inflection x^3-2x^2-4x

解答

拐点

x^{3} - 2 x^{2} - 4 x

(\frac{2}{3}, - \frac{88}{27})

求解步骤

f^{''} (x) = 6 x - 4

Find intervals:

向下凹

: - \infty < x < \frac{2}{3},

向上凹

: \frac{2}{3} < x < \infty

将

x = \frac{2}{3}

代入

x^{3} - 2 x^{2} - 4 x : - \frac{88}{27}

(\frac{2}{3}, - \frac{88}{27})

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