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How do you solve f^-1(3)= when f(x)=2x-7 /3?
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First you need to find the inverse of \(f(x)\).
\(f(x) = \Large\frac{2x-7}{3}\) or \(f(x) = 2x-\Large\frac{7}{3}\) ?
The first one
Ok. Take \(f(x)\) as \(y=\Large\frac{2x-7}{3}\) . Now swap \(x\) and \(y\) and solve for \(y \) again. Give a try.
y=2x-7/3 x=2y-7/3 3x=2y-7 3x+7=2y 3x+7/2=y Is this right?
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Correct. Now \(f^{-1}(x) = \large\frac{3x+7}{2}\). Substitute 3 for \(x\) and solve. :)
f^-1(x)=3x+7/2 f^-1(3)=3(3)+7/2 f^-1(3)=16/2 f^-1(3)= 8 Is this right?
That's right!
Thank you:)
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