Question

what is the domain and range of sqrt of {(x-1)/(x+2)}

Answer 1

Answer on Question #51627 - Math - Real analysis

What is the domain and range of $f(x) = \sqrt{\frac{x - 1}{x + 2}}$ ?

Solution:

Let's find domain of $f(x)$ :

$$\frac {x - 1}{x + 2} \geq 0$$

Thus, the domain of $f(x)$ is $D(f) = (-\infty, -2) \cup [1, +\infty)$ .

Let's find the derivative of $f(x)$ :

$$f ^ {\prime} (x) = \frac {1}{2} \sqrt {\frac {x + 2}{x - 1}} \frac {x + 2 - x + 1}{(x + 2) ^ {2}} = \sqrt {\frac {x + 2}{x - 1}} \frac {3}{2 (x + 2) ^ {2}} \geq 0 \quad \text {on the domain} \quad D (f) = (- \infty , - 2) \cup [ 1, \infty).$$

Therefore $f(x)$ is monotonic increasing on $(-\infty, -2)$ and $f(x)$ is monotonic increasing on $(1, \infty)$ .

Let's find limits:

$$\lim _ {x \rightarrow - \infty} f (x) = \lim _ {x \rightarrow - \infty} \sqrt {\frac {x - 1}{x + 2}} = \lim _ {x \rightarrow - \infty} \sqrt {\frac {x}{x} \cdot \frac {1 - \frac {1}{x}}{1 + \frac {2}{x}}} = 1;$$ $$\lim _ {x \rightarrow - 2 - 0} f (x) = \lim _ {x \rightarrow - 2 - 0} \sqrt {\frac {x - 1}{x + 2}} = \left\{\sqrt {\frac {- 3}{- 0}} \right\} = + \infty ;$$ $$f (1) = \sqrt {\frac {1 - 1}{1 + 2}} = 0;$$ $$\lim _ {x \rightarrow + \infty} f (x) = \lim _ {x \rightarrow + \infty} \sqrt {\frac {x - 1}{x + 2}} = \lim _ {x \rightarrow + \infty} \sqrt {\frac {x}{x} \cdot \frac {1 - \frac {1}{x}}{1 + \frac {2}{x}}} = 1.$$

Besides, $f(x) = \sqrt{\frac{x - 1}{x + 2}} = \sqrt{\frac{x + 2 - 3}{x + 2}} = \sqrt{1 - \frac{3}{x + 2}} \neq 1$ for every $x$ .

Since, $f(x)$ is continuous function on the domain $D(f) = (-\infty, -2) \cup [1, +\infty)$ , then we obtain the range of $f(x): E(f) = [0,1) \cup (1,\infty)$ .

Answer: $D(f) = (-\infty, -2) \cup [1, \infty)$ , $E(f) = [0, 1) \cup (1, +\infty)$ .

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