What is (f\circ g^{-1})^{-1} ?

The solution to (f\circ g^{-1})^{-1} is g/(\circ f)

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simplify (f\circ g^{-1})^{-1}

simplify

(f \circ g^{- 1})^{- 1}

\frac{g}{\circ f}

Solution steps

(f \circ g^{- 1})^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{\circ f g ^{- 1}}

Multiply

f \circ g^{- 1} : \frac{\circ f}{g}

= \frac{1}{\frac{\circ f}{g}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{g}{f \circ}

e^{x} \times e^{- x}

e^{- 12.394}

e^{- 2.79}

3 \cdot 140000 \cdot e^{- 3 \cdot 2}

\frac{x ^{a - 1}}{1 ^{a - 1}}

What is (f\circ g^{-1})^{-1} ?
The solution to (f\circ g^{-1})^{-1} is g/(\circ f)