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Passage: Let f:[0,1]->R be twice differentiable with f(0)=f(1)=0 and satisfying f''(x) - 2 f'(x) + f(x) >= e^x for all x in [0,1]. Suppose the function e^(-x) f(x) attains its minimum on [0,1] at x = 1/4. Which of the following is true?
- f'(x) < f(x) for 1/4 < x < 3/4
- f'(x) > f(x) for 0 < x < 1/4
- f'(x) < f(x) for 0 < x < 1/4
- f'(x) < f(x) for 3/4 < x < 1
Correct answer: f'(x) < f(x) for 0 < x < 1/4
Solution
With g(x) = e^(-x)f(x), g'(x) = e^(-x)(f'(x)-f(x)); the interior minimum at 1/4 forces g to be decreasing on (0,1/4), so g' < 0 there, i.e. f'(x) < f(x) for 0 < x < 1/4.
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