anonymous
  • anonymous
Find the inverse of the function f(x)=x^{2}-8x+4, where <4. State its domain.
Calculus1
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[where x \le 4\]
anonymous
  • anonymous
hartnn
  • hartnn
can you complete the square ? know how to ?

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hartnn
  • hartnn
\(f(x)=x^2-8x+4= (x-...)^2-...\)
anonymous
  • anonymous
so the answer is inversef(x)=4+sqrt{x+12} . domain is x bigger or equal -12 ?
hartnn
  • hartnn
to be more precise, \(f^{-1}(x)=4\pm \sqrt{x+12} \) the domain \(x \ge -12 \) is correct.
hartnn
  • hartnn
wait, its given that x<4
hartnn
  • hartnn
so, it'll be only \(f^{-1}(x)=4 - \sqrt{x+12}\) with same domain.
anonymous
  • anonymous
how can we understand that it is minus when x<4 ?
hartnn
  • hartnn
4+ anything means >4 4- anything means <4
hartnn
  • hartnn
does that clear it up ? or you have different doubt altogether ? :P
anonymous
  • anonymous
a little bit :D because it tells about x but we take the function <4 :/
hartnn
  • hartnn
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
hartnn
  • hartnn
***range of \(f^{-1}(x)\) is \(f^{-1}(x)\)<4
hartnn
  • hartnn
***maximum value of 4
hartnn
  • hartnn
sorry for the typos.
anonymous
  • anonymous
oh. okey. thank youu :)
hartnn
  • hartnn
welcome ^_^

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