Respuesta :

HomertheGenius
f ( x ) = ( 2 x + 1 )³
y = ( 2 x + 1 )³
[tex]2 x + 1 = \sqrt[3]{y} \\ 2 x = \sqrt[3]{y}- 1 \\ x = \frac{ \sqrt[3]{y}-1}{2} \\ f^{-1}(x)= \frac{ \sqrt[3]{x}-1 }{2} = g ( x )[/tex]
g` ( x ) = [tex] \frac{1}{6 \sqrt[3]{ x^{2} } } [/tex]
g`(1)= 1/6

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