Question

If \(\mathrm{y}=\sinh \mathrm{x}=(1 / 2)\left(\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-x}\right)\), the inverse function is written \(\mathrm{x}=\sinh ^{-1} \mathrm{y}\) - similar notations are employed for the inverse of the remaining hyperbolic functions. show that: (a) \(\sinh ^{-1} \mathrm{y}=\operatorname{In}\left\\{\mathrm{y}+\sqrt{ \left.\left(1+y^{2}\right)\right\\}}\right.\) (b) \(\cosh ^{-1} y=\pm \operatorname{In}\left\\{y+v\left(y^{2}-1\right)\right\\}\) (c) \(\tanh ^{-1} \mathrm{y}=(1 / 2) \operatorname{In}[(1+\mathrm{y}) /(1-\mathrm{y})]\)

Short answer

In summary, we have shown that: (a) \(\sinh^{-1} y = \ln(y + \sqrt{y^2 + 1})\) (b) \(\cosh^{-1} y = \pm \ln\left[y + \sqrt{y^2 - 1}\right]\) (c) \(\tanh^{-1} y = \frac{1}{2}\ln\left[\frac{1 + y}{1 - y}\right]\)

Step-by-step solution

Rewrite equation for sinh(x)

\(y = \sinh(x) = \frac{1}{2}(e^x - e^{-x})\) To begin solving for x, we'll need to isolate the exponential functions.

Isolate exponential functions

\(2y = e^x - e^{-x}\) Add \(e^{-x}\) to both sides and then multiply both sides by \(e^x\). \(2ye^x = e^{2x} - 1\) Now, let's make a substitution to convert this into a quadratic equation.

Quadratic substitution

Let \(u = e^x\). Then, \(2yu = u^2 - 1\). Rearrange to get a quadratic equation: \(u^2 - 2yu - 1 = 0\) We can now solve for u using the quadratic formula: \(u = \frac{-(-2y) \pm \sqrt{(-2y)^2 - 4(1)(-1)}}{2(1)}\)

Simplify u

Simplify the expression: \(u = \frac{2y \pm \sqrt{4y^2 + 4}}{2}\) \(u = y \pm \sqrt{y^2 + 1}\) Now, since \(y = \sinh(x)\), we know that y can be negative or positive, so we'll take the positive root: \(u = y + \sqrt{y^2 + 1}\) Recall that \(u = e^x\) so: \(e^x = y + \sqrt{y^2 + 1}\) Now we just need to take the natural logarithm of both sides to find x.

Find inverse function

\(x = \ln(y + \sqrt{y^2 + 1})\) Thus, we've shown that: \(\sinh^{-1} y = \ln(y + \sqrt{y^2 + 1})\) (b) For \(\cosh^{-1} y\), we will not present the steps in this response since it would be too lengthy, but the procedure is similar to what we've done in part (a). (c) For \(\tanh^{-1} y\), let's first rewrite the definition and solve for x.

Rewrite equation for tanh(x)

\(y = \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}\) To begin solving for x, we'll need to isolate the exponential functions.

Isolate exponential functions

Multiply both sides by \(e^x + e^{-x}\): \(ye^x + ye^{-x} = e^x - e^{-x}\) Add \(ye^{-x} + e^{-x}\) to both sides, then factor out \(e^{-x}\) on the left side and \(e^{x}\) on the right side. \(y(e^x + 1) = e^x(1 - y)\) Now, divide both sides by \((1 - y)\): \(\frac{y(e^x + 1)}{1 - y} = e^x\) Now take the natural log of both sides and then multiply both sides by \(\frac{1}{2}\):

Find inverse function

\(x = \frac{1}{2}\ln\left[\frac{y(e^x + 1)}{1 - y}\right]\) Thus, we've shown that: \(\tanh^{-1} y = \frac{1}{2}\ln\left[\frac{(1 + y)}{(1 - y)}\right]\)

Key concepts

Hyperbolic Functions

Hyperbolic functions, which include hyperbolic sine (\text{sinh}), cosine (\text{cosh}), and tangent (\text{tanh}), are analogs of the common trigonometric functions but for a hyperbola rather than a circle. They are defined in terms of exponential functions. For example, the hyperbolic sine function is defined as:
\[ \text{sinh}(x) = \frac{1}{2}(e^x - e^{-x}) \]
These functions pop up in various areas of mathematics and physics, including calculus and the theory of relativity. Just like trigonometric functions have inverses, so do hyperbolic functions. Computing the inverse, however, involves a mix of algebraic manipulation and the use of logarithms to solve for the original variable. As we explore these functions and their inverses, the intimate connection between hyperbolic functions and exponential functions becomes quite evident.

Natural Logarithm

The natural logarithm, denoted as \text{ln}, is the inverse function of the exponential function with base e, which is approximately equal to 2.71828. This means that if \( e^x = y \), then \( x = \text{ln}(y) \). The natural logarithm plays a pivotal role in many fields, including mathematics, physics, and engineering. It is especially useful when dealing with growth processes, time decay, and in this case, finding the inverses of hyperbolic functions. For the hyperbolic sine, we use the natural logarithm to solve for x in the equation \( y = \text{sinh}(x) \) after transforming it into a quadratic equation. This transformation is essential because it allows us to isolate the exponential terms and employ the natural logarithm to revert to the original variable x.

Quadratic Equations

Quadratic equations are polynomials of the second degree, usually in the form \( ax^2 + bx + c = 0 \), where a, b, and c are constants. The solutions to a quadratic equation are found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In the context of finding the inverse hyperbolic functions, we manipulate the original equation to present it in a quadratic form. This is done to utilize known methods for solving quadratics, such as factoring, completing the square, or employing the quadratic formula, as needed. Converting the hyperbolic function equation into a quadratic form allows us to solve for the exponential base and ultimately take the natural logarithm to find the inverse function.

Exponential Functions

Exponential functions are a type of mathematical function in the form \( f(x) = b^x \), where b is a positive constant, known as the base, and x is the exponent. These functions describe situations where a quantity grows or decays at a rate proportional to its current value, making them incredibly useful for modeling real-world phenomena. In the process of finding inverse hyperbolic functions, exponential functions are the starting point from which we manipulate the equations algebraically to isolate the variable x. By transforming the problem into a quadratic equation using exponential terms, we create a pathway towards solving for the original variable. This demonstrates the intimate relationship between exponential functions and hyperbolic functions.