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?
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?
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).
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.