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Let f: R -> R be differentiable with f'(x) > 2 f(x) for every real x and f(0) = 1. Which conclusion holds?
- f(x) > e^(2x) in (0, infinity)
- f(x) is decreasing in (0, infinity)
- f(x) is increasing in (0, infinity)
- f'(x) < e^(2x) in (0, infinity)
Correct answer: f(x) > e^(2x) in (0, infinity)
Solution
With g(x) = f(x) e^(-2x), g'(x) = e^(-2x)(f'(x) - 2f(x)) > 0, so g is increasing; since g(0) = 1, g(x) > 1 for x > 0, i.e. f(x) > e^(2x).
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