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what is the domain and range for the function : f(x)= 1/√9-x^2

 Jan 7, 2015

Best Answer 

 #1
avatar+130560 
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f(x) = 1 / √(9 - x^2)

Since the denominator can't  = 0 , x cannot be 3 or -3.  And since we can't take the square root of a negative number, any x values < -3 or > 3 make the square root non-real

So....the domain is  just (-3, 3)

The range is a little trickier.....as "small" as y can be is when x = 0 ...and this is 1/3....and as x approaches either -3 or 3 in the domain, y approaches infimity...so, the range is [1/3, ∞)

Here's a graph.....https://www.desmos.com/calculator/9t9yvcfn2t

 

 

 Jan 7, 2015
 #1
avatar+130560 
+5
Best Answer

f(x) = 1 / √(9 - x^2)

Since the denominator can't  = 0 , x cannot be 3 or -3.  And since we can't take the square root of a negative number, any x values < -3 or > 3 make the square root non-real

So....the domain is  just (-3, 3)

The range is a little trickier.....as "small" as y can be is when x = 0 ...and this is 1/3....and as x approaches either -3 or 3 in the domain, y approaches infimity...so, the range is [1/3, ∞)

Here's a graph.....https://www.desmos.com/calculator/9t9yvcfn2t

 

 

CPhill Jan 7, 2015

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