SOLUTION: Im in Pre-Calc. What is the domain of: h(x) = square root of (4-x) ALL OVER (x+1)(x squared+1) give answers like (-3, 1) U (1, infinity) - that is just an answer from a prev

Algebra ->  Functions -> SOLUTION: Im in Pre-Calc. What is the domain of: h(x) = square root of (4-x) ALL OVER (x+1)(x squared+1) give answers like (-3, 1) U (1, infinity) - that is just an answer from a prev      Log On


   

Question 486971: Im in Pre-Calc. What is the domain of:
h(x) = square root of (4-x) ALL OVER (x+1)(x squared+1)
give answers like (-3, 1) U (1, infinity) - that is just an answer from a previous problem
im having trouble with this current one because not much makes sense - thanks for the help
Found 3 solutions by John10, jim_thompson5910, MathLover1:
Answer by John10(297) About Me  (Show Source):
You can put this solution on YOUR website!
Try to find the RESTRICTED value which make the denominator to be zero
set the denominator to be zero:
(x + 1)(x^2 + 1) =0
Since x^2 + 1 is NEVER to be zero.
so x + 1 = 0
x = -1
Thus the domain is all REAL NUMBERS EXCEPT -1
D = {x|(-inf, -1)U (-1, inf)}
John10:)

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You cannot divide by zero. So the denominator cannot be equal to zero. If it were, then


%28x%2B1%29%28x%5E2%2B1%29=0


which means that


x%2B1=0 or x%5E2%2B1=0


Solve the first equation to get x=-1. Solve the second equation to get x=i or x=-i.


Ignore the last two answers (since we're only concerned about real numbers).


So if x=-1, then the entire denominator is zero.


So we must exclude this value from the domain.


So the domain is


Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
h%28x%29+=+sqrt%28%284-x%29%29%2F%28%28x%2B1%29%28x%5E2%2B1%29%29
denominator %28%28x%2B1%29%28x%5E2%2B1%29%29 cannot be equal to zero
so, if either %28x%2B1%29 or %28x%5E2%2B1%29 equal to zero denominator will be zero
what values for x are NOT solutions, we will find out if we make denominator equal to zero
if
x%2B1=0....->...x=-1
or
if
x%5E2%2B1=0....->...x%5E2=-1....>...x=+-i...imaginary roots

so, you have real solution x=-1
(-infinity, -1) U (-1, infinity)