Problem 6 - Calculus Early Transcendentals

Question

What is the domain of \(\operatorname{sech}^{-1} x ?\) How is \(\operatorname{sech}^{-1} x\) defined in terms of the inverse hyperbolic cosine?

Step-by-Step Solution

Answer

Answer: The domain of the inverse hyperbolic secant function, \(\operatorname{sech}^{-1} x\), is \((0, 1]\). It can be defined in terms of the inverse hyperbolic cosine function as \(\operatorname{sech}^{-1} x = \operatorname{cosh}^{-1}\left(\frac{1}{x}\right)\).

Step 1: Determine the range of the hyperbolic secant function

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 \((0, 1]\).

Step 2: Find the domain of the inverse hyperbolic secant function

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 \((0, 1]\).

Step 3: Define the inverse hyperbolic secant in terms of inverse hyperbolic cosine

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: \(\operatorname{sech}^{-1} x = \operatorname{cosh}^{-1}\left(\frac{1}{x}\right)\).