# Question aea6c

Jul 11, 2017

"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.

$\left(4 - x\right) \left(x - 8\right) > 0$

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

Therefore the domain is 4≤x≤8

Jul 11, 2017

Domain :$4 \le x \le 8$ , in interval notation: $\left[4 , 8\right]$

#### Explanation:

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

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

$\ge 0 \therefore - {x}^{2} + 12 x - 32 \ge 0 \mathmr{and} {x}^{2} - 12 x = 32 \le 0$ or

$\left(x - 4\right) \left(x - 8\right) \le 0$ When  x= 4 or x=8, x-4)(x-8) = 0#

critical points are $x - 4 = 0 \mathmr{and} x = 4 \mathmr{and} x - 8 = 0 \mathmr{and} x = 8$

Sign Change:

when $x < 4$ sign of $\left(x - 4\right) \left(x - 8\right) = \left(-\right) \cdot \left(-\right) = + \therefore > 0$

when $4 < x < 8$ sign of $\left(x - 4\right) \left(x - 8\right) = \left(+\right) \cdot \left(-\right) = - \therefore < 0$

when $x > 8$ sign of $\left(x - 4\right) \left(x - 8\right) = \left(+\right) \cdot \left(+\right) = + \therefore > 0$

So $4 \le x \le 8$ for $\left(x - 4\right) \left(x - 8\right) \le 0$

Domain :$4 \le x \le 8$ , in interval notation $\left[4 , 8\right]$

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

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