For the curve y=2^[sqrt(9-x2)], find the range of the function.
I think the range is [0,8] but I don't know why. Could you show me how to get to that answer? Thanks.
I 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
Last sentence is a typo.
Thanks! 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
Sure! To find the range of the function y = 2^sqrt(9-x^2), we need to determine all possible values that y can take on.
First, let's consider the domain of the function. The expression under the square root, 9 - x^2, must be non-negative since the square root of a negative number is undefined in the real number system. So, we have the inequality 9 - x^2 ≥ 0.
To solve this inequality, we can rearrange it as x^2 ≤ 9, which gives us -3 ≤ x ≤ 3.
Now, we can focus on the function y = 2^sqrt(9-x^2). Since the base of the exponential function is 2, the function will only output positive values.
We can observe that when x = -3 or x = 3, the expression under the square root becomes zero, which means the output y will be 2^0, or 1.
Furthermore, as x varies from -3 to 3 within the domain, the square root term √(9-x^2) will always yield a real value.
Therefore, the range of the function y = 2^sqrt(9-x^2) is [1, ∞) or (1, ∞), as it can take on any positive value larger than or equal to 1.
It seems there was a mistake in your initial response. The correct range is not [0,8], but rather [1, ∞) or (1, ∞).