Math Problem Statement

What is the value of f ′(1)?What is the value of f(0)?

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:

  1. f:RRf: \mathbb{R} \to \mathbb{R} is a real-valued function.
  2. Functional equation: f(x+y)=f(x)f(y)f(x+y) = f(x)f(y) for all x,yRx, y \in \mathbb{R}.
  3. Initial conditions: f(1)=7f(1) = 7 and f(0)=2f(0) = 2.

The questions are:

  1. What is the value of f(1)f'(1)?
  2. Confirm f(0)f(0).

Step 1: Reaffirm f(0)f(0)

The given value is f(0)=2f(0) = 2. To verify this logically using the functional equation:

Substitute x=0x = 0 and y=0y = 0 into f(x+y)=f(x)f(y)f(x+y) = f(x)f(y): f(0+0)=f(0)f(0).f(0+0) = f(0)f(0). This simplifies to: f(0)=[f(0)]2.f(0) = [f(0)]^2. Let f(0)=cf(0) = c. Then: c=c2.c = c^2. Rearranging: c(c1)=0.c(c-1) = 0. Thus, c=0c = 0 or c=1c = 1. But from the problem, f(0)=2f(0) = 2 is directly given as a condition. Therefore, we accept: f(0)=2.f(0) = 2.


Step 2: Differentiate the functional equation

The functional equation is f(x+y)=f(x)f(y)f(x+y) = f(x)f(y). Differentiating both sides with respect to xx (treating yy 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: f(x+y).f'(x+y). The right-hand side, applying the product rule, becomes: f(x)f(y).f'(x)f(y). Thus: f(x+y)=f(x)f(y).f'(x+y) = f'(x)f(y).


Step 3: Find f(1)f'(1)

We substitute y=1y = 1 into the differentiated functional equation: f(x+1)=f(x)f(1).f'(x+1) = f'(x)f(1). From the initial condition f(1)=7f(1) = 7, this simplifies to: f(x+1)=7f(x).f'(x+1) = 7f'(x).

Now, consider x=0x = 0: f(0+1)=7f(0).f'(0+1) = 7f'(0). This simplifies to: f(1)=7f(0).f'(1) = 7f'(0).


Step 4: Determine f(0)f'(0)

From the earlier analysis of f(0)=2f(0) = 2 and the functional equation f(x+y)=f(x)f(y)f(x+y) = f(x)f(y), we can differentiate to find f(0)f'(0). Since f(0)=2f(0) = 2, and differentiating a constant gives zero, f'(0) = 0.


Step 5: Final Answer

Substitute f(0)=0f'(0) = 0 into f(1)=7f(0)f'(1) = 7f'(0): f(1)=70=0.f'(1) = 7 \cdot 0 = 0.

Thus:

  • The value of f(1)f'(1) is 0\mathbf{0}.
  • The value of f(0)f(0) is 2\mathbf{2}.

Follow-Up Questions:

  1. What is the general solution to the functional equation f(x+y)=f(x)f(y)f(x+y) = f(x)f(y)?
  2. How would the result change if f(1)f(1) were 11 instead of 77?
  3. Prove that f(x)f(x) must be an exponential function under certain conditions.
  4. How does differentiability influence the behavior of ff in such functional equations?
  5. What happens if f(0)f(0) is assumed to be 00 instead of 22?

Tip:

When dealing with functional equations like f(x+y)=f(x)f(y)f(x+y) = f(x)f(y), exponential functions f(x)=axf(x) = a^x 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)