Question:medium

If \(f(x+y)=f(x)f(y),\ \forall x,y\), suppose \(f(3)=3\) and \(f'(0)=11\), then \(f'(3)=\)

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For \(f(x+y)=f(x)f(y)\), differentiating with respect to \(y\) and putting \(y=0\) gives \(f'(x)=f(x)f'(0)\).
  • \(22\)
  • \(33\)
  • \(28\)
  • \(9\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The given functional equation \(f(x+y)=f(x)f(y)\) is the property of exponential functions. Functions that satisfy this are of the form \(f(x)=a^x\). We are asked to find the value of the derivative at a point, given the value of the function at that point and the value of the derivative at the origin.

Step 2: Key Formula or Approach:

1. Use the definition of the derivative for \(f'(x)\): \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] 2. Apply the given functional equation \(f(x+h) = f(x)f(h)\) to simplify the expression for \(f'(x)\). 3. Use the given value of \(f'(0)\) to relate \(f'(x)\) to \(f(x)\). 4. Substitute \(x=3\) to find the final answer.

Step 3: Detailed Explanation:

Let's find the general expression for \(f'(x)\). \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Using the functional equation, \(f(x+h) = f(x)f(h)\): \[ f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h} \] Since \(f(x)\) does not depend on \(h\), we can take it out of the limit: \[ f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h} \] Now, let's look at the limit term. It looks like the definition of a derivative. Let's see what \(f'(0)\) is: \[ f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \] From the original equation, \(f(x+y)=f(x)f(y)\), let \(x=x\) and \(y=0\). Then \(f(x) = f(x)f(0)\). Since this must hold for a non-trivial function \(f(x)\), we must have \(f(0)=1\). (The question states \(f'(0)=11\), implying differentiability, which implies continuity. If \(f\) is continuous and not identically zero, then \(f(0)=1\).) So, the derivative at 0 is: \[ f'(0) = \lim_{h \to 0} \frac{f(h) - 1}{h} \] This is exactly the limit term in our expression for \(f'(x)\). Therefore, we have the relationship: \[ f'(x) = f(x) \cdot f'(0) \] We are given \(f(3)=3\) and \(f'(0)=11\). We want to find \(f'(3)\). Using the relationship we just derived: \[ f'(3) = f(3) \cdot f'(0) \] \[ f'(3) = 3 \cdot 11 = 33 \]

Step 4: Final Answer:

The value of \(f'(3)\) is 33, which corresponds to option (B).
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