SOLUTION: Let f(x) = (2x + 3)/(x - 7) + sqrt(-x) + 1/(x^2 - 4). Find the domain of f. Give your answer using interval notation.

Algebra ->  Functions -> SOLUTION: Let f(x) = (2x + 3)/(x - 7) + sqrt(-x) + 1/(x^2 - 4). Find the domain of f. Give your answer using interval notation.      Log On


   

Question 1209311: Let
f(x) = (2x + 3)/(x - 7) + sqrt(-x) + 1/(x^2 - 4).
Find the domain of f. Give your answer using interval notation.
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


The domain is the set of x values for which the function is defined.

The function is the sum of three distinct expressions. Look at each of them to determine what values of x, if any, that one produces.

Restricted values of x are those for which either
(a) the square root of the expression is negative; or
(b) the denominator of a fraction is 0

(1) %282x%2B3%29%2F%28x-7%29 --> the single value x=7 is excluded from the domain

(2) sqrt%28-x%29 --> the value of "-x" can't be negative --> x can't be positive --> x must be negative or zero

(3) 1%2F%28x%5E2-4%29=1%2F%28%28x%2B2%29%28x-2%29%29 --> the values x=2 and x=-2 are excluded from the domain

The values x=7 from (1) and x=2 from (3) are already excluded by (2), so the only excluded value of x not covered by (2) is the x=-2 from (3). So the domain is all numbers that are less than or equal to 0, except for x=-2.

ANSWER: (-infinity,-2) U (-2,0]

Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

The portion sqrt%28-x%29 has domain x+%3C=+0 to yield the interval notation (-infinity, 0]
The square bracket includes the endpoint 0.
This is the set of negative numbers with 0 involved as well.
I suppose you could call this the set of non-positive numbers, but I don't think that's an actual math term (I could be wrong).


If x = -2 then the denominator x%5E2-4 becomes zero.
Division by zero is not allowed, so we must remove x = -2 from the domain.
(-infinity, 0] will update to (-infinity, -2) U (-2, 0] which is the final answer.
You can think of it like poking at hole at -2 on the number line.

Replace the word "infinity" with the symbol if needed.
The positive solutions to x-7=0 and x%5E2-4=0 were never in the original domain to begin with, so we don't have to worry about them.