Find range, domain, and Determine of the function.

a.
y = 5 - square root of (9 - x^2),
max value of square root of (9 - x^2) = 3 when x=0
min value of square root of (9 - x^2) = -3 when x=0
thus, y = [2,8]

b.
(9 - x^2) is inside the square root,
therefore 9-x^2 >= 0
9>=x^2
x = [-3,3]

c-
let f(x)=y=5 - square root of (9 - x^2)
f(-x)=5 - square root of (9 - (-x)^2)
= 5 - square root of (9 - x^2)
=f(x)
the function is even.

y = 5 - square root of (9 - x^2)

a. since min of square root of (9 - x^2)=0
so, max of y=5.
thus, the range is every real number less than of equal to 5.

b. since square root of (9 - x^2), (9 - x^2) must>=0,
so, x should belongs to [-3,3]
thus, the domain is as above.

c. the y=f(x)=5 - square root of (9 - x^2)
f(-x)=5 - square root of (9 - (-x)^2)=5 - square root of (9 - x^2)=f(x)
hence, it is an even function.

2008-05-29 14:59:05 補充：
in part a,
since min of square root of (9 - x^2)=0
and
max of square root of (9 - x^2)=3 when x=0
so, max of y=5, min y=2
thus, the range is every real number less than or equal to 5 and greater than or equal to 2.

2008-05-29 14:59:10 補充：
ONE THING SHOULD BE NOTICED is that...
only consider square root of (9 - x^2) is non-negative real number.
If it can be NEGATIVE, it should be written as +/- square root of (9 - x^2)

2008-05-31 09:52:32 補充：
我諗呢位朋友喺SQUARE ROOT同埋FUNCTION嘅concept上有d特別...

part a, 如果square root一定係+/-，咁點解呢條題目square root前會只係得"-"？
只有當你要拆除square時，用嘅square root先有+/-,
e.g. x^2 = 9, x=+/- square root of 9 = +/-3,
如果題目係x= square root of 9, 你唔可以話x=+/-3.

所以你嘅答案不能苟同。