91Ó°ÊÓ

Recall that the hyperbolic secant function is defined as \(\operatorname{sech} x = \frac{2}{e^x + e^{-x}}\). We can see that as \(x\) approaches infinity or negative infinity, the function converges to 0. Moreover, when \(x=0\), the function equals \(\frac{2}{e^0 + e^{-0}} = \frac{2}{1+1} = 1\). Thus, the range of \(\operatorname{sech} x\) is \((0, 1]\).

As explained in step 1, we found the range of the original hyperbolic secant function to be \((0, 1]\). Since the domain of an inverse function is the range of the original function, the domain of the inverse hyperbolic secant function, \(\operatorname{sech}^{-1} x\), is \((0, 1]\).

By definition, the inverse hyperbolic secant function is the inverse function of the hyperbolic secant function. We can rewrite the hyperbolic secant function as \(\operatorname{sech} x = \frac{1}{\operatorname{cosh} x}\), where \(\operatorname{cosh}x\) is the hyperbolic cosine function. From here, we can find the inverse function: 1. Let \(y = \operatorname{sech}^{-1} x\) 2. Applying the hyperbolic secant function to both sides: \(\operatorname{sech} y = x\) 3. Rewriting in terms of the hyperbolic cosine function: \(\frac{1}{\operatorname{cosh} y} = x\) 4. Solving for \(\operatorname{cosh} y\): \(\operatorname{cosh} y = \frac{1}{x}\) 5. Applying the inverse hyperbolic cosine function to both sides: \(y = \operatorname{cosh}^{-1}\left(\frac{1}{x}\right)\) So, we can define the inverse hyperbolic secant function in terms of the inverse hyperbolic cosine function as follows: \(\operatorname{sech}^{-1} x = \operatorname{cosh}^{-1}\left(\frac{1}{x}\right)\).