Question

What is the range of √(9-x^2). Why cannot it be (-3,3)???

Answer 1

we can't take the square root of a negative number, any x values < -3 or > 3 make the square root non-real

So....

Answer 2

to find range ,first find domain

since the root value is always +ve so the given no. is ≥ {grater than or equal to 0}

⇒9-x²≥0

⇒9≥x²

⇒x²≤9

⇒x²-9≤0

⇒{x+3}{x-3}≤0

⇒x ∈ [-3,3]

hence domain is [-3,3]

range:-

let y=f{x}

y=√9-x²

sq. on both side.....

y²=9-x²

x²+y²=9

x²=9-y²

x=√9-y²

proceed as same as in upper case you got,

x∈[-3,3]

since root value can never in negative ,so you neglect the -ve value,

so range is [0,3] { all +ve form 0 to 3 {including}