Finding Tangent Points on a Graph Using the Chain Rule

[Saterial]

1) The problem.

Determine the coordinates of the points on the graph of f(x) = (x^3 - 3x^2)^2 at which the graph is tangent to the x-axis.

2) Relevant Equations
Chain Rule of Derivatives ?

3) My attempt.
I used chain rule to get f'(x) = (2)(x^3 - 3x^2)(3x^2 - 6x
I don't know where to go from here what does it mean by the graph is tangent to the x-axis?

 

[JonF]

Tangent in this context means touches it at one point but doesn't cross. So if your curve touches the tangent line at one point, but doesn't cross, what type of point would that need to be?

 

[Saterial]

So I would need a vertex of the curve at that point?

In that case would it mean that I need to set the derivative equal to zero and solve for x? Take that x and plug it into the original equation to find y?

 

[JonF]

yup, first find the critical points, second plug them in and see if y=0.

 

[arkajad]

Tangent to the x-axis means derivative is zero and the function is zero (touchs the x axis). First solve for derivative, when it is zero, then check if among those there is an x at which also f(x)=0.

 

[JonF]

Oh I almost forgot! Make sure they are max/min not inflection points too.