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anonymous
 one year ago
If f(x)=sqrt2x and g(x) = x − 1, then find the domain of h(x) = f(x) g(x).
anonymous
 one year ago
If f(x)=sqrt2x and g(x) = x − 1, then find the domain of h(x) = f(x) g(x).

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0you wrote `h(x) = f(x) g(x)` but there's a gap between f(x) and g(x). Is there a symbol missing?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0is f(x) this? \[\Large f(x) = \sqrt{2x}\] OR is it this? \[\Large f(x) = \sqrt{2}x\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0so h(x) would be \[\Large h(x) = \frac{f(x)}{g(x)}\] \[\Large h(x) = \frac{\sqrt{2x}}{x1}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0Question: which value of x makes the denominator, x1, equal to 0?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0if x1 = 0, then what must x be?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0since x = 1 makes x1 = 0 true, that means x = 1 makes the denominator equal to zero. Agreed?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0to avoid dividing by zero, we have to kick out x = 1 from the domain

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0now onto the next part

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0when you solve \(\Large 2  x \ge 0\) for x, what do you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.0why not add x to both sides?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436580427007:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436580444120:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0what happens?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well dont the x by the 0 just cancel cause its 0

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436580524713:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436580534366:dw make sense?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0what you mean as long as x is some number that satisfies those conditions above, then that number is allowed in the domain

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0if x is less than or equal to 2 AND x is not 1 then x is allowed in the domain

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0those are the two conditions I'm talking about

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.0the problem is done at this point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but there is no answer in answer choices?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0what are the choices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0( , 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
 one year ago
Best ResponseYou've already chosen the best response.0it would be \[\Large (\infty, 1) \cup (1, 2]\] if you wrote those conditions above in interval notation

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0start with (infinity, 2] and then kick out 1 from that interval
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