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Prove that the line $y=2x$ intersects the cubic curve $y = x^3 - x + 1$ in at least three different points

This is a homework question and I don't know where to begin, how would I go about proving this?

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    So, you want to find if the cubic $x^3-3x+1$ intersects the horizontal axis at least thrice?2011-12-05
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    In any event: [since $4(-3)^3+27(1)^2<0$](https://en.wikipedia.org/wiki/Discriminant#Cubic), you certainly do have three roots...2011-12-05
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    It should be noted that there cannot be _more_ than three intersection points, because a third-degree algebraic equation cannot have more than three solutions. So "at least three" implies _exactly_ three.2011-12-06

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Solve the equation $$\tag{1}(x^3-x+1)-2x=0$$ The intersection points of the two curves correspond to the solutions of this equation.

Equation (1) is equivalent to the equation $x^3-3x+1=0$.

Now, if you set $f(x)=x^3-3x+1$, then $f$ is continuous and:

$$ \eqalign{ f(-10)&<0\cr f(0)&>0\cr f(1)&<0\cr f(10)&>0\cr}$$

So, by the Intermediate Value Theorem, the equation $f(x)=0$, and hence (1), has solutions in each of the intervals $(-10,0)$, $(0,1)$, and $(1,10)$ (note the strict inequalities above).

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    I am thinking why do you thought of checking the points $-10,0,1$?2011-12-05
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    @Max They were good guesses :) Actually, a "big" $x>0$ would give a positive value and a "big" $x<0$ would give a negative value. $0$ and $1$ are easy to compute...2011-12-05
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    Indeed,but I probably would go for $\pm2$ instead of $\pm 10$ and $0,1$ are obvious :)2011-12-05
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Facing such cases, first thing comes to my mind is: $$ y = y $$ $$ 2x=x^3-x+1$$ $$ \tag{1}x^3 -3x+1=0$$ The roots of equation (1) is equivalent to the number of points the line crosses the cubic curve, it's a monic cubic polynomial without quadratic term ($x^3+px+q$ ), which has discriminant:

$$ \Delta = -4p^3 -27q^2$$

$\Delta > 0 $ and According to nature of the roots, the equation has 3 distinct real roots.Hence, the line intersects the cubic curve in at least three different points.