How do you find the domain and range of $f(x) = sqrt x / (x^2 + x - 2)$ ?

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Answer 1 · 2017-01-29T14:11:43

Domain is $x in [0, oo)$ , sans x = 1.
Range is $f in (-oo, oo)$ .
Asymptotes : $uarr x = 1 darr and y = 0 rarr$ .

To make f real, $x >=0$ .

$f=sqrtx/((x+2)(x-1))$

As $x to 0_(+-), f to +-oo$ .

As $x to oo, f to 0$ .

The asymptotes keep the two branches of the graph, in #Q_1 and

Q_4, respectively.

So, domain is $x in [0, oo)$ , sans x = 1.

Range is $f in (-oo, oo)$ .

See the illustrative Socratic graph.

graph{(sqrtx/(x^2+x-2)-y)(x-.99+.01y) =0 [-10, 10, -5, 5]}

How do you find the domain and range of f(x) = sqrt x / (x^2 + x - 2)f(x) = sqrt x / (x^2 + x - 2)?