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Is it possible to construct a quartic function with two local maxima and one local minima at any three arbitrary points, making some assumptions? If so, how can I do it?

  • The x-coordinate of the local minima lies between that of the maxima
  • The coordinates of the local minima and maximas are not the same

If I have a quartic function $f(x) = ax^4+bx^3+cx^2+dx+e$, then stationary points of a quartic are located when $f'(x)=0$. However, given I have three points that the function must pass through, we have three system of equations each with five coefficient variables. Then, we have three more linear equations (this time without the $e$ term) by setting $f'(x) = 0$.

However, it doesn't seem like there is a solution to these linear equations that I construct using this approach. Is there another way to do it?

Monolith
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    Hint: $f'(x) = A(x-\alpha)(x-\beta)(x-\gamma)$. What is $f(x)$? – Calvin Lin May 23 '23 at 21:52
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    Note: By requiring the x and y coordinates of all 3 turning points, you might have an over-specified problem. (Like you realized, you have 6 linear equations with 5 unknowns, and so it sometimes doesn't yield a solution.) $\quad$ As a specific example, if you know that the 2 local minima are $(-1, 0) $ and $(1, 0)$, then $f(x) = A ( x-1)^2(x+1)^2$, and we can conclude (EG by differentiation) that the local maxima must occur at $ x = 0$. So we can't have a quartic with a maxima at (say) $ (0.5, 1)$. – Calvin Lin May 23 '23 at 21:55

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