Đáp án:
`a, min_{[1;3]} y= 0, max_{[1;3]} y=\sqrt{6}`
`b,max y= 2\sqrt{3}, min y =\sqrt{6}`
Giải thích các bước giải:
`a, y= \sqrt{x²-x}`
`D= (-\infty;0)∪[1;+\infty)`
`y' = (2x-1)/(2\sqrt{x²-x})`
Xét `y' =0 => x = 1/2 \notin [1;3]`
Ta có:
`y (1)=0, y(3)=\sqrt{6}`
`=> min_{[1;3]} y= 0, max_{[1;3]} y=\sqrt{6}`
`b, y= \sqrt{3-x}+\sqrt{x+3}`
`D =[-3;3]`
`y' = (-\sqrt{x+3} +\sqrt{3-x})/(2\sqrt{9-x²}) `
Xét `y' =0 `
`<=> \sqrt{3-x}=\sqrt{x+3}`
`<=> -2x =0 <=> x=0 \in [-3;3]`
Ta có:
`y(±3) = \sqrt{6}, y(0) = 2\sqrt{3}`
`=>max y= 2\sqrt{3}, min y =\sqrt{6}`
