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, denoted by \(\operatorname{sech}^{-1} x\), is \((0, 1]\). It can be expressed in terms of the inverse hyperbolic cosine function as \(\operatorname{sech}^{-1} x = \operatorname{cosh}^{-1}\left(\frac{1}{x}\right)\).

Step 1: Find the domain of \(\operatorname{sech}^{-1} x\)

To find the domain of the inverse function, we need to find the range of the original function, which is the hyperbolic secant function, \(\operatorname{sech}(x)\). The hyperbolic secant can be defined in terms of the hyperbolic cosine function, such that \(\operatorname{sech}(x) = \frac{1}{\operatorname{cosh}(x)}\). The hyperbolic cosine function is always non-negative, and it increases without bound as the input value approaches positive or negative infinity. Therefore, the smallest value of \(\operatorname{sech}(x)\) occurs when the denominator, \(\operatorname{cosh}(x)\), is at its minimum (\(1\)). Thus, the range of \(\operatorname{sech}(x)\) is \((0, 1]\). Since the domain of the inverse function is the range of the original function, the domain of \(\operatorname{sech}^{-1} x\) is \((0, 1]\).

Step 2: Define \(\operatorname{sech}^{-1} x\) in terms of the inverse hyperbolic cosine

We know that \(\operatorname{sech}(x) = \frac{1}{\operatorname{cosh}(x)}\). To define the inverse hyperbolic secant function in terms of the inverse hyperbolic cosine function, we need to express \(x\) in terms of \(\operatorname{cosh}^{-1}\). Let \(y = \operatorname{sech}^{-1}(x)\). Then, we have \(\operatorname{sech}(y) = x\). By substituting the definition of \(\operatorname{sech}(y)\), we get \(\frac{1}{\operatorname{cosh}(y)} = x\). Now, we solve for \(y\): $$\operatorname{cosh}(y) = \frac{1}{x}$$ Taking the inverse hyperbolic cosine on both sides: $$y = \operatorname{cosh}^{-1}\left(\frac{1}{x}\right)$$ Since \(y\) is defined as \(\operatorname{sech}^{-1}(x)\), we can rewrite the expression as: $$\operatorname{sech}^{-1} x = \operatorname{cosh}^{-1}\left(\frac{1}{x}\right)$$ Thus, we have defined \(\operatorname{sech}^{-1} x\) in terms of the inverse hyperbolic cosine function.