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Answer 1 · 2017-07-11T12:04:43

$Dom (f) = 4 \leq x \leq 8$ $"Dom"(f) = 4 ≤ x ≤ 8$

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.

$f (x) = \sqrt (- x^{2} + 12 x - 32)$ $f(x)=√(-x^2+12x-32)$

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

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

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

Therefore the domain is $4 \leq x \leq 8$ $4≤x≤8$

Answer 2 · 2017-07-11T12:40:18

Domain : $4 \leq x \leq 8$ $4 <= x <=8$ , in interval notation: $[4, 8]$ $[4,8]$

$f (x) = \sqrt{- x^{2} + 12 x - 32}$ $f(x) =sqrt( -x^2 +12x -32)$

Domain means the possible value of input $x$ $x$ , 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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