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thewinterfawn Group Title

Find the domain of the function.

  • one year ago
  • one year ago

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  1. thewinterfawn Group Title
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    \[f(x)=\sqrt[4]{x^2+3x}\]

    • one year ago
  2. myininaya Group Title
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    Hint: Inside cannot be negative.

    • one year ago
  3. thewinterfawn Group Title
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    I don't understand how to start the problem. :/

    • one year ago
  4. myininaya Group Title
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    \[f(x)=\sqrt[4]{g(x)}\] Solve the following inequality: \[g(x) \ge 0\] Solving this will give you your domain.

    • one year ago
  5. myininaya Group Title
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    Because g can be zero and it can be positive, but it cannot be negative.

    • one year ago
  6. myininaya Group Title
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    Like you know whatever g is. In this case your g is x^2+3x

    • one year ago
  7. thewinterfawn Group Title
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    Um. I'm not good at math so I'm having trouble understanding what you are saying. sorry.

    • one year ago
  8. myininaya Group Title
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    Can you do fourth root of a negative number? Would that exist?

    • one year ago
  9. thewinterfawn Group Title
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    I don't think you can do that.

    • one year ago
  10. myininaya Group Title
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    like for real numbers anyways, no it wouldn't you are right that is why I'm allowing g to be positive or zero.

    • one year ago
  11. thewinterfawn Group Title
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    oh. so I just solve \[x^2+3x \ge0\]?

    • one year ago
  12. myininaya Group Title
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    That is right.

    • one year ago
  13. thewinterfawn Group Title
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    thank you. :D

    • one year ago
  14. myininaya Group Title
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    Np. Do you know how to solve it?

    • one year ago
  15. thewinterfawn Group Title
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    I think I do haha. Let me try it...

    • one year ago
  16. thewinterfawn Group Title
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    Um. Okay. I don't know how to. ):

    • one year ago
  17. myininaya Group Title
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    Ok. Just think domain are the x values where the function can exist for. If you talk about range later, just ask yourself what are the y values where the function exist. --- Ok... So do you know how to solve x^2+3x=0?

    • one year ago
  18. thewinterfawn Group Title
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    yeah it would be x=0, -3... i think.

    • one year ago
  19. myininaya Group Title
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    Yep yep...You did x(x+3)=0 which implies x=0 or x+3=0 great job... Now make a number line (on this number line you would include the zeros, -3 or 0 , but you would also include what x values made the expression not existing which we don't have any for x^2+3x) so here we go ---------------|-------------|----------- -3 0 Test all 3 intervals. You are looking for what intervals make x^2+3x positive since we are trying to solve x^2+3x>0 (We already solved when x^2+3x=0 ) (-4)^2+3(-4) (-1)^2+3(-1) (1)^2+3(1) = = = ---------------|-------------|----------- -3 0 Now all I'm doing is plugging in numbers around my zeros (-3 and 0) This is to test the intervals Remember we are looking for positive output Which one of those expressions I wrote above those intervals gives us positive output?

    • one year ago
  20. thewinterfawn Group Title
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    Sorry. The site was updating. ): The first and last intervals?

    • one year ago
  21. myininaya Group Title
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    Yes so the domain is the first and last intervals include x=-3, 0. So you would state it as (-inf, -3] U [0, inf) We don't want to include anything in between -3 and 0 because like you said that gave us negative output. And yeah sorry os went down earlier.

    • one year ago
  22. thewinterfawn Group Title
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    Thank you! Your explanation was really helpful. (:

    • one year ago
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