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maryannlluch
 one year ago
determine the domain of a function Determine the domain of the function.
f as a function of x is equal to the square root of x plus three divided by x plus eight times x minus two.
maryannlluch
 one year ago
determine the domain of a function Determine the domain of the function. f as a function of x is equal to the square root of x plus three divided by x plus eight times x minus two.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The domain of the function is the set of all values that you can "plug into" the function and it still remains valid. Now I am assuming here by function you mean a function of a real variable... meaning it takes a real number and returns a real number. So the domain would be the set of all real numbers that you can "plug into" the function while retaining this definition (meaning it returns a real number). With this in mind: A function of x: f(x) Is equal to the square root of: f(x)=sqrt( x plus three divided by x plus eight times x minus two: \[f(x)=\sqrt{( \frac{x+3}{x+8}) (x2)}\] Now be warned I am not sure how the square root was applied. It could have meant this: \[f(x)=\frac{\sqrt{x}+3}{x+8}(x2)\] It all depends on where the square root part is ended. I would ask you teacher for a clarification. Sometime pure algebra problems don't translate easily into words unless you make an effort to be extraordinarily precise. So assuming the problem refers to the first equation. Clearly the domain includes any possible real number > or = 2, since the last factor will be negative if x<2. If this term is negative then (since the other term is strictly positive) the quantity under the radical will be negative and thus the function will cease to return a real value (the answer will be a complex number). Additionally, any number x between 3 and 8 (8<x<=3) will make both the (x2) term and the (x+3) term negative while keeping the (x+8) term positive. The negatives will cancel and there will be no issue. If x=8, then we have a zero in the denominator which is unacceptable so we ruled it out; however if x=3 we have a zero in the numberator which means the function returns a zero which is perfectly acceptable. Therefore: Domain: \[\left\{ x \in \mathbb{R} \ \ x \ge 2 \ \& 8<x \le 3\right\}\] Read as: The domain is the set of all real numbers x such that x is greater than or equal to 2 and greater than 8 but less than or equal to 3. There are no doubt other ways to state this, but this is one way I recall so hopefully it conforms to your class's requirements.
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