Question

求y=(x-5)^(1/2)+(24-3x)^(1/2)的值域

Answer 1

y=(x-5)^(1/2)+(24-3x)^(1/2)
for real y
x-5 >=0 and 24-3x >=0
=> x>=5 and x<=8
=>5 <= x <=8

y=(x-5)^(1/2)+(24-3x)^(1/2)
y' = 1/[2(x-5)^(1/2)]- 3/[2(24-3x)^(1/2)] =0
1/[2(x-5)^(1/2)]= 3/[2(24-3x)^(1/2)]
1/[4(x-5)] = 9/(4(24-3x))
3(x-5) = 8-x
4x=23
x=23/4
y(23/4)=2根3
y(5)=3
y(8)=根号3
值域为[根3，2根3】

Answer 2

y=(x-5)^(1/2)+(24-3x)^(1/2)
for real y
x-5 >=0 and 24-3x >=0
=> x>=5 and x<=8
=>5 <= x <=8

y=(x-5)^(1/2)+(24-3x)^(1/2)
y' = 1/[2(x-5)^(1/2)]- 3/[2(24-3x)^(1/2)] =0
1/[2(x-5)^(1/2)]= 3/[2(24-3x)^(1/2)]
1/[4(x-5)] = 9/(4(24-3x))
3(x-5) = 8-x
4x=24
x = 6
y(6) = (6-5)^(1/2)+(24-3(6))^(1/2) = 1+√6
y(5) = (5-5)^(1/2)+(24-3(5))^(1/2) = 3
y(8) = (8-5)^(1/2)+(24-3(8))^(1/2) = √3
y的值域 =[√3,1+√6]

Answer 3

由题知5=<x<=8,对y求导知上式单调递增，所以
x=5时取得最小值y=3
x=8时取得最大值y=根号3
y=(x-5)^(1/2)+(24-3x)^(1/2)
for real y
x-5 >=0 and 24-3x >=0
=> x>=5 and x<=8
=>5 <= x <=8

y=(x-5)^(1/2)+(24-3x)^(1/2)
y' = 1/[2(x-5)^(1/2)]- 3/[2(24-3x)^(1/2)] =0
1/[2(x-5)^(1/2)]= 3/[2(24-3x)^(1/2)]
1/[4(x-5)] = 9/(4(24-3x))
3(x-5) = 8-x
4x=24
x = 6
y(6) = (6-5)^(1/2)+(24-3(6))^(1/2) = 1+√6
y(5) = (5-5)^(1/2)+(24-3(5))^(1/2) = 3
y(8) = (8-5)^(1/2)+(24-3(8))^(1/2) = √3
y的值域=【根号3，2倍根号3】

Answer 4

由题知5=<x<=8,对y求导知上式单调递增，所以
x=5时取得最小值y=3
x=8时取得最大值y=根号3