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Find the equation for the horizontal tangents to the curve y=x^3-3x-2
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take the derivative, set it equal to zero, and solve for \(x\)
clear or no?
i am going to take that as a "no" the derivative is \(3x^2-3\) et it equal to zero and solve for \(x\) i.e. solve \[3x^2-3=0\] or even \(x^2-1=0\)
sorry I was working on the other parts of this problem. How did you find the derivative?
Ok forget it, I figured it out. Thanks.
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So the tangents are x=1 and x= -1?
No it's just 1.
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