Mathematics
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OpenStudy (anonymous):
If f(x)=sqrt2-x and g(x) = x − 1, then find the domain of h(x) = f(x) g(x).
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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
\[\Large h(x) = \frac{f(x)}{g(x)}\]
\[\Large h(x) = \frac{\sqrt{2-x}}{x-1}\]
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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?
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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
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jimthompson5910 (jim_thompson5910):
why not add x to both sides?
jimthompson5910 (jim_thompson5910):
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jimthompson5910 (jim_thompson5910):
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