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Let $f:\mathbf R \to \mathbf R$ be a differentiable function. Let another function $g:\mathbf R \to \mathbf R$ be defined as $$g(x)=f\bigl(x-x^2\bigr)$$

It is given that $g(x)$ has a local maximum at $x=1/2$ but the absolute maxima exist elsewhere. What is the minimum number of solutions in $x$ that the equation, $g'(x)=0$, must have?

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  • $\begingroup$ @AlexProvost $g'(x)=(1-2x)f'(x-x^2)$. Let $x=a$ be an absolute maximum. Then $g'(a)=0 \implies f'(a-a^2)=0$. But since $x-x^2$ is a quadratic, there exists $b$ such that $b-b^2=a-a^2$. So, $f'(b-b^2)=0$ and there exists another extremum at $b$ so that $g'(b)=0$. This gives us three minimum solutions. I am interested to know if there are more. $\endgroup$ – Karan Karan May 11 '17 at 18:30
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There must be at least $ 5 $, and this is easy to see if you picture the graph of $ g $. The graph of $ g $ is symmetric across the line $ x = 1/2 $ and has a local maximum at $ x = 1/2 $. We are given that there is a global maximum elsewhere, so there must actually be at least two global maxima, since if a global maximum occurs at $ x = a $, another one must occur at $ x = 1 - a $ by symmetry. Moreover, since $ x = 1/2 $ is a local maximum, it follows that between $ x = 1/2 $ and the two global maxima, there must be two local minima as well. Therefore, this gives us at least $ 5 $ points on which the derivative of $ g $ vanishes. (You can make this argument precise using continuity and Rolle's theorem.)

To see that this bound is tight, note that $ f(x) = (x+1)(x+1/2)(x+2/3) $ satisfies the desired conditions, and $ g'(x) $ is a polynomial of degree five, so it can have no more than five roots; and the five roots come from the two local minima and the three local maxima of $ f(x - x^2) = g(x) $.

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