What is the inverse of f(x)=2cos(3x)+1 ?

The inverse of f(x)=2cos(3x)+1 is (arccos((x-1)/2))/(3)

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inverse of f(x)=2cos(3x)+1

inverse

f (x) = 2 cos (3 x) + 1

\frac{arccos ( \frac{x - 1}{2} )}{3}

Solution steps

Solve for

cos (3 x), y = 2 cos (3 x) + 1

cos (3 x) = \frac{y - 1}{2}

cos (a) = b \Rightarrow a = arccos (b)

3 x = arccos (\frac{y - 1}{2})

Solve for

x, 3 x = arccos (\frac{y - 1}{2})

\frac{arccos ( \frac{y - 1}{2} )}{3}

Substitute

y = x

\frac{arccos ( \frac{x - 1}{2} )}{3}

\frac{1}{6} x^{3} - 3

y = 5 x - 3

f (x) = 2^{- x}

\frac{5}{9} (x - 32)

f (x) = \frac{1}{- 2 x}

What is the inverse of f(x)=2cos(3x)+1 ?
The inverse of f(x)=2cos(3x)+1 is (arccos((x-1)/2))/(3)