Question

Question #aea6c

Answer 1

`"Dom"(f) = 4 ≤ x ≤ 8`

Explanation:

Since we have a radical function, we know straight away that the domain of the function will only be the values for which the radical is real.

`f(x)=√(-x^2+12x-32)`

`f` is real for -x^2+12x-32 > 0.

`(4-x)(x-8) > 0`

Since `(f)^2` is a decreasing function, we know that it will be positive between the roots.

Therefore the domain is `4≤x≤8`

Answer 2

Domain :`4 <= x <=8` , in interval notation: `[4,8]`

Explanation:

`f(x) =sqrt( -x^2 +12x -32)`

Domain means the possible value of input `x`, under root should be

`>=0 :. -x^2 +12x -32 >= 0 or x^2 -12x =32 <= 0` or

`(x-4)(x-8)<= 0` When `x= 4 or x=8, x-4)(x-8) = 0`

critical points are `x-4=0 or x=4 and x-8 =0 or x=8`

Sign Change:

when `x <4` sign of `(x-4)(x-8) = ( -) *(-) = + :. >0`

when `4 < x <8` sign of `(x-4)(x-8) = ( +) *(-) = - :. <0`

when `x >8` sign of `(x-4)(x-8) = ( +) *(+) = + :. >0`

So `4 <= x <=8` for `(x-4)(x-8)<= 0`

Domain :`4 <= x <=8` , in interval notation `[4,8]`

graph{(-x^2+12x-32)^0.5 [-9.96, 9.96, -4.98, 4.98]} [Ans]

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