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Mathematics 17 Online
OpenStudy (anonymous):

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

jimthompson5910 (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?

OpenStudy (anonymous):

division symbol

jimthompson5910 (jim_thompson5910):

is f(x) this? \[\Large f(x) = \sqrt{2-x}\] OR is it this? \[\Large f(x) = \sqrt{2}-x\]

OpenStudy (anonymous):

the first

jimthompson5910 (jim_thompson5910):

so h(x) would be \[\Large h(x) = \frac{f(x)}{g(x)}\] \[\Large h(x) = \frac{\sqrt{2-x}}{x-1}\]

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

so confused

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

1

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

yes

jimthompson5910 (jim_thompson5910):

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

jimthompson5910 (jim_thompson5910):

now onto the next part

jimthompson5910 (jim_thompson5910):

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

OpenStudy (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

jimthompson5910 (jim_thompson5910):

why not add x to both sides?

jimthompson5910 (jim_thompson5910):

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jimthompson5910 (jim_thompson5910):

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