Question

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`?

Answer 1

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]}