If f is differentiable, then what is f'(0.5) equal to?
Consider the following for the next two (02) items that follow: Let a function f be defined on \mathbb{R}-\{0\} and 2f(x)+f\left(\frac{1}{x}\right)=x+3.
- A. \frac{1}{4}
- B. \frac{2}{3}
- C. 2 ✓
- D. 4
Correct Answer: C. 2
Explanation
Replace x with \frac{1}{x} in the original equation to get 2f(\frac{1}{x}) + f(x) = \frac{1}{x} + 3. Multiply the original equation by 2: 4f(x) + 2f(\frac{1}{x}) = 2x + 6. Subtracting the substituted equation from this yields: 3f(x) = 2x - \frac{1}{x} + 3, so f(x) = \frac{2}{3}x - \frac{1}{3x} + 1. Differentiating gives f'(x) = \frac{2}{3} + \frac{1}{3x^2}. At x=0.5=\frac{1}{2}, f'(0.5) = \frac{2}{3} + \frac{1}{3(1/4)} = \frac{2}{3} + \frac{4}{3} = 2.
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