Question

求函数的导数的题

设函数f(x)=sinx-cosx+x+1 , 0<x<2派 , 求函数f(x)的单调区间与极值

麻烦给个详细的步骤, , 谢谢啊, 好了追加50分 在线等

Answer 1

f(x)=sinx-cosx+x+1
f '(x) = cosx+sinx+1 = √2 sin(x+π/4) + 1
f '(x)=0 => sin(x+π/4) = -√2 /2 => x+π/4= 2kπ-π/4 或 x+π/4 = 2kπ-3π/4
驻点 x = 2kπ-π/2 或 x = 2kπ-π
由f '(x)<0 => 2kπ-3π/4 < x+π/4 < 2kπ-π/4, f(x)的单减区间 [2kπ-π < x-π/2]
由f '(x)>0 => 2kπ-π/4 < x+π/4 < 2kπ+5π/4, f(x)的单增区间 [2kπ-π/2 < x+π]
f(x)的极大值 f (kπ) =2+kπ,
f(x)的极小值 f (2kπ-π/2) =2kπ-π/2
对于 0<x<2π, f(x)的单减区间 [π, 3π/2],
f(x)的单增区间 (0, π ) 及 (3π/2, 2π),
f(x)的极大值 f (π) =2+π,
f(x)的极小值 f (3π/2) = 3π/2

Answer 2

f'(x)=cosx+sinx+1=

Answer 3

f'(x)=cosx+sinx+1=)>-√2sin(x+π/4)+1
f'(x)>0时，sin(x+π/4)>-√2/2,又π/4<x+π/4<9π/4,结合该区间上正弦函数的图像可知，不等式的解为0<x<π或3π/2<x<2π，所以函数的单调增区间为(0,π),(3π/2,2π),单调减区间为(π,3π/2)
所以函数的极小值为f(3π/2)=3π/2,极大值为f(π)=2+π.