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anewbie

  • 3 years ago

Find the inverse of the function f(x)=x^{2}-8x+4, where <4. State its domain.

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  1. anewbie
    • 3 years ago
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    \[where x \le 4\]

  2. anewbie
    • 3 years ago
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    @hartnn

  3. hartnn
    • 3 years ago
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    can you complete the square ? know how to ?

  4. hartnn
    • 3 years ago
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    \(f(x)=x^2-8x+4= (x-...)^2-...\)

  5. anewbie
    • 3 years ago
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    so the answer is inversef(x)=4+sqrt{x+12} . domain is x bigger or equal -12 ?

  6. hartnn
    • 3 years ago
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    to be more precise, \(f^{-1}(x)=4\pm \sqrt{x+12} \) the domain \(x \ge -12 \) is correct.

  7. hartnn
    • 3 years ago
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    wait, its given that x<4

  8. hartnn
    • 3 years ago
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    so, it'll be only \(f^{-1}(x)=4 - \sqrt{x+12}\) with same domain.

  9. anewbie
    • 3 years ago
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    how can we understand that it is minus when x<4 ?

  10. hartnn
    • 3 years ago
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    4+ anything means >4 4- anything means <4

  11. hartnn
    • 3 years ago
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    does that clear it up ? or you have different doubt altogether ? :P

  12. anewbie
    • 3 years ago
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    a little bit :D because it tells about x but we take the function <4 :/

  13. hartnn
    • 3 years ago
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    oh, .... domain of f(x) means range of its inverse function \(f^{-1}(x)\) here, x<4 is domain of f(x), so range of \(f^{-1}(x)\) is x<4 , means \(f^{-1}(x)\) has least value of 4, so we discard 4+sqrt {x+12} because that will be >4

  14. hartnn
    • 3 years ago
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    ***range of \(f^{-1}(x)\) is \(f^{-1}(x)\)<4

  15. hartnn
    • 3 years ago
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    ***maximum value of 4

  16. hartnn
    • 3 years ago
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    sorry for the typos.

  17. anewbie
    • 3 years ago
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    oh. okey. thank youu :)

  18. hartnn
    • 3 years ago
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    welcome ^_^

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