How do you find the points of intersection between #x=y^(2/3)-1# and #y=x^(1/3)+1#?

1 Answer
May 9, 2018

Please see below.

Explanation:

.

#x=y^(2/3)-1#

#y=x^(1/3)+1#

#y^(2/3)=x+1#

#y^2=(x+1)^3#

#y^2=(root3x+1)^2#

#(x+1)^3=(root3x+1)^2#

Let #u=root3x#

#(u^3+1)^3=(u+1)^2#

#u^9+3u^6+3u^3+1=u^2+2u+1#

#u^9+3u^6+3u^3-u^2-2u=0#

#u(u^8+3u^5+3u^2-u-2)=0#

#u=0, :. root3x=0, :. x=0, y=1#

#(0,1)# is one intersection point.

#u^8+3u^5+3u^2-u-2=0#

This equation can be solved by a using rational roots theorem and Newton-Raphson method; and will yield the following results:

#u=-1, 0.77541, -0.95795#

Setting each one equal to #root3x# yields the three #x# values:

#x=-1, 0.46622, and -0.87907#

Therefore, we have four points of intersection:

#(0,1)#, #(-1, 0)#, #(0.46622, 1.77541)#, and #(-0.87907, 0.04205)#

The graph below shows these points:

enter image source here

#x=-1# and #x=-0.87907# are very close together and do not show up clearly on the graph. The following is the close up of the area that shows them better:

enter image source here