How do you find out the coordinates of the points where these two curves (2x-2 & x³-x) meet?

Answer 1

you need to solve the equation:
2x-2 = x^3 -x
x^3 -3x +2 =0
x=1 is a solution, since 1 -3 + 2 =0
so x^3 -3x +2 = ( x-1) (x^2 +x -2) = (x-1) (x +2) ( x-1)
so the curves meet when x= 1 and when x =-2

Answer 2

By solving the system of equations:

y=2x - 2
and y=x^3 -x

To find the x coordinates just plug y=2x - 2
into y=x^3 - x

2x - 2 = x^3 - x

2(x - 1) = x(x^2 - 1)

2(x - 1) = x(x + 1)(x - 1)

You can see that x=1 satisfies the equation.

Let's divide both sides by (x - 1) and find two more roots

2 = x(x + 1)

2 = x^2 + x

x^2 + x - 2 = 0

Completing the square:
x^2 + 2*0.5x + 0.5^2 - 0.5^2 - 2 = 0

(x + 0.5)^2 - 0.25 - 2 = 0

(x + 0.5)^2 - 2.25=0

(x + 0.5)^2 - 1.5^2 = 0

(x + 0.5 - 1.5)(x + 0.5 + 1.5) = 0

(x - 1)(x + 2) = 0

The other roots are
x=1 (double)
and x=-2

Lets find relevant y coordinates:
x=1 -> y= 2x-2 = 2*1-2 = 0

x=-2 -> y=2x-2 = 2*(-2)-2 = -4 - 2 = -6

The points are (1,0) and (-2, -6)

Answer 3

♦ y1=2x-2; y2=x^3-x;
♠ y1=y2; 2x-2=x^3-x; 2(x-1) =x(x-1)(x+1), hence x1=1 and 2=x(x+1);
or x^2 +x -2=0, hence x=(-1±√(1^2 +4*1*2))/2 =(-1±3)/2 =-2, +1;
thus x1=x2=1, x3=-2; this means they touch at x=1 and cross at x=-2;
♥ why touch?
y1’(x)=2, y2’(x)=3x^2-1;
y1’(1) = 2 = y2’(1);