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I had a question that goes like this:

Let $m$ be the number of local minima and $M$ be the number of local maxima. Can you create a function where $M > m + 2$ ? Graph.

I tried graphing it using piecewise function and I improvised it by doing 3 parabolas opening downward and 1 parabola opening upward with each horizontal line separating them.

I was just wondering if there is an easier way in creating a graph given such condition?

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Such a function is not possible if you want a continuous function, because between 2 local minumums has to be a local maximum, and vice versa. So local minumums and maximums always alternate (if you exclude cases where $f(x)$ is constant on an interval).

If you allow discontinuous functions, you can use

$$f(x) = \lfloor x\rfloor -x$$

It has a local maximum at each integer ($f(x) = 0, \forall x \in \mathbb Z$), but no local minimum at all: for a given $x_0 \in \mathbb R$ with $\lfloor x_0\rfloor = k$, choose any $x \in (x_0, k+1): f(x_0) = k - x_0 > k - x = f(x)$, so $x_0$ can't be a local minimum.

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  • $\begingroup$ it makes perfect sense because a function does not need to be continuous to have local minima and maxima $\endgroup$ – Codex Oct 19 '18 at 14:56
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Because we need a local maxima and minima, we can choose a third degree polynomial because its derivative has two roots So can't we create a function that has two roots then take its antiderivative? Let y and z be roots of the function. $z < M-2$. So lets get $z=M-3. f(x)=(x-M)*(x-(M-3)). f(x)=x²+3x-xM-xM-3M+M² =x²+3x-2xM-3M+M².$ antiderivative$(x²+3x-2xM-3M+M²)=(x³/3)+(3x²/2)-Mx²-3Mx+M²x + C$. I need help! First time using this, don't know how to set up equations.

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  • $\begingroup$ it cant be only a third degree polynomial since you need larger maxima than minima so the least amount that you can do is 3 maxima and no minima. $\endgroup$ – Codex Oct 19 '18 at 4:42
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    $\begingroup$ I think you misunderstand the question. $m$ and $M$ are the number of maxima/minima, not the $x$-values on which they are taken. $\endgroup$ – Ingix Oct 19 '18 at 7:21
  • $\begingroup$ Oh, I completely misunderstood the question. M is the number of maximas, not the maxima. Sorry $\endgroup$ – Matheus Andrade Barreto Oct 19 '18 at 14:35

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