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Determine the points in which the given function has a horizontal tangent line.

y =
x^(3) + x

1 Answer

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\bf y=x^3+x\implies \cfrac{dy}{dx}=3x^2+1 \\\\ \textit{now setting the derivative to 0} \\\\ 0=3x^2+1\implies \cfrac{-1}{3}=x^2\implies \sqrt{\cfrac{-1}{3}}=x \\\\\\ \cfrac{√(3)}{3}\ i=x

meaning, it doesn't have one :)
User Farnabaz
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