Respuesta :
From the identity:
[tex]sec(x)= \frac{1}{cos(x)} [/tex]
[tex]f(x)=sec(x)= \frac{1}{cos(x)} [/tex]
the inverse of f is g such that f(g(x))=x,
we must find g(x), such that [tex] \frac{1}{cos[g(x)]}=x [/tex]
thus, [tex]cos[g(x)]= \frac{1}{x} [/tex]
[tex]g(x)=cos^{-1} (\frac{1}{x}) [/tex]
Answer: b. g(x)=cos^-1(1/x)
[tex]sec(x)= \frac{1}{cos(x)} [/tex]
[tex]f(x)=sec(x)= \frac{1}{cos(x)} [/tex]
the inverse of f is g such that f(g(x))=x,
we must find g(x), such that [tex] \frac{1}{cos[g(x)]}=x [/tex]
thus, [tex]cos[g(x)]= \frac{1}{x} [/tex]
[tex]g(x)=cos^{-1} (\frac{1}{x}) [/tex]
Answer: b. g(x)=cos^-1(1/x)
