Questions: Is the function (h(x)=frac1ln left(frac1x+1right)+1) invertible? If so, find an expression for its inverse. (h^-1(x)=e^(1-x) / x-1) (h^-1(x)=frac1e^1 /(x+1)+1) (h^-1(x)=frac1-e^(1-x) / xe^(1-x) / x) (h^-1(x)=frac1-e^1 / ye^1 / y) (h^-1(x)=frace^1 / y+2e^1 / y+1) (h(x)) does not have an inverse.

Solution Steps

To determine if the function \( h(x) = \frac{1}{\ln \left(\frac{1}{x+1}\right)+1} \) is invertible, we need to check if it is a one-to-one function. A function is one-to-one if it is either strictly increasing or strictly decreasing. We can check this by analyzing the derivative of the function. If the derivative is always positive or always negative, the function is one-to-one and thus invertible. If the function is invertible, we can find its inverse by solving the equation \( y = h(x) \) for \( x \).

To check if the function \( h(x) = \frac{1}{\ln \left(\frac{1}{x+1}\right)+1} \) is invertible, we first find its derivative. The derivative is given by:

\[ h'(x) = \frac{1}{(x + 1) \left(\ln \left(\frac{1}{x+1}\right) + 1\right)^2} \]

The derivative \( h'(x) \) is positive for all \( x \) in the domain of \( h(x) \). This indicates that the function is strictly increasing, and therefore, it is one-to-one and invertible.

Since the function is invertible, we solve the equation \( y = \frac{1}{\ln \left(\frac{1}{x+1}\right)+1} \) for \( x \). Solving this equation gives us the inverse function:

\[ h^{-1}(y) = e^{\frac{y - 1}{y}} - 1 \]

Final Answer