Question

Can you check my work please?

Answer 1

Find the domain and range of:
\[y=\sqrt{x}+\sqrt{4-x}\]

Answer 2

Domain:
\[\sqrt{x} = 0\] then x is greater than or equal to 0
\[\sqrt{4-x}\] then x is less than or equal to 4

Answer 3

therefore, Domain is [0, 4]

Answer 4

Range:
???? [2] ????

Answer 5

yes, very very good !!!!

Answer 6

but i've got confused on the range :(

Answer 7

What about the range, it is not just 2.

Answer 8

yes... i used 0 and 4 but ended up with "2"... so....

Answer 9

\(\large\color{blue }{ \rm y=\sqrt{x} +\sqrt{4-x} }\)

\(\large\color{ green }{ \rm y=\sqrt{0} +\sqrt{4-0} ~~~~-->~~y=2 }\)

\(\large\color{ red }{ \rm y=\sqrt{1} +\sqrt{4-1} ~~~~-->~~y=\sqrt{3}+1 }\)
\(\large\color{ red }{ \rm y=\sqrt{3} +\sqrt{4-3} ~~~~-->~~y=\sqrt{3}+1 }\)

\(\large\color{blue }{ \rm y=\sqrt{4} +\sqrt{4-4} ~~~~-->~~y=2 }\)

Answer 10

\(\large\color{blue }{ \rm y=\sqrt{2} +\sqrt{4-2} ~~~~-->~~y=\sqrt{2}+2 }\)

Answer 11

Range ` [ 1 + √3 , 2 + √2 ] `

Answer 12

I think...

Answer 13

excuse me sir... just a question

Answer 14

about \[y=\sqrt{2}+2\]
wouldn't it be \[y=2\sqrt{2}\] ???
since squareroot of 2 plus squareroot of 2 is 2 sqaure root of 2?

Answer 15

yes, if x=2, then 2√2. My bad

Answer 16

Well, range ` [ 1 + √3 , 2√2 ] `

Answer 17

i see :D thank you very much sir

Answer 18

If I was sir, then I would not have made that mistake-:9 just a teenager .

Answer 19

>_< hahah. it's ok. nobody's perfect everyone can make a mistake

Answer 20

true :)