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]}