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hugsandkisses

  • 3 years ago

f(x)= -√x=1 Graph the functions and describe the domain and range of each function. anybody?

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  1. Hero
    • 3 years ago
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    your function has two equal signs in it.

  2. hugsandkisses
    • 3 years ago
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    f(x)= -√x+1

  3. Hero
    • 3 years ago
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    domain x >=-1 range y <= 0

  4. rickjbr
    • 3 years ago
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    If you take the square root of a negative number, it is not a member of the real number system. x cannot be negative. The domain of f(x) is \[[0,\infty)\]

  5. Hero
    • 3 years ago
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    ^not true, x can be negative

  6. rickjbr
    • 3 years ago
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    Yes, but it will not yield a real number

  7. rickjbr
    • 3 years ago
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    I'm assuming he wants real numbers

  8. Hero
    • 3 years ago
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    x + 1 cannot be less than zero

  9. hugsandkisses
    • 3 years ago
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    Umm, how about I just post the thing that has all these on them

  10. rickjbr
    • 3 years ago
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    It says \[-\sqrt{x}=1\]

  11. rickjbr
    • 3 years ago
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    How can x be negative?

  12. Hero
    • 3 years ago
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    sqrt(x+1) not sqrt(x)

  13. rickjbr
    • 3 years ago
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    Oh i was going by the original post. Didn't see the updated one.

  14. cinar
    • 3 years ago
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    \[f(x)=-\sqrt{x+1} \,\,\text {or} f(x)=-\sqrt{x}+1?\]

  15. rickjbr
    • 3 years ago
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    That's what I thought cinar

  16. hugsandkisses
    • 3 years ago
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    that was a typo. sorry you guys!

  17. rickjbr
    • 3 years ago
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    So, which is it?

  18. Hero
    • 3 years ago
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    Use parentheses

  19. hugsandkisses
    • 3 years ago
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    f(x)= -√x+1

  20. rickjbr
    • 3 years ago
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    so 1 is not under the square root?

  21. hugsandkisses
    • 3 years ago
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    it is. x+1 are under the square root

  22. rickjbr
    • 3 years ago
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    OK, use parentheses next time \[\sqrt{x+1}>0, x>=-1\] Domain is \[[-1,\infty)\]

  23. cinar
    • 3 years ago
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    then domain is [-1,infinite) range is [0, infinite)

  24. hugsandkisses
    • 3 years ago
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    thank you!

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