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

The inverse of f(x)=2cos(ln(1-x))+1 is -e^{arccos((x-1)/2)}+1

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inverse of f(x)=2cos(ln(1-x))+1

inverse

f (x) = 2 cos (ln (1 - x)) + 1

- e^{a r c c o s (\frac{x - 1}{2})} + 1

Solution steps

Solve for

cos (ln (1 - x)), y = 2 cos (ln (1 - x)) + 1

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

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

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

Solve for

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

- e^{a r c c o s (\frac{y - 1}{2})} + 1

Substitute

y = x

- e^{a r c c o s (\frac{x - 1}{2})} + 1

f (x) = 4 x^{3} + 2 x^{2}

(- 2, - 3), (1, - 9)

lo g_{2} (n)

f (x) = 9 - 9 x

What is the inverse of f(x)=2cos(ln(1-x))+1 ?
The inverse of f(x)=2cos(ln(1-x))+1 is -e^{arccos((x-1)/2)}+1