Determine the values of x so that the tangent to the functiony=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]}