If f(x)=sqrt2-x and g(x) = x − 1, then find the domain of h(x) = f(x) g(x).

- anonymous

If f(x)=sqrt2-x and g(x) = x − 1, then find the domain of h(x) = f(x) g(x).

- schrodinger

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- 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

division symbol

- 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

the first

- 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

Question: which value of x makes the denominator, x-1, equal to 0?

- anonymous

so confused

- jim_thompson5910

if x-1 = 0, then what must x be?

- anonymous

1

- jim_thompson5910

since x = 1 makes x-1 = 0 true, that means x = 1 makes the denominator equal to zero. Agreed?

- anonymous

yes

- jim_thompson5910

to avoid dividing by zero, we have to kick out x = 1 from the domain

- jim_thompson5910

now onto the next part

- jim_thompson5910

when you solve \(\Large 2 - x \ge 0\) for x, what do you get?

- 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

why not add x to both sides?

- jim_thompson5910

|dw:1436580427007:dw|

- jim_thompson5910

|dw:1436580444120:dw|

- jim_thompson5910

what happens?

- anonymous

2>0x

- anonymous

well dont the x by the 0 just cancel cause its 0

- jim_thompson5910

|dw:1436580524713:dw|

- jim_thompson5910

|dw:1436580534366:dw|
make sense?

- anonymous

kinda

- 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

confused

- jim_thompson5910

where?

- 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

if x is less than or equal to 2
AND
x is not 1
then x is allowed in the domain

- jim_thompson5910

those are the two conditions I'm talking about

- anonymous

okay

- 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

the problem is done at this point

- anonymous

but there is no answer in answer choices?

- jim_thompson5910

what are the choices

- 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

it would be \[\Large (-\infty, 1) \cup (1, 2]\] if you wrote those conditions above in interval notation

- jim_thompson5910

start with (-infinity, 2] and then kick out 1 from that interval

- anonymous

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