Let f(x) = y
So:
\[y = 6x^3 + 3\]
\[6x^3 = y - 3\]
Divide by 6 both the sides..
\[x^3 = \frac{y}{6} - \frac{1}{2} \implies x = \sqrt[3]{\frac{y}{6} - \frac{1}{2}} \]
OpenStudy (anonymous):
Just flip x and y now and that will be your \(f^{-1}(x) ..\)
OpenStudy (anonymous):
so the final answer is
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OpenStudy (anonymous):
f^-1 (x) =3 sq x-3/6
OpenStudy (anonymous):
? is that right?
OpenStudy (anonymous):
Don't you think that will be x/6 ??
OpenStudy (mimi_x3):
abbey is right
OpenStudy (anonymous):
Rest is fine...
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OpenStudy (anonymous):
Yes you are right if you put that in brackets..
OpenStudy (mimi_x3):
\[\large f^{-1}x=\sqrt[3]{\frac{x-3}{6}} \]
OpenStudy (anonymous):
If you mean to say this :
\[f^{-1}(x) = \sqrt[3]{\frac{x - 3}{6}}\]
OpenStudy (anonymous):
Well Done...
OpenStudy (anonymous):
thank you so much guys!!!!!! i owe u BIG time
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