Determine the stationary points, A and B, and point of inflection, G, for each of the following cubic polynomials.

(a) $y=x^3 -3x^2 -9x+7$

(b) $y=x^3 -12x^2 +21x-14$

(c) $y=x^3 +9x^2 -12$

Is there any common relationship between A, B and G?

  • 1
    To augment the previous comment: Explaining what you know about a problem, and where exactly you got stuck, helps answerers tailor their responses to your skill level, while avoiding wasting time telling you things you already know. (Plus, it helps convince people that you aren't simply trying to get them to do your homework for you.) In this case, do you know what stationary and inflection points are? Can you find them? (If so, tell us what you think they are, even if you're wrong.) Are you only having trouble finding the "common relationship"? – Blue Oct 12 at 6:56

The first derivative of a cubic polynomial is a quadratic polynomial $q$, and as such is even with respect to its stationary point $\xi$. It follows that the real zeros of $q$, if there are any, are at equal distance on both sides of $\xi$. This implies $G={A+B\over2}$.

For any cubic, by translating the independent variable, you can make the quadratic term disappear and obtain a form

$$x^3+px+q.$$

You can also translate vertically and reduce to

$$x^3+px.$$

Then by rescaling the variable and the function,

$$\frac{(sx)^3+psx}{s^3}=x^3+\frac p{s^2}x=x^3\pm \,x$$ by setting $s^2=|p|$. And if $p=0$, we simply have $x^3$.

These transformation are affine and preserve ratios, and in particular the relative positions of the extrema and the inflection.

Without further computation, as the function is odd the single inflection must lie midway of the two extrema.


We can in fact say that, as regards the remarkable points, there are essentially three cubics: $x^3,x^3-x,x^3+x$.

enter image description here

(If the cubic coefficient was negative, the graph must be flipped and one may distinguish six cases, $\pm x^3,\pm x^3\pm x$.)

You are in fact asking if there is a relation between the real roots and the extremum of a quadratic function. By symmetry of the parabola, yes, $G$ is the mid-point of $AB$.

enter image description here

Hint: $\frac{\mathrm dy}{\mathrm dx} = 0$ to find the $x$-values of the stationary points, then sub these back into the original equation to find the y-values. As for the points of inflection, these occur when $\frac{\mathrm d^2y}{\mathrm dx^2} = 0$, similarly find the x-values and sub these back into the original equation for the y-values of the points of inflection.

Eg for a) $$\frac{\mathrm dy}{\mathrm dx} = 3x^2 - 6x - 9 = 0$$ $$x^2-2x-3=0$$ $$(x-3)(x+1)=0 \implies x=-1, x=3$$

$$y(-1) = 12$$ $$y(3) = -20$$ $$\frac{\mathrm d^2y}{\mathrm dx^2} = 6x-6 = 0 \implies x = 1$$ $$y(1) = -4$$

So the two stationary points are $(-1,12)$ and $(3,-20)$ and the point of inflection is $(1,-4)$

  • 1
    I am afraid that this does not answer the question. – Yves Daoust Oct 12 at 8:22
  • Ah indeed haha, I didn't notice he asked about the relationship between the different points. Thanks for pointing it out – Patrick Jankowski Oct 12 at 10:35

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.