2}+|[\mathrm{g}(\mathrm{x} / 2)]|^{2}\) and \(F(5)=5\) then \(F(10)=\) (a) 5 (b) 10 (c) (d) 15

Short answer

The short answer is: \(F(10) = 5\).

Step-by-step solution

Express f(5) and g(5) in terms of f'(5)

By using the given relationships f'(x) = -f(x) and g(x) = f'(x), we can express f(5) and g(5) in terms of f'(5) as follows: f(5) = -f'(5) g(5) = f'(5)

Find F(5) using f(5) and g(5)

Now, we can use the expressions for f(5) and g(5) in terms of f'(5) to find F(5): F(5) = |f(5/2)|^2 + |g(5/2)|^2 = |-f'(5/2)|^2 + |f'(5/2)|^2 We are given that F(5) = 5. So, 5 = |-f'(5/2)|^2 + |f'(5/2)|^2 5 = 2 * |f'(5/2)|^2 Now, let's find the value of |f'(5/2)|^2: |f'(5/2)|^2 = 5/2 Now we will find the expressions for f(10) and g(10) in terms of f'(10).

Express f(10) and g(10) in terms of f'(10)

By using the given relationships f'(x) = -f(x) and g(x) = f'(x), we can express f(10) and g(10) in terms of f'(10) as follows: f(10) = -f'(10) g(10) = f'(10)

Find F(10) using f(10) and g(10)

Now, we can use the expressions for f(10) and g(10) in terms of f'(10) to find F(10): F(10) = |f(10/2)|^2 + |g(10/2)|^2 = |-f'(10/2)|^2 + |f'(10/2)|^2 Since f'(5/2) and f'(10/2) are just f'(5) and f'(10) respectively, we can use the result from Step 2 to find F(10): F(10) = |-f'(5)|^2 + |f'(5)|^2 = |f'(5)|^2 * 2 = (5/2) * 2 = 5 So, F(10) = 5, which is option (a).

Key concepts

Calculus Problems

When we dive into calculus problems, we often deal with functions and their derivatives. In this exercise, our primary function is represented as \( f(x) \). We are given a relationship: \( f'(x) = -f(x) \). This simple yet significant equation is a differential equation, showing the relationship between a function and its derivative. Solving such differential equations is a common calculus problem because it allows us to understand how functions behave and change.

In this context, tackling the problem involves finding relationships among several functions occurring at different points, specifically at \( x=5 \) and \( x=10 \). Calculus problems often involve a sequence of logical steps to derive these relationships and evaluate them effectively, providing a part of the puzzle that can lead to deeper insights into the behavior of functions.

Function Relationships

Understanding function relationships is core to solving differential equations. In our problem, we see how different functions, \( f(x) \), \( g(x) \), and \( F(x) \), interconnect:

• \( g(x) = f'(x) \): means \( g(x) \) is immediately derived from \( f(x) \).
• \( F(x) = |f(x/2)|^2 + |g(x/2)|^2 \): combines both \( f(x) \) and \( g(x) \) into a singular value at any \( x \).

These expressions imply that variables are interdependent and rely on one another for accurate computation and evaluation.

This relationship shows how the derivative, \( g(x) \), can influence the ultimate function \( F(x) \). Understanding how derivatives relate to the original function is key to comprehending the nature of the function, especially when predicting future values of the function at new points.

Step-by-Step Solutions

Step-by-step solutions are essential for tackling complex calculus problems, making each stage easier to grasp and solve. This process transforms the mysterious calculus challenges into more manageable tasks:

• Express Functions: First, establish clear expressions for functions based on given relationships (e.g., \( f(10) = -f'(10) \)). This step-by-step approach builds the groundwork for further calculation.
• Substitute and Simplify: Replace and compute values at specific points (e.g., \( x=5 \)) which simplifies the overall task, breaking it into smaller and actionable parts.
• Evaluate Values: After substitution, compare and find consistency in values (like \( F(5)=5 \)) to ensure calculations are progressing courageously through each stage.

These steps help in systematically solving problems, making the sophisticated world of calculus approachable even when dealing with multiple interconnected functions.