What are the domain and range of the function
f(x)=1/(sq. root of(9-x^2))???
2007-08-21 16:53:21 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
f(x)=1/(sq. root of(9-x^2))
The domain of f is the set of all x where f(x) is defined.
f(x) is defined on all x where 9 - x^2 >0
or x^2 < 9
|x| < 3
The range of f is the set of all y=f(x) = 1/(9 - x^2)^(1/2)
which is the set of all y >0 such that:
For each y>0, we can find an x s.t. y = f(x)
namely s.t.
9 - x^2 = 1/y^2
x^2 = 9 - 1/y^2
or |x| = (9 - 1/y^2)^(1/2) exits
which is
the set of all y >0 s.t. 9 - 1/y^2 > 0
or 1/y^2 < 9
or y^2 > 1/9
i.e. all y > 1/3 is the range
2007-08-21 17:03:39 · answer #1 · answered by vlee1225 6 · 0⤊ 0⤋
you cannot have 9-x^2<=0 because that would mean that either the divider is zero or the sqrt root is negative. both are impossible.
the roots are 3 and -3 so the domain is -3 < x < 3.
2007-08-21 17:02:26 · answer #2 · answered by Christophe G 4 · 0⤊ 0⤋
domain is -3 to +3. range is +1/3 to infinity.
2007-08-21 17:08:12 · answer #3 · answered by William B 4 · 0⤊ 0⤋
The domain (x-values) is from -3 to +3
The range (y-values) is from -3 to +3
2007-08-21 17:00:38 · answer #4 · answered by cattbarf 7 · 0⤊ 0⤋
since the x is under a radical, the quantity (9-x^2) cannot be negative. Since it's in the denominator, it cannot equal 0 (you cannot divide by 0)
D: x is between -3 and 3.
2007-08-21 17:00:36 · answer #5 · answered by Abby 2 · 0⤊ 0⤋