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Find f^-1 for the function f(x)=^3sqrt x-2 +8. A)f^-1(x)=(x-8)^3+2 B)f^-1(x)=(x+8)^3+2 C)f^-1(x)=^3sqrt x-8 +2 D)f^-1(x)=(x-8)^3-2
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could this be \[f(x)=\sqrt[3]{x-2}+8\]?
yes satellite73
First, your presentation needs work, as satellite73 is hinting by asking for clarification.
i would set \[y=\sqrt[3]{x-2}+8\] then switch \(x\) and \(y\) because that is what the inverse does, and solve \[x=\sqrt[3]{y-2}+8\] for \(y\) in three steps
clear how to do that? if not, try solving \[\sqrt[3]{y-2}+8=5\] the steps will be the same
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