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Context:

Let $f(x)$ be a polynomial of degree $4$ with $2$ inflection points. A line is draw through the inflection points and three regions are made. What inference can you make about these regions?

  1. Two regions have equal area;

  2. Area of one is equal to sum of other two;

  3. area of one is double the sum of other two;

  4. area of one region is square of the sum of the other two;

What I found already:

https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-5/a/behavior-of-antiderivative-of-f-from-graph-of-f

This articles tells that increasing function suggest concave up integral but not sure if that is what I need to solve this problem.

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    $\begingroup$ First of all, I suggest you choose a specific quartic with two points of inflection, and see what the areas of the resulting regions are. Unless you are unlucky, this will tell you which of the four options is the correct one. Then you can try to prove it for a general quartic. $\endgroup$ – TonyK Apr 11 at 11:35
  • $\begingroup$ I tried the function I got was $x^4 -x^3\\$; $f''(x)= 12x^2 - 6x\\$ inflection is at x=.5 and x=0 I think 1. is the correct option now what? $\endgroup$ – Buraian Apr 11 at 11:45
  • $\begingroup$ Your chosen polynomial is highly symmetric. That is a recipe for being "unlucky" in a problem like this. Is the equality of areas actually true in general, or is it an artifact of the symmetry? $\endgroup$ – Paul Sinclair Apr 11 at 21:50
  • $\begingroup$ urghh this is very frustrating... I can't figure this out at all like where to start even... Some more hints would be appreciated $\endgroup$ – Buraian Apr 11 at 22:27
  • $\begingroup$ @PaulSinclair: We can transform the quartic so that it is symmetrical, without changing the areas of the three regions. See my answer. $\endgroup$ – TonyK Apr 12 at 12:28
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Let our quartic be $P(x)=x^4+bx^3+cx^2+dx+e$ (we may suppose after scaling that the coefficient of $x^4$ is $1$). Suppose we add a linear term of the form $ux+v$ to get $Q(x)=x^4+bx^3+cx^2+(d+u)x+(e+v)$. Then the $x$-coordinates of the points of inflection are unchanged (because $\frac{d^2}{dx^2}(ux+v)=0$), and the areas of the three regions are unchanged (because the graph has been transformed by a vertical shear transform).

But now we can choose $u$ and $v$ so that $Q(x)=0$ at the two points of inflection.

The points of inflection of $P$ and $Q$ are at $x$-coordinates $$-\frac{b}{4}\pm\sqrt{9b^2-48c}$$ We now transform the graph by shifting it right, to get the polynomial

$$R(x)=Q(x-b/4)=x^4+Bx^3+Cx^2+Dx+E$$

This transformation preserves the shapes of the three regions, and therefore their areas. And its points of inflection lie on the $x$-axis, symmetrical about the $y$-axis. $R(x)$ can differ from $P(x)$ in all terms except the first, but the areas of the three regions cut off by the line joining the points of inflection are unchanged.

Its points of inflection are at $$-\frac{B}{4}\pm\sqrt{9B^2-48C}$$ But these points are symmetrical about the $y$-axis. Therefore $B=0$, which means that the points of inflection are $(\pm\alpha,0)$, where $\alpha=\sqrt{-48C}$.

And they lie on the curve $R(x)=0$, so $R(\alpha)=R(-\alpha)$, or

$$\alpha^4+C\alpha^2+D\alpha+E=\alpha^4+C\alpha^2-D\alpha+E$$

which gives us $D=0$.

So now we are left with $R(x)=x^4+Cx^2+E$, which is symmetrical about the $y$-axis. Hence the first and third regions have the same area.

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  • $\begingroup$ "But these points are symmetrical about the y-axis. Therefore B=0, which means that the points of inflection are (±α,0), where α=−48C−−−−−√." What does this statement mean? Also if you could , some pictures would be nice, I'm finding it hard to visualize this problem $\endgroup$ – Buraian Apr 12 at 14:41
  • $\begingroup$ @DDD4C4U - The points of inflection are given by $-\frac B4\pm\sqrt{9B^2-48C}$. The only way for these two points to be symmetric about the $y$-axis is for the first term to be $0$. Therefore $B = 0$. And when $B = 0$, the expression reduces to $\pm\sqrt{-48C}$. (The points are symmetric about the $y$-axis because he specifically chose transformations to make them so.) $\endgroup$ – Paul Sinclair Apr 12 at 14:51
  • $\begingroup$ Why is it a must that this expression should be symmetric again? $\endgroup$ – Buraian Apr 12 at 14:52
  • $\begingroup$ @DDD4C4U: Which expression? $\endgroup$ – TonyK Apr 12 at 14:53
  • $\begingroup$ Like why should the roots of second derivative be symmetric ?/ how did you transform it to become symeteric? $\endgroup$ – Buraian Apr 12 at 14:54

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