What is f^{-1}(2x+cos(x)) ?

The solution to f^{-1}(2x+cos(x)) is (2x+cos(x))/f

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simplify f^{-1}(2x+cos(x))

simplify

f^{- 1} (2 x + cos (x))

\frac{2 x + cos ( x )}{f}

Solution steps

f^{- 1} (2 x + cos (x))

Apply exponent rule:

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

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

= \frac{1}{f} (2 x + cos (x))

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ( 2 x + cos ( x ) )}{f}

Apply rule:

1 \cdot a = a

1 \cdot (2 x + cos (x)) = (2 x + cos (x))

= \frac{2 x + cos ( x )}{f}

x^{2} (x + 1)^{2} (2 x^{2} + 2 x + 1) (2 x^{2} - 2 x + 1)

(3)^{2} \cdot 3

- e^{(- 1)}

2 x (x)^{3}