y=√(25-x^2)+lg(cosx)的定义域
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/09 02:26:05
y=√(25-x^2)+lg(cosx)的定义域
![y=√(25-x^2)+lg(cosx)的定义域](/uploads/image/z/19106430-6-0.jpg?t=y%3D%E2%88%9A%EF%BC%8825-x%5E2%EF%BC%89%2Blg%28cosx%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F)
y=√(25-x²) + Lg(cosx) 的定义域
① (25 - x²) ≥ 0 ===> x² ≤ 25 ===> x ∈ [- 5 ,+5]
② cosx >0 ===> x ∈(-π/2 ,π/2)±2kπ = ((2k-1/2)π ,+(2k+1/2)π)
估算:
k 小数 精确值
k = -1 -7.85 → -4.71 → x ∈[ - 5/2 ,-3π/2]
k = -0 -1.57 → 1.57 → x ∈[ - π/2 ,π/2]
k = +1 4.71 → 7.85 → x ∈[ 3π/2 ,5π/2]
答案:x ∈[ - 5,-3/2π] 或者 [ - π/2,1/2π] 或者 [ 3π/2,5]
① (25 - x²) ≥ 0 ===> x² ≤ 25 ===> x ∈ [- 5 ,+5]
② cosx >0 ===> x ∈(-π/2 ,π/2)±2kπ = ((2k-1/2)π ,+(2k+1/2)π)
估算:
k 小数 精确值
k = -1 -7.85 → -4.71 → x ∈[ - 5/2 ,-3π/2]
k = -0 -1.57 → 1.57 → x ∈[ - π/2 ,π/2]
k = +1 4.71 → 7.85 → x ∈[ 3π/2 ,5π/2]
答案:x ∈[ - 5,-3/2π] 或者 [ - π/2,1/2π] 或者 [ 3π/2,5]