At what points do the functions #y = x^2-x# and #y = sin pix# intersect?

1 Answer
Jan 24, 2017

Answer:

These two equations intersect at the points #(0, 0)# and #(1, 0)#

Explanation:

First, let us take a look at:

#y = x^2-x#

We can factor this as:

#y = x(x-1)#

so this quadratic has #x# intercepts at #(0, 0)# and #(1, 0)#

It has minimum value at the midpoint of these two #x# coordinates, where #x=1/2#:

#y = color(blue)(1/2)(color(blue)(1/2)-1) = -1/4#

So note that #y=x^2-x < 0# for all #x in (0, 1)#

The intersections of #y = x^2-x# with the horizontal line #y=1# are at the points given by solving:

#0 = x^2-x-1#

#color(white)(0) = x^2-x+1/4-5/4#

#color(white)(0) = (x-1/2)^2-(sqrt(5)/2)^2#

#color(white)(0) = (x-1/2-sqrt(5)/2)(x-1/2+sqrt(5)/2)#

That is:

#x = 1/2+-sqrt(5)/2#

Note that #1/2+sqrt(5)/2 ~~ 1.618 in (1, 2)#

Note that #1/2-sqrt(5)/2 ~~ -0.618 in (-1, 0)#

So #x^2-x in (0, 1]# when #x in [1/2-sqrt(5)/2, 0)# or #x in (1, 1/2+sqrt(5)/2]#

Outside these intervals, #x^2-x > 1# so cannot be equal to #sin(pix)#

Now consider #y = sin(pix)# in each of these intervals:

  • If #x in [1/2-sqrt(2), 0)# then #sin(pix) < 0#

  • If #x=0# then #sin(pix) = 0 = x^2-x#

  • If #x in (0, 1)# then #sin(pix) > 0#

  • If #x = 1# then #sin(pix) = 0 = x^2-x#

  • If #x in (1, 1/2+sqrt(5)/2)# then #sin(pix) < 0#

So in each of the intervals #[1/2 - sqrt(5)/2, 0)#, #(0, 1)# and #(1, 1/2+sqrt(5)/2]# the two functions have opposite signs and cannot intersect.

So the only two points of intersection are:

#(x, y) = (0, 0)#

#(x, y) = (1, 0)#

graph{(y-x^2+x)(y - sin(pix)) = 0 [-2.105, 2.895, -1.19, 1.31]}