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It is known that the derivative of y = f (x) defined on R is f '(x), satisfying f' (x)
Constructor g (x) = e ^ (- x) f (x)
From the known f '(x)
It is known that f (x) is a differentiable function defined on (- ∞, + ∞), and f (x) < f '(x) holds for X ∈ R
What is the size relationship between F (2012) and f (2011) e? I know the answer is that the front is greater than the back
But it still needs a process. The most important thing is that if you use the construction method, please tell me what you think carefully. I don't quite understand that. Of course, if it's not the construction method, it's OK!
Let f (x) = e ^ (- x) f (x), then f '(x) = e ^ (- x) [f' (x) - f (x)] > 0, so f (x) increases monotonically, f (2012) > F (2011), that is e ^ (- 2012) f (2012) > e ^ (- 2011) f (2011), so f (2012) > F (2011) e
Given that FX is an even function defined on R, and f (1) = 0, Let f'x be the derivative of the function FX
Let f'x be the derivative of function FX. When x is greater than 0, if XF'x-fx / x ^ 2 is less than 0, then the inequality x ^ 2 (e ^ x + 1) FX is greater than 0
A:
The even functions f (x) defined on R are: F (- x) = f (x)
So: F (- 1) = f (1) = 0
Because: [XF '(x) - f (x)] / x ^ 2
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