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