calculate the coordinates of the critical path points of the curve y = x^3 - 6x^2 and sketch the graph . anyone can help me?

Question

anonymous · Answer

critical points -> points where the "derivative" becomes 0
so, first differentiate the function.

cucacula · Answer

can you show me the 1st step? @electrokid

anonymous · Answer

$$y=x^3-6x^2\
{dy\over dx}=y'={d\over dx}(x^3-6x^2)=?$$

cucacula · Answer

isit d/dx ( 3x^2 - 12x ) @electrokid

anonymous · Answer

good.. without the d/dx part :) since you already evaluated it,

anonymous · Answer

now, set it equal to zero and solve for ":x"

cucacula · Answer

can give me the hint? @electrokid

anonymous · Answer

$$3x^2-12x=0$$
solve for "x"

cucacula · Answer

3x^2 = 12x
x^2 = 12x/3
x^2 = 4x
x     = 4 

isit true? @electrokid

anonymous · Answer

no.. you missed one possible solution...
$$3x^2-12x=0\
3x(x-4)=0\\boxed{x=0}\qquad{\rm OR}\qquad\boxed{x=4}$$

cucacula · Answer

so the x = 0 or x = 4 ? @electrokid can u sketch the graph?

anonymous · Answer

those are the critical points.
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