Given $f (x) = \frac{1}{8} x - 3$ $f(x)=1/8x-3$ and $g (x) = x^{3}$ $g(x)=x^3$ , how do you find $(g^{- 1} o g^{- 1}) (- 4)$ $(g^-1og^-1)(-4)$ ?

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Answer 1 · 2017-02-25T14:17:27

$- \sqrt[9]{4}$ $-root(9) 4$

What is the inverse of $g$ $g$ ?

$g (x) = y \Rightarrow x = g^{- 1} (y)$ $g(x) = y => x = g^-1(y)$

$x^{3} = y = \Rightarrow x = \sqrt[3]{y} \Rightarrow g^{- 1} (x) = \sqrt[3]{x}$ $x^3 = y ==> x = root(3) y => g^-1(x) = root(3) x$

$g^{- 1} (- 4) = \sqrt[3]{- 4}$ $g^-1(-4) = root(3) {-4}$

$g^{- 1} (g^{- 1} (- 4)) = g^{- 1} (\sqrt[3]{- 4}) = \sqrt[3]{\sqrt[3]{- 4}}$ $g^-1(g^-1(-4)) = g^-1(root(3) {-4}) = root(3) {root(3) {-4}}$

Given f(x)=1/8x-3f(x)=18x−3f(x)=1/8x-3 and g(x)=x^3g(x)=x3g(x)=x^3, how do you find (g^-1og^-1)(-4)(g−1og−1)(−4)(g^-1og^-1)(-4)?