周期性 y=3cot(1/2 x)-2

解答

周期性

y = 3 cot (\frac{1}{2} x) - 2

2 π

十进制

6.28318 \dots

求解步骤

周期

co t (b x + c) + d = \frac{cot ( x ) 的周期}{∣ b ∣}

cot (x)

的周期是

π

= \frac{π}{\frac{1}{2}}

化简

= 2 π

d o main \frac{1}{1 + 5 ( \frac{1 - x}{5 x} )}

r an g e X^{3}

e x t re m e y = \frac{x}{x ^{2} + 1}

r an g e f (x) = 1 + (8 + x)^{\frac{1}{2}}

e x t re m e f (x) = 3 x + \frac{9}{x}