Limits, Continuity & Differentiability
Differentiability and Functional Equations
Grade 12
Question:
<p><strong>Ex. 9</strong> A function $f: \mathbb{R} \to \mathbb{R}$ satisfies the equation $f(x+y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$, $f(x) \geq 0$. Suppose that the function is differentiable at $x = 0$ and $f'(0) = 2$. Then,</p>
<p>(a) $f'(x) = 2f(x)$</p>
<p>(b) $f'(x) = f(x)$</p>
<p>(c) $f'(x) = f(x) + 2$</p>
<p>(d) $f'(x) = 2f(x) + x$</p>
Step-by-Step Solution
Key Concept: Use the functional equation property and the derivative definition to relate $f'(x)$ to $f'(0)$. The key is recognizing that $f(x+h) = f(x) + f(h)$ allows us to factor out $f(x)$.
<p><strong>Solution:</strong> We are given that $f(x+y) = f(x) + f(y)$ ... (i)</p><p>$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = 2$ ... (ii)</p><p>Now, $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$</p><p>$= \lim_{h \to 0} \frac{f(x) + f(h) - f(x) - f(0)}{h}$ [using Eq. (i)]</p><p>$= f(x) \lim_{h \to 0} \frac{f(h) - f(0)}{h}$</p><p>$= f(x) \cdot 2f(x)$ [using Eq. (ii)]</p><p>$\therefore f'(x) = 2f(x)$</p>
Correct Answer: A