monotone sqrt(x+2)

解答

单调区间

x + 2

递增 : - 2 < x < \infty

求解步骤

f^{'} (x) = \frac{1}{2 x + 2}

Find intervals:

递增

: - 2 < x < \infty

递增 : - 2 < x < \infty

d o main \frac{x + 1}{∣ x ∣}

e x t re m e f (x) = x^{4} - 4 x^{3} + 10

d o main f (x) = \frac{1}{2 x + 1}

d o main \frac{6}{x - 5}

in v erse g (x) = \frac{5}{2} x + 4