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What is the inverse of h? h(x) = 6x + 1
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substitute y for h(x) and rewrite y = 6x + 1. now swap x with y... so x = 6y+1 now solve for y
can u do the rest?
im stuck with x/6-1/6
x = 6y + 1 -1 = 6y \[y =-\frac{ 1 }{ 6 }\]
would the answer be.... 1/6(x-1)?
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am i wrong?
right?
\[h ^{-1}(x)=-\frac{ 1 }{ 6 }x+1\]
sorry was trying to get my kids to bed.
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