The slope of the tangent line will be given by inserting a point #x= a# into the derivative. Hence, it makes sense to start by finding the derivative of each function.
Let #f(x) = x^3 - 3x + 4# and #g(x) = 3x^2 - 3x#.
#f'(x) = 3x^2 - 3 # and #g'(x) = 6x - 3#
We are looking for the points of intersection, where the same input of an #x# value equals the same slope when inserted inside the derivative. So, let's solve the system of equations written below:
#3x^2 -3 = y#
#y = 6x - 3#
#3x^2 - 3 = 6x - 3#
#3x^2 - 6x = 0#
#3x(x - 2) = 0#
#x = 0 and 2#
So, the functions will share tangent lines at the points #x = 0# and #x= 2#. Now, we determine the corresponding #y#-coordinate that each tangent passes through.
For #y = x^3 - 3x + 4#:
#y = 0^3 - 3(0) + 4 = 4#
AND
#y = 2^3 - 3(2) + 4 = 8 - 6 + 4 = 6#
The points are #(0, 4)# and #(2, 6)#.
For #y = 3x^2 - 3x#
#y = 3(0)^2 - 3(0) = 0#
AND
#y = 3(2)^2 - 3(2) = 12 - 6 = 6#
The points are #(0, 0)# and #(2, 6)#.
The point in common for the two functions would be #(2, 6)#.
Hopefully this helps!
