Question

函数f(x)=√(x+3)+√(1-x)的最大值和最小值为何?

Answer 1

f(x) = √[√(x+3) + √(1-x)]^2 (-3 ≦ x ≦ 1) = √{4+2√[(3+x)(1-x)]} = √{4+2√[4-(x+1)^2]} 最大值发生在 x+1 = 2
即 f(-1) = √8 = 2√2. 最小值发生在端点 x = -3
1. f(-3) = f(1) = 2.
f(x)=√(x+3)+√(1-x) f'(x)=0.5(x+3)^-0.5 +(-1)0.5(1-x)^-0.5 f'(x)=0.5(x+3)^-0.5 -0.5(1-x)^-0.5 f'(x)=[(1-x)^0.5 -(x+3)^0.5]/{2[(x+3)(1-x)]^0.5} f'(x)=0 (1-x)^0.5 -(x+3)^0.5=0 x+3=1-x 2x=-2 x=-1 f(1)=√(1+3)+√(1-1) f(1)=2 最小值=2