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find the inverse f^-1 of f(x) = 3x^2, where x ≥ 0. Then find f^-1(12)

 Mar 28, 2020
 #2
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To find the inverse, switch x and y in the equation, making it \(x=3y^2\) and plugging in 12, you get 2 or -2. 

 Mar 28, 2020
 #3
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First, we solve f^-1, which can be done by setting f(y)=x...

 

Thus, we have 3y^2=x

\(y=\frac{\sqrt{3x}}{3}\).

 

Thus, f^-1(12) is \(y=\frac{\sqrt{3*12}}{3}=\pm2.\)

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 Mar 28, 2020
edited by tertre  Mar 28, 2020

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