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

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

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

simplify

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

\circ x

Solution steps

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

f^{- 1} f = f^{- 1 + 1}

= \circ f^{- 1 + 1} x

Add/Subtract the numbers:

- 1 + 1 = 0

= \circ f^{0} x

Apply rule

a^{0} = 1, a \neq = 0

f^{0} = 1

= 1 \circ x

Multiply:

1 \cdot x = x

= \circ x

2^{2^{2^{2}}}

1.02 (lo g_{10} (1228000))^{- 2.5}

(x + 4)^{2} (x - 5)^{2}

(2 x + 3)^{2} (2 x - 3)^{2}

\frac{3}{2 w}