Question

determine a função inversa f(f^-1), em cada caso:

a)f(x)= 3x^3   1


b)


c) $f(x)= \frac{x}{x+1}$


d) $f(x)= \frac{1}{ \sqrt[3]{x 2} }$

720ce08994a0b425dca95f060aadb8db.pdf

Answer 1

a) $f(x) = 3x^3-1\\\\ y = 3x^3-1\\\\ x = 3y^3-1\\\\ x+1 = 3y^3\\\\ y^3 = \frac{x+1}{3}\\\\ y = \sqrt[3]{\frac{x+1}{3}}\\\\\ \boxed{f^-(x) =\sqrt[3]{\frac{x+1}{3}}}$

b) $f(x) = \sqrt[5]{4x+2} \\\\ y = \sqrt[5]{4x+2} \\\\ x = \sqrt[5]{4y+2}\\\\ (x)^5 = (\sqrt[5]{4y+2})^5\\\\ x^5 = 4y+2\\\\ y = \frac{x^5-2}{4}\\\\ \boxed{f(^-x)=\frac{x^5-2}{4}}$

c) $f(x) = \frac{x}{x+1}\\\\ y = \frac{x}{x+1}\\\\ x = \frac{y}{y+1}\\\\ x(y+1) = y\\\\ xy+x = y\\\\ xy-y = -x\\\\ y(x-1) = -x\\\\ y = \frac{-x}{x-1}\\\\ \boxed{f(^-x) = \frac{x}{1-x}}$

d) $f(x) =\frac{1}{ \sqrt[3]{x-2} }\\\\ y = \frac{1}{ \sqrt[3]{x-2} }\\\\ x = \frac{1}{ \sqrt[3]{y-2} }\\\\ x( \sqrt[3]{y-2}) = 1\\\\ (\sqrt[3]{y-2})^3 = (\frac{1}{x})^3\\\\ y-2 = \frac{1}{x^3}\\\\ y = x^-^3+2\\\\ \boxed{f(-^x) = x^-^3+2}$