anonymous
  • anonymous
If f(x)=sqrt2-x and g(x) = x − 1, then find the domain of h(x) = f(x) g(x).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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jim_thompson5910
  • jim_thompson5910
you wrote `h(x) = f(x) g(x)` but there's a gap between f(x) and g(x). Is there a symbol missing?
anonymous
  • anonymous
division symbol
jim_thompson5910
  • jim_thompson5910
is f(x) this? \[\Large f(x) = \sqrt{2-x}\] OR is it this? \[\Large f(x) = \sqrt{2}-x\]

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More answers

anonymous
  • anonymous
the first
jim_thompson5910
  • jim_thompson5910
so h(x) would be \[\Large h(x) = \frac{f(x)}{g(x)}\] \[\Large h(x) = \frac{\sqrt{2-x}}{x-1}\]
jim_thompson5910
  • jim_thompson5910
Question: which value of x makes the denominator, x-1, equal to 0?
anonymous
  • anonymous
so confused
jim_thompson5910
  • jim_thompson5910
if x-1 = 0, then what must x be?
anonymous
  • anonymous
1
jim_thompson5910
  • jim_thompson5910
since x = 1 makes x-1 = 0 true, that means x = 1 makes the denominator equal to zero. Agreed?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
to avoid dividing by zero, we have to kick out x = 1 from the domain
jim_thompson5910
  • jim_thompson5910
now onto the next part
jim_thompson5910
  • jim_thompson5910
when you solve \(\Large 2 - x \ge 0\) for x, what do you get?
anonymous
  • anonymous
i dont even know how to solve that, ive had sucky math teachers in high school they passed you based on if they liked you
jim_thompson5910
  • jim_thompson5910
why not add x to both sides?
jim_thompson5910
  • jim_thompson5910
|dw:1436580427007:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1436580444120:dw|
jim_thompson5910
  • jim_thompson5910
what happens?
anonymous
  • anonymous
2>0x
anonymous
  • anonymous
well dont the x by the 0 just cancel cause its 0
jim_thompson5910
  • jim_thompson5910
|dw:1436580524713:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1436580534366:dw| make sense?
anonymous
  • anonymous
kinda
jim_thompson5910
  • jim_thompson5910
so we know that \(\Large x \le 2\) AND x cannot equal 1 as long as x is some number that satisfies those conditions above, then that number is allowed in the domain
anonymous
  • anonymous
confused
jim_thompson5910
  • jim_thompson5910
where?
anonymous
  • anonymous
what you mean as long as x is some number that satisfies those conditions above, then that number is allowed in the domain
jim_thompson5910
  • jim_thompson5910
if x is less than or equal to 2 AND x is not 1 then x is allowed in the domain
jim_thompson5910
  • jim_thompson5910
those are the two conditions I'm talking about
anonymous
  • anonymous
okay
anonymous
  • anonymous
i just want to be done with this, ive had 3 hours of sleep, and been working since 6 this morning, and still have more to do ugh
jim_thompson5910
  • jim_thompson5910
the problem is done at this point
anonymous
  • anonymous
but there is no answer in answer choices?
jim_thompson5910
  • jim_thompson5910
what are the choices
anonymous
  • anonymous
( , 1) (-1, 2] ( , 1) (1, 2] ( , 2) (1, 2] ( , -1) (1, 2] ( , -2) (1, 2] the first thing is a little infinity sign thing
jim_thompson5910
  • jim_thompson5910
it would be \[\Large (-\infty, 1) \cup (1, 2]\] if you wrote those conditions above in interval notation
jim_thompson5910
  • jim_thompson5910
start with (-infinity, 2] and then kick out 1 from that interval
anonymous
  • anonymous
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