设f(x)=(x²+4)/(x+1)
f'(x)=(2x(x+1)-(x²+4)*1)/(x+1)²
=(x²+2x-4)/(x+1)²
f'(x)=0时,f(x)有极值
(x²+2x-4)/(x+1)²=0
x²+2x-4=0
(x+1)²=5
x1=-1-√5
x2=-1+√5
f(x1)=((-1-√5)²+4)/(-1-√5+1)=-(10+2√5)/√50
∴(x^2+4)/(x+1)的最大值:2(√5-1)
4x+y=1
y=1-4x
设f(x)=1/x+1/y
=(x+y)/(xy)
=(1-3x)/(x-4x²)
=(1-3x)/(-4(x²-x/4)
=(1-3x)/(-4(x-1/8)²+1/16)
对于分母,当x=1/8时,分母有最大值1/16,此时分式取得最小值:f(x)min=(1-3/8)/(1/16)=10
即1/x+1/y的最小值为10