inflection x^3-2x^2-4x+6

解答

拐点

x^{3} - 2 x^{2} - 4 x + 6

(\frac{2}{3}, \frac{74}{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 + 6 : \frac{74}{27}

(\frac{2}{3}, \frac{74}{27})

a sy m pt o t es f (x) = \frac{x ^{2} - 11 x - 24}{x ^{2} - 9}

r an g e f (x) = 5 x^{3} + 6 x^{2} - 1

in v erse f (x) = \frac{5 x - 1}{2 x}

d o main f (x) = x^{3} - 3 x^{2} + 2

r an g e f (x) = \frac{x ^{2} - 5}{x ^{2} - 9}