I take your expression to be y = 1/(sqrt(9 - x^2)) written with parenthesis for clarity.
The range is [1/3, +infinity) -- The graph "in blue" on http://www.wolframalpha.com/input/?i=Graph+y+%3D+1... makes it clear that y goes to +infinity. Also the single minimum of 1/3 is obtained at x = 0 because 9 - x^2 is the largest at x = 0.
Note that the domain is (-3, +3) and the function is symmetric in the y-axis with vertical asymptotes at x = +/- 3. Taking one-sided limits near these points will justify that the function will indeed take on very large positive values. There are no horizontal asymptotes here, since x cannot go to infinity.
In general if you wish to see the range, it is best to look at the graph either drawn by hand employing the calculus techniques to find the maximum/minimum value(s) and inflection points, or with a graphing device which has a trace function.
IF x=3 or -3 then x^2=9 thus making your function 1/0= positive infinity
at x=0, f(0)=1/3
IF x>4, then you end up with looking for the square root of -7, which doesn't exist as a real number, likewise of x<-4, so no possible y value outside the domain of -3<x<3.
so the range of possible y values is +1/3 to infinity.
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I take your expression to be y = 1/(sqrt(9 - x^2)) written with parenthesis for clarity.
The range is [1/3, +infinity) -- The graph "in blue" on http://www.wolframalpha.com/input/?i=Graph+y+%3D+1... makes it clear that y goes to +infinity. Also the single minimum of 1/3 is obtained at x = 0 because 9 - x^2 is the largest at x = 0.
Note that the domain is (-3, +3) and the function is symmetric in the y-axis with vertical asymptotes at x = +/- 3. Taking one-sided limits near these points will justify that the function will indeed take on very large positive values. There are no horizontal asymptotes here, since x cannot go to infinity.
In general if you wish to see the range, it is best to look at the graph either drawn by hand employing the calculus techniques to find the maximum/minimum value(s) and inflection points, or with a graphing device which has a trace function.
IF x=3 or -3 then x^2=9 thus making your function 1/0= positive infinity
at x=0, f(0)=1/3
IF x>4, then you end up with looking for the square root of -7, which doesn't exist as a real number, likewise of x<-4, so no possible y value outside the domain of -3<x<3.
so the range of possible y values is +1/3 to infinity.
The range is ALL possible values of y.
[1/3 , infinity)
(-infiniti, 1/â9]
[0, infinity]]