SOLUTION: Find the critical numbers/points of the function: g(x) = x-5/(x^2)
I know that I first have to take the derivative and then set that to zero to find it's zeros. But I am getting
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I know that I first have to take the derivative and then set that to zero to find it's zeros. But I am getting
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Question 995459: Find the critical numbers/points of the function: g(x) = x-5/(x^2)
I know that I first have to take the derivative and then set that to zero to find it's zeros. But I am getting tripped up somewhere.
g'(x) = 1+10/(x^3)
*Side question: When we take the derivative and have another number like here 1 do we get like denominators for it?
anywho, after 1+10/(x^3)=0 I am unsure how to go about solving for the zeros.
Please explain
Thank you Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! g'(x) = 1+10/(x^3)
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1 + 10/x^3 = 0
10/x^3 = -1
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x^3 = -10
x = cbrt(-10) = -2.1544
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Cheers,
Stan H.
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