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anonymous

  • one year ago

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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  1. jim_thompson5910
    • one year ago
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    you wrote `h(x) = f(x) g(x)` but there's a gap between f(x) and g(x). Is there a symbol missing?

  2. anonymous
    • one year ago
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    division symbol

  3. jim_thompson5910
    • one year ago
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    is f(x) this? \[\Large f(x) = \sqrt{2-x}\] OR is it this? \[\Large f(x) = \sqrt{2}-x\]

  4. anonymous
    • one year ago
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    the first

  5. jim_thompson5910
    • one year ago
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    so h(x) would be \[\Large h(x) = \frac{f(x)}{g(x)}\] \[\Large h(x) = \frac{\sqrt{2-x}}{x-1}\]

  6. jim_thompson5910
    • one year ago
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    Question: which value of x makes the denominator, x-1, equal to 0?

  7. anonymous
    • one year ago
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    so confused

  8. jim_thompson5910
    • one year ago
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    if x-1 = 0, then what must x be?

  9. anonymous
    • one year ago
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    1

  10. jim_thompson5910
    • one year ago
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    since x = 1 makes x-1 = 0 true, that means x = 1 makes the denominator equal to zero. Agreed?

  11. anonymous
    • one year ago
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    yes

  12. jim_thompson5910
    • one year ago
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    to avoid dividing by zero, we have to kick out x = 1 from the domain

  13. jim_thompson5910
    • one year ago
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    now onto the next part

  14. jim_thompson5910
    • one year ago
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    when you solve \(\Large 2 - x \ge 0\) for x, what do you get?

  15. anonymous
    • one year ago
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    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

  16. jim_thompson5910
    • one year ago
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    why not add x to both sides?

  17. jim_thompson5910
    • one year ago
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    |dw:1436580427007:dw|

  18. jim_thompson5910
    • one year ago
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    |dw:1436580444120:dw|

  19. jim_thompson5910
    • one year ago
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    what happens?

  20. anonymous
    • one year ago
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    2>0x

  21. anonymous
    • one year ago
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    well dont the x by the 0 just cancel cause its 0

  22. jim_thompson5910
    • one year ago
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    |dw:1436580524713:dw|

  23. jim_thompson5910
    • one year ago
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    |dw:1436580534366:dw| make sense?

  24. anonymous
    • one year ago
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    kinda

  25. jim_thompson5910
    • one year ago
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    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

  26. anonymous
    • one year ago
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    confused

  27. jim_thompson5910
    • one year ago
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    where?

  28. anonymous
    • one year ago
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    what you mean as long as x is some number that satisfies those conditions above, then that number is allowed in the domain

  29. jim_thompson5910
    • one year ago
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    if x is less than or equal to 2 AND x is not 1 then x is allowed in the domain

  30. jim_thompson5910
    • one year ago
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    those are the two conditions I'm talking about

  31. anonymous
    • one year ago
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    okay

  32. anonymous
    • one year ago
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    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

  33. jim_thompson5910
    • one year ago
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    the problem is done at this point

  34. anonymous
    • one year ago
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    but there is no answer in answer choices?

  35. jim_thompson5910
    • one year ago
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    what are the choices

  36. anonymous
    • one year ago
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    ( , 1) (-1, 2] ( , 1) (1, 2] ( , 2) (1, 2] ( , -1) (1, 2] ( , -2) (1, 2] the first thing is a little infinity sign thing

  37. jim_thompson5910
    • one year ago
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    it would be \[\Large (-\infty, 1) \cup (1, 2]\] if you wrote those conditions above in interval notation

  38. jim_thompson5910
    • one year ago
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    start with (-infinity, 2] and then kick out 1 from that interval

  39. anonymous
    • one year ago
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