Đáp án:
`max\text( )y=0` khi `x∈{0;4}`
`min\text( ) y=-3` khi `x=2`
Giải thích các bước giải:
`y=1/4x^2-x-\sqrt{4x-x^2}`TXĐ: `D=[0;4]`
`y'=1/2 x -1 -(4-2x)/(2\sqrt{4x-x^2})(x∈(0;4))`
`=1/2 x -1- (2-x)/(\sqrt{4x-x^2})`
`=(x-2)/(2)+(x-2)/(\sqrt{4x-x^2})`
`=(x-2).(1/2 +1/(\sqrt{4x-x^2}))`
`y'=0<=>(x-2).(1/2 +1/(\sqrt{4x-x^2}))=0`
`<=>x-2=0` (do `1/2+1/(\sqrt{4x-x^2})>0∀x∈(0;4)`)
`<=>x=2` (Thỏa mãn)
Ta có: `y(0)=0`
`y(2)=-3`
`y(4)=0`
Vậy `max\text( )y=0` khi `x∈{0;4}`
`min\text( ) y=-3` khi `x=2`
