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April 1, 2015

April 1, 2015

Posted by **jane** on Monday, September 3, 2012 at 8:11pm.

I think the range is [0,8] but I don't know why. Could you show me how to get to that answer? Thanks.

- precalc review/calc -
**Damon**, Monday, September 3, 2012 at 8:30pmI assume you mean:

y=2^[sqrt(9-x^2)]

we do not want to take the sqrt of a negative number so domain is |x|<3

or

-3 < x < +3

eg

if x = -3 or + 3, y = 2^0 = 1

if x = -1 or + 1, y = 2^sqrt 8 = 7.1

if x = -2 or + 2, y = 2^sqrt 5 = 4.7

if x = 0 , y = 2^3 = 8

So

I think the range of y is 1 to 8, not 0 to 8

that corresponds to y between 0 and 8 so I agree with you

- do NOT agree with you -
**Damon**, Monday, September 3, 2012 at 8:31pmLast sentence is a typo.

- precalc review/calc -
**jane**, Monday, September 3, 2012 at 10:34pmThanks! So because the smallest and biggest numbers +/- 3 and the middle number 0 result in the smallest and largest y values, 1 and 8 respectively, the range is therefore [1, 8]. That makes sense. :D

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