How do you find the domain of $f (x) = \frac{{(x + 8)}^{\frac{1}{2}}}{(x + 3) (x - 2)}$ $f(x) = ((x + 8)^(1/2)) /( (x + 3)(x - 2))$ ?

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How do you find the domain of $f (x) = \frac{{(x + 8)}^{\frac{1}{2}}}{(x + 3) (x - 2)}$ $f(x) = ((x + 8)^(1/2)) /( (x + 3)(x - 2))$ ?

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Answer 1 · 2016-12-27T07:13:03

$x \in [- 8; - 3) \cup (- 3; 2) \cup (2; + \infty)$ $x in [-8;-3)uu(-3;2) uu (2;+oo)$

Explanation:

You can rewrite the given expression as:

$\frac{\sqrt{x + 8}}{(x + 3) (x - 2)}$ $sqrt(x+8)/((x+3)(x-2)$

Then, since you cannot calculate the square root of a negative number and you cannot divide by zero, the domain is the solution of the following conditions:

$x + 8 \geq 0 and x + 3 \neq 0 and x - 2 \neq 0$ $x+8>=0 and x+3!=0 and x-2!=0$

that's

$x \geq - 8 and x \neq - 3 and x \neq 2$ $x>=-8 and x!=-3 and x!=2$

that can be written as:

$x \in [- 8; - 3) \cup (- 3; 2) \cup (2; + \infty)$ $x in [-8;-3)uu(-3;2) uu (2;+oo)$

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How do you find the domain of $f (x) = \frac{{(x + 8)}^{\frac{1}{2}}}{(x + 3) (x - 2)}$ $f(x) = ((x + 8)^(1/2)) /( (x + 3)(x - 2))$ ?

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How do you find the domain of $f (x) = \frac{{(x + 8)}^{\frac{1}{2}}}{(x + 3) (x - 2)}$ $f(x) = ((x + 8)^(1/2)) /( (x + 3)(x - 2))$ ?