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anonymous

  • 4 years ago

Explain what do you mean by a one to one funtion? b) Explain briefly in steps how you would determine the inverse of a function f(x) c) Find the inverse of the function f(x)= x-1 the x-1 is in the square root formation

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  1. anonymous
    • 4 years ago
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    a) as i think , I'm not sure .. each value of x gives one value to y and each value of y gives one value of x . b) solution is ... Put x on one side and y on the other side of equation .. and to make the function details by y only c) f(x)=(x−1)2 y=(x−1)2 take root for both sides √y=(x−1) Put x on part and y on the other part .. and to make the function details by y only. x=√y+1

  2. precal
    • 4 years ago
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    I believe the originial function is \[f(x)=\sqrt{x-1}\] to find the inverse rewrite as \[y=\sqrt{x-1}\] switch the x and y \[x=\sqrt{y-1}\] Solve for y \[x=\sqrt{y-1}\] \[\left( x \right)^2=\left( \sqrt{y-1} \right)^2\] x=y-1 add 1 to both sides x+1=y your inverse is \[ f \left( x \right)^{-1}=x+1\]

  3. nikvist
    • 4 years ago
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    \[f^{-1}(x)=x^2+1\quad,\quad x\ge 0\]

  4. anonymous
    • 4 years ago
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    NikVist ? how you Find that answer ?

  5. precal
    • 4 years ago
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    yes I am sorry. You are correct in correcting my last step I dropped the 2nd power all of my steps are correct. \[x^2=y-1\] add 1 to both sides \[x^2+1=y\] This is the inverse \[f \left( x \right)^{-1}=x^2+1\]

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