Given function $f(x)$ and its derivative $f'(x)$. If $f(x)$ is convex in $(a,b)$ and $f(x)>0$ for all $a Find $x$ within $(a,b)$ which minimize $$\frac{f(x)}{f'(x)}$$
Minimize the ratio of a convex function and its derivative
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calculus
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0See if you can justify that $x=a$ is the minimum (provided that $f'(a)\neq0$). – 2017-01-02
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0Do you want the interval to be closed? For instance take $f(x)=x^2$ on some positive open interval and this ratio has no minimum on that interval. – 2017-01-02
1 Answers
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Hint: Differentiate $f/f'$ and equate to zero, do you have any roots? Try given convexity and boundary conditions.