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anonymous

  • 4 years ago

let f(x)= 1/x+9 what is f^-1(x)=?

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  1. nenadmatematika
    • 4 years ago
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    x+9

  2. Mertsj
    • 4 years ago
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    \[x=\frac{1}{y+9}\]

  3. Mertsj
    • 4 years ago
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    \[x(y+9)=1\]

  4. nenadmatematika
    • 4 years ago
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    oh you're true Mertsj...it's the inverse function :D

  5. Mertsj
    • 4 years ago
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    \[y+9=\frac{1}{x}\]

  6. Mertsj
    • 4 years ago
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    \[y=\frac{1}{x}-9\]

  7. Mertsj
    • 4 years ago
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    \[y=\frac{1-9x}{x}\]

  8. anonymous
    • 4 years ago
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    thanks guys!

  9. cristiann
    • 4 years ago
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    This question is either tricky or just plain sloppy ... First: it's not clear the form of the function \[f_{1}(x)=\frac{1}x+9\] or \[f_{2}(x)=\frac{1}{x+9}\] Then: no matter which function, you still have problems, because of the domain and codomain of the functions: For f1: \[x \neq 0, y \neq 9\] so only if the function is in the form \[f_{1}:\mathbb{R}^{*}\rightarrow \mathbb{R} \setminus (9) \] it is bijective (one-to-one/injective and onto/surjective) For the second function, you have the same type of problem, with \[x \neq -9, y \neq 0\] Only now you may talk about the expression of the inverse.

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