Find the critical numbers.
1. g(x) = x^3 - 3x
So far I have g'(x) = 3x^2 - 3 = 0
My question is do I now take the three out so it goes 3(x^2) or is it 3(x^2 -1) ?
1. g(x) = x^3 - 3x
So far I have g'(x) = 3x^2 - 3 = 0
My question is do I now take the three out so it goes 3(x^2) or is it 3(x^2 -1) ?
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3x^2 - 3 = 3(x^2 - 1)
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When you pull out the three, it must be pulled out of every term, so the second one would be a correct derivative. But leaving the three in would still be correct as well. Which would make x=0 at 1 and -1