inflection x^3-3x^2-9x+10

解答

拐点

x^{3} - 3 x^{2} - 9 x + 10

(1, - 1)

求解步骤

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

Find intervals:

向下凹

: - \infty < x < 1,

向上凹

: 1 < x < \infty

将

x = 1

代入

x^{3} - 3 x^{2} - 9 x + 10 : - 1

(1, - 1)

(1, - 19), (8, - 14)

x = 8

arctan (x + 2)

f (x) = - 2 x^{5} - x^{4} - 9

\frac{x - 5}{3 x - 1}