Determine the values of x so that the tangent to the function#y=3(root(3)x)# is parallel to the line #x+16y+3=0#?

1 Answer
Feb 24, 2018

No solutions

Explanation:

Start with our line: #x+16y=3#

#16y=-x+3#
#y=-1/16x+3#

So the gradient of the tangent to the curve is #-1/16#

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

#dy/dx=x^(-2/3)#

Let #dy/dx=-1/16#

#x^(-2/3)=-1/16#

#1/x^(2/3)=-1/16#

#x^(2/3)=-16#

#x=(-16)^(3/2)#

#x=(sqrt(-16))^3#

This is imaginary, so there are no solutions

graph{(y-3x^(1/3))(x+16y+3)=0 [-22.18, 23.43, -11.6, 11.2]}