find all points on the curve y=x^3-3X where the tangent is parallel to the x-axis.

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find all points on the curve y=x^3-3X where the tangent is parallel to the x-axis.

Mathematics
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this should be intuitive because tangent parallel to the x-axis implies that its value is 0. And then you find the prime of y, and set it to 0
tangent is parallel to the x-axis so its slope equals to 0. on the other hand slope of tangent line|dw:1349077584783:dw| equals to derivative of function at the given point
|dw:1349077683740:dw|

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find the first derivative... = 3x^2 -3=0 so at x= +1 , -1 the slope of tanget is zero..
the points r (1,-2),(-1,2)
http://www.wolframalpha.com/input/?i=plot&a=*C.plot-_*Calculator.dflt-&a=FSelect_**Plot-.LogPlotCalculator-&f3=x^3-3x&x=5&y=9&f=Plot.plotfunction_x^3-3x&a=*FVarOpt.1-_**-.***Plot.plotvariable-.*Plot.plotlowerrange-.*Plot.plotupperrange--.**Plot.ymin-.*Plot.ymax---.*--
hmm its correct and mine answer too ..
@WilsonWorla Hi, \(\huge \color{red}{\text{Welcome to Open Study}}\ddot\smile\) Are you following the explanation, if doubts feel free to ask them.
See attached screen shot of my answer, done in Geogegebra

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