# If $g(x)=f(x-x^2)$ has local maximum at $x=1/2$ but the absolute maxima exist elsewhere, find the minimum number of solutions of $g'(x)=0$.

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?

• @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. – Karan Karan May 11 '17 at 18:30

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 $f(x - x^2) = g(x)$.