Hello, Can you please verify if I have the correct answer? y=2x^3+3 the inverse should be x=Cuberoot(y-3/2)
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Originally Posted by l flipboi l Hello, Can you please verify if I have the correct answer? y=2x^3+3 the inverse should be x=Cuberoot(y-3/2) well yes but no, you swap x and y and solve for y. You mean to say $\displaystyle y = \sqrt[3]{\frac{x-3}{2}}$ Your answer could be mistaken for $\displaystyle y = \sqrt[3]{x-\frac{3}{2}}$
Last edited by pickslides; Jan 14th 2010 at 11:49 AM. Reason: wrong variable
Follow This : $\displaystyle y=2(x^3)+3 \rightarrow \frac {y-3}{2} = x^3 \rightarrow x=\sqrt[3]{\frac {y-3}{2}} \rightarrow f^(-1)(x) = \sqrt[3]{\frac {x-3}{2}} $
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