SOLUTION: Without graphing, find the domain of the following functions: a){{{f(x) = sqrt((x - 3)/(2x^2 - 8))}}}, and b){{{f(x) = root(3,(x + 5)/(x^2 - 9))}}}.

Algebra ->  Functions -> SOLUTION: Without graphing, find the domain of the following functions: a){{{f(x) = sqrt((x - 3)/(2x^2 - 8))}}}, and b){{{f(x) = root(3,(x + 5)/(x^2 - 9))}}}.      Log On


   



Question 389690: Without graphing, find the domain of the following functions:
a)f%28x%29+=+sqrt%28%28x+-+3%29%2F%282x%5E2+-+8%29%29, and
b)f%28x%29+=+root%283%2C%28x+%2B+5%29%2F%28x%5E2+-+9%29%29.

Found 2 solutions by josmiceli, robertb:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
a)f%28x%29+=+sqrt%28%28x+-+3%29%2F%282x%5E2+-+8%29%29
f%28x%29+=+sqrt%28x+-+3%29%2Fsqrt%282x%5E2+-+8%29%29
if x+=+3
The numerator is 0 and the denominator is (+)
and
If x+%3C+3, The numerator is imaginary
-------------------
If x+%3E+3, numerator is (+)
and denominator is (+)
-------------------
If x+=+2
The denominator is 0
and the numerator is imaginary
If x+%3C+2, the denominator is imaginary
-------------------
The restriction x+%3E+=+3 includes
the restriction x+%3E+2, so that is the domain
x+%3E=+3 answer
b) f%28x%29+=+root%283%2C%28x+%2B+5%29%2F%28x%5E2+-+9%29%29.
f%28x%29+=+root%283%2C%28x+%2B+5%29%29%2Froot%283%2C%28x%5E2+-+9%29%29
If x+=+-5 the numerator is 0
and the denominator is (+)
If x+%3C+-5 the numerator is (-)
and the denominator is (+)
If x+%3E+-5 the numerator is (-)
and the denominator is (+) or (-)
---------------------------
If x+=+3, the denominator is 0
If x+%3C+3 or x+%3E+3 denominator is (+)
-----------------------------
x+=+3 not allowed is the only restriction
The domain is x+%3E+3 and x+%3C+3

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
a) Since there is a square root symbol, we have to ensure that %28x+-+3%29%2F%282x%5E2+-+8%29%3E=0. To find the critical numbers, determine all x values that make either the top or the bottom equal to 0. By inspection they are -2, 2, and 3. These critical numbers partition the real number line into 4 parts. Choosing the test numbers -3, 0, 2.5, and 4, and checking for the signs:
For (-infinity, -2), %28x+-+3%29%2F%282x%5E2+-+8%29+%3C+0.
For (-2, 2), %28x+-+3%29%2F%282x%5E2+-+8%29%3E0.
For (2,3), %28x+-+3%29%2F%282x%5E2+-+8%29%3C+0.
For (3, infinity), %28x+-+3%29%2F%282x%5E2+-+8%29%3E+0.
Of the critical numbers, only 3 can be accepted, and not 2 and -2. Therefore the domain is (-2, 2)U [3, infinity).

b) The radical symbol is a cube-root symbol, and every real number has a cube root (whether positive or negative). Then the only thing we have to ensure is that the denominator won't be 0. The x-values that make the denominator 0 are -3 and 3. Hence the domain is the set of all real numbers except -3 and 3, or R\{-3, 3}.