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Please help! Medal And Fan! Will post equation below. I need to go to sleep but have been stumped on this for hours, hopefully someone can help me through the process while I sleep tonight. Thank you!!!
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The one-to-one function f is defined below. \[f(x)=\sqrt{5x-5}\] Find f^-1, the inverse of f. Then give the domain of f^-1 using interval notation.
hi
to find the inverse of f(x) = y swap x and y, then solve for y
x = sqrt(5y-5)
x^2 = 5y - 5 \[f ^{-1}(x)= \frac{ 1 }{ 5 }x ^{2} + 1\]
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\[x \ge 0 \] is a restriction from the original function
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