Does there exists a positive function (i.e $f(x) > 0$) such that for all $x > 0$ it is less than it's derivative (ie $f'(x)$)? Intuition says this would not be possible, but is there a more formal proof of this?
A positive function that is less than its derivative for all x?
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calculus
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6not sure what you mean? How about $\exp(2x)$? – 2017-01-30
2 Answers
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One thought might be to naively start drawing a picture. As long as the picture is hyper-increasing, one might intuitively recognize that this is possible.
You could also think to find a solution to a differential equation. For instance, is there a solution to $f'(x) = 2 f(x)$ that is always positive? If so, then it satisfies your condition by design (the derivative is always twice the value of the function). And yes, there is a solution: $f(x) = e^{2x}$.
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Such a function does exist. Take $f(x) = \exp(2x) > 0$. By the chain rule,
\begin{align*} f'(x) &= \frac{\mathrm{d}}{\mathrm{d}x} \exp 2x \\&= \exp'(2x) \cdot \left(\frac{\mathrm{d}}{\mathrm{d}x} 2x \right) \\&= \exp(2x) \cdot 2 \\&> f(x). \end{align*}