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find the points where the tangent line to the function is horizontal y=-x^3+x^2-2

 Jan 5, 2015

Best Answer 

 #1
avatar+102355 
+5

y=-x^3+x^2-2.....the tangent line will be horizontal wherever the derivative = 0

So, the derivative is given by

y ' = -3x^2 + 2x

Set this to 0...so we have.....

-3x^2 + 2x = 0    factor

x(-3x+ 2 ) = 0     set each factor to  0

x = 0   and   x = 2/3

Here's the graph showing the points of interest......https://www.desmos.com/calculator/aewacr9y2d

 

 

 Jan 5, 2015
 #1
avatar+102355 
+5
Best Answer

y=-x^3+x^2-2.....the tangent line will be horizontal wherever the derivative = 0

So, the derivative is given by

y ' = -3x^2 + 2x

Set this to 0...so we have.....

-3x^2 + 2x = 0    factor

x(-3x+ 2 ) = 0     set each factor to  0

x = 0   and   x = 2/3

Here's the graph showing the points of interest......https://www.desmos.com/calculator/aewacr9y2d

 

 

CPhill Jan 5, 2015
 #2
avatar+102355 
0

Thanks to Alan for pointing out that I inadvertantly omitted the leading "-" sign in my previous answer.......I have edited the answer.....(I hope I finally got it correct...!!!)

 

 Jan 5, 2015

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