Use the functions $f ( x ) = \frac { 1 } { 125 } x - 5$ and $g ( x ) = x ^ { 3 }$ to find $( f \circ g ) ^ { - 1 }$.
A) $( f \circ g ) ^ { - 1 } = \frac { x ^ { 3 } + 5 } { 5 }$
B) $( f \circ g ) ^ { - 1 } = \frac { x ^ { 3 } - 625 } { 125 }$
C) $( f \circ g ) ^ { - 1 } = \frac { \sqrt [ 3 ] { x + 5 } } { 5 }$
D) $( f \circ g ) ^ { - 1 } = 5 x + 5$
E) $( f \circ g ) ^ { - 1 } = 5 \sqrt [ 3 ] { x + 5 }$

Question

Quizplus · Accepted Answer

First, we need to find $f(g(x))$, which is equal to $f(x^3) = \frac{1}{125}(x^3) - 5 = \frac{x^3 - 625}{125}$.
Now we need to find the inverse of $f(g(x))$, which means we need to solve for $x$ in the equation $\frac{x^3 - 625}{125} = y$.
Multiplying both sides by 125 and then adding 625, we get $x^3 = 125y + 625$.
Taking the cube root of both sides, we get $x = \sqrt[3]{125y + 625} = 5\sqrt[3]{y + 5}$.
Therefore, $(f \circ g)^{-1} = 5\sqrt[3]{x + 5}$.

Use the Functions f(x)=1125x−5f ( x ) = \frac { 1 } { 125 } x - 5f(x)=1251​x−5

Use the Functions $f ( x ) = \frac { 1 } { 125 } x - 5$