# Finding existence of zeros of cubic by multiplying y values?

1. Jul 31, 2016

### Celestion

1. The problem statement, all variables and given/known data

I can do the question, but in a different way to the worked solution which I don't understand. So my question is can anyone explain the worked solution which is in point 3 below.

The question was to show there is exactly one zero to the function f(x) = Ax^3 - Ax + 1, with A>0, when A < 3sqrt(3) / 2.

2. Relevant equations

The part above this asks to find the stationary points, and there are two of them, at + and - 1/sqrt(3).

3. The attempt at a solution

My reasoning was that since A>0, which is the coefficient of x^3 (i.e. the highest degree term), the graph goes up into the top right of the first quadrant as x and y approach positive infinity. And therefore to have only one root, the y value of the stationary point at x = 1/sqrt(3) must be >0. (I also considered the case where the stationary point at 1/sqrt(3) was <0 and the one at -1/sqrt(3) was also <0, but this is impossible given the constant term of +1 and from playing with graphing software, and having A>0). Solving this, i.e. y = f(x) being >0 at x=1/sqrt(3) gives the correct answer of A < 3sqrt(3) / 2 from point 1 above.

HOWEVER, the worked solution calculates the y values at both + and - 1/sqrt(3), and then multiplies the y values together, giving a quadratic in A (i.e. containing A^2), and then uses the condition that the two y values multiplied together must be >0 for there to be only one zero. Which I don't understand at all...

i.e. f(1/sqrt(3)) * f(-1/sqrt(3)) = (1 - 2A/3sqrt(3))(1 + 2A/3sqrt(3)) = 1 - 4A^2 / 27

and when this is >0, i.e. when A < 3sqrt(3)/2, there will be only one zero of f(x).

Last edited: Jul 31, 2016
2. Jul 31, 2016

### andrewkirk

Draw a graph, without scale or axes, of a cubic of this sort. It'll come from $-\infty$ at the left, rising but curving downwards, then reach a maximum at $x=x_1=-\frac1{\sqrt 3}, y=y_1$, curve down, go through an inflexion point at $x=0$, start curving up, reach a minimum at $x=x_2=\frac1{\sqrt 3},y=y_2$, then keep curving up towards $+\infty$.

Now imagine that you can draw the x axis anywhere on the page. The number of roots is the number of times the axis cuts the curve.

If the x axis is tangent to the curve at the local maximum point $(x_1,y_1)$, there are two roots. If it is higher than that, there is only one. If it is tangent to the curve at the local min point $(x_2,y_2)$ there are two roots. If it is lower than that, there is only one. If it is between the heights of the local minimum and the local maximum it cuts the curve three times.

So there is one root iff the x axis cuts the curve below $y_2$ or above $y_1$, which is the case iff the signs of $y_1$ and $y_2$ are the same, which is the case iff $y_1\times y_2>0$.

I think your solution is just as good though.

3. Aug 1, 2016

### Celestion

Thanks Andrew, that's great, it makes perfect sense. I think their/your solution is more elegant than mine :) now that I understand it