y=x+√(5-x²)
5-x²≥0,定义域x∈【-√5,√5】
令x=√5sint,其中t∈【-π/2,π/2】,易知cost≥0
将x=√5sint代入原式得:
y=√5sint+√5cost = √10(sintcosπ/4+costsinπ/4) = √10sin(t+π/4)
t∈【-π/2,π/2】
t+π/4∈【-π/4,3π/4】
t=-π/4时,取最小值ymin=√10*(-√2/2) = -√5
t=π/2时,取最大值ymax=√10*1 = √10
值域【-√5,√10】
y=x+√(5-x²)
5-x²≥0,定义域x∈【-√5,√5】
令x=√5sint,其中t∈【-π/2,π/2】,易知cost≥0
将x=√5sint代入原式得:
y=√5sint+√5cost = √10(sintcosπ/4+costsinπ/4) = √10sin(t+π/4)
t∈【-π/2,π/2】
t+π/4∈【-π/4,3π/4】
t=-π/4时,取最小值ymin=√10*(-√2/2) = -√5
t=π/2时,取最大值ymax=√10*1 = √10
值域【-√5,√10】