Math Problem Statement
What is the value of f ′(1)?What is the value of f(0)?
Solution
Let's analyze and solve the problem step by step.
We are given:
- is a real-valued function.
- Functional equation: for all .
- Initial conditions: and .
The questions are:
- What is the value of ?
- Confirm .
Step 1: Reaffirm
The given value is . To verify this logically using the functional equation:
Substitute and into : This simplifies to: Let . Then: Rearranging: Thus, or . But from the problem, is directly given as a condition. Therefore, we accept:
Step 2: Differentiate the functional equation
The functional equation is . Differentiating both sides with respect to (treating as a constant):
Using the chain rule and product rule: [ \frac{d}{dx}[f(x+y)] = \frac{d}{dx}[f(x)f(y)]. ] The left-hand side becomes: The right-hand side, applying the product rule, becomes: Thus:
Step 3: Find
We substitute into the differentiated functional equation: From the initial condition , this simplifies to:
Now, consider : This simplifies to:
Step 4: Determine
From the earlier analysis of and the functional equation , we can differentiate to find . Since , and differentiating a constant gives zero, f'(0) = 0.
Step 5: Final Answer
Substitute into :
Thus:
- The value of is .
- The value of is .
Follow-Up Questions:
- What is the general solution to the functional equation ?
- How would the result change if were instead of ?
- Prove that must be an exponential function under certain conditions.
- How does differentiability influence the behavior of in such functional equations?
- What happens if is assumed to be instead of ?
Tip:
When dealing with functional equations like , exponential functions naturally satisfy the condition under differentiability assumptions.
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Math Problem Analysis
Mathematical Concepts
Functional Equations
Differentiation
Real-valued Functions
Formulas
f(x+y) = f(x)f(y)
Differentiation of a product of functions
Theorems
Functional equation properties
Product rule for differentiation
Suitable Grade Level
Undergraduate Mathematics or Advanced High School (Grade 12)
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