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Precalculus Study Guide

Chapter 9: The Elementary Transcendental Functions

Introduction

The elementary transcendental functions include the six trigonometric functions and their inverses, the six hyperbolic functions and their inverses, the exponential function and its inverse, the logarithmic function.

A listing of these functions appears in the next section. This chapter provides tools for exploring the behavior of an elementary function $f (x)$ under transformations of the form $a f (b x + x) + d$ , where the parameters $a, b, c$ , and $d$ are real numbers.

The hyperbolic functions satisfy identities that parallel those satisfied by the trigonometric functions. These identities can be explored with one of the tools explained in this chapter. In addition, there are tools for solving equations containing exponential and logarithmic functions, and even a tool for solving more general transcendental equations.

Chapter Glossary

The following terms in Chapter 9 are linked to the Maple Math Dictionary.

amplitude

arc cosecant

arc cosine

arc cotangent

arc hyperbolic cosecant

arc hyperbolic cosine

arc hyperbolic cotangent

arc hyperbolic secant

arc hyperbolic sine

arc hyperbolic tangent

arc secant

arc sine

arc tangent

complex number

converge

cosecant

cosine

cotangent

domain

exponential

elementary function

exponential function

exponentiate

floating-point

hyperbolic cosecant

hyperbolic cosine

hyperbolic cotangent

hyperbolic function

hyperbolic secant

hyperbolic sine

hyperbolic tangent

inverse function

irrational number

iteration

midpoint

natural logarithm

one-to-one

quadratic equation

rational number

real number

reflection

secant

sequence

sine

tangent

The Elementary Transcendental Functions

The following 26 functions are generally considered to be the elementary transcendental functions.

Trigonometric Functions		Inverse Trigonometric Functions
sine	$\sin (x)$	arcsine	$\arcsin (x)$
cosine	$\cos (x)$	arccosine	$\arccos (x)$
tangent	$\tan (x)$	arctangent	$\arctan (x)$
cotangent	$\cot (x) = \frac{1}{\tan (x)}$	arccotangent	$arccot (x)$
secant	$\sec (x) = \frac{1}{\cos (x)}$	arcsecant	$arcsec (x)$
cosecant	$\csc (x) = \frac{1}{\sin (x)}$	arccosecant	$arccsc (x)$
The Exponential Function: $e^{x}$
The Natural Logarithm: $\ln (x)$

Hyperbolic Functions		Inverse Hyperbolic Functions with Logarithmic Equivalents
hyperbolic sine	$\sinh (x) = \frac{e^{x} - e^{- x}}{2}$	inverse hyperbolic sine	$arcsinh (x) = \ln (x + \sqrt{x^{2} + 1})$
hyperbolic cosine	$\cosh (x) = \frac{e^{x} + e^{- x}}{2}$	inverse hyperbolic cosine	$arccosh (x) = \ln (x + \sqrt{x^{2} - 1})$	$x \geq 1$
hyperbolic tangent	$\tanh (x) = \frac{\sinh (x)}{\cosh (x)}$	inverse hyperbolic tangent	$arctanh (x) = \ln (\sqrt{\frac{1 + x}{1 - x}})$	$\|x\| < 1$
hyperbolic cotangent	$\coth (x) = \frac{1}{\tanh (x)}$	inverse hyperbolic cotangent	$arccoth (x) = \ln (\sqrt{\frac{x + 1}{x - 1}})$	$\|x\| > 1$
hyperbolic secant	$sech (x) = \frac{1}{\cosh (x)}$	inverse hyperbolic secant	$arcsech (x) = \ln (\frac{1 + \sqrt{1 - x^{2}}}{x})$	$0 < x < 1$
hyperbolic cosecant	$csch (x) = \frac{1}{\sinh (x)}$	inverse hyperbolic cosecant	$arccsch (x) = \ln (\frac{1}{x} + \sqrt{1 + \frac{1}{x^{2}}})$	$x \neq 0$

Typical Problems

9.1. Graph $\sin (x)$ and $3 \sin (2 x - \frac{π}{4}) + 5$ on the same set of axes.

9.2. State the domain on which $\cos (x)$ is inverted to the principal branch of the arccosine function.

9.3. Prove the identity: $\cosh^{2} x - \sinh^{2} x = 1$ .

9.4. Derive the formula $arcsinh (x) = \ln (x + \sqrt{x^{2} + 1})$ .

9.5. Solve the equation $2 e^{3 x} = 5$ for $x$ .

9.6. Solve the equation $\ln (2 x + 5) = 3$ for $x$ .

9.7. Solve the equation $\cos (x) = 2 - x$ for $x$ .

Maple Initializations

Before accessing the Maple version of the solutions to the typical problems stated above, initialize Maple by pressing the button provided on the right.

Solutions

Problem 9.1

9.1 - Mathematical Solution

The lower curve in Figure 9.1.1 is a graph of $\sin (x)$ whereas the upper curve is the graph of

$g (x) = 3 \sin (2 x - \frac{π}{4}) + 5$

The amplitude of $\sin (x)$ is 1, whereas the amplitude of $g (x)$ is 3.

The angular frequency of $\sin (x)$ is 1, but the angular frequency of $g (x)$ is 2.

The phase angle for $\sin (x)$ is 0, but for $g (x)$ , it is $\frac{π}{4}$ . The center-line for $\sin (x)$ is $y = 0$ , but for $g (x)$ , it is $y = 5$ .

The required graph could have been drawn from this information, and from a knowledge of how to graph the function $\sin (x)$ .

Figure 9.1.1 Graphs of $\sin (x)$ (red) and $3 \sin (2 x - \frac{π}{4}) + 5$ (black)

9.1 - Maplet Solution

A graph of $\sin (x)$ and $3 \sin (2 x - \frac{π}{4}) + 5$ can be obtained with the Transcendental Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 9.1.2.

The Transcendental Function Tutor modifies an elementary function $f (x)$ to $a f (b x + c) + d$ . In this problem, $a = 3, b = 2, c = - \frac{π}{4}$ , and $d = 5$ , values that are entered interactively in the appropriate windows. Clicking the Graph button then produces the desired graph.

Figure 9.1.2 Thumbnail image of the Transcendental Function Tutor

Maple unexpectedly exploits the trigonometric identity

$\cos (A + \frac{π}{2}) = \cos (A) \cos (\frac{π}{2}) - \sin (A) \sin (\frac{π}{2})$ = $- \sin (A)$

Taking $A = 2 x - \frac{π}{4}$ allows Maple to write

$3 \sin (2 x - \frac{π}{4}) + 5 = - 3 \cos (2 x + \frac{π}{4}) + 5$

To launch the Transcendental Function Tutor, click the following link: Transcendental Function Tutor

9.1 - Interactive Solution

Re-initialize Maple by clicking the button to the right.

•	Type $f (x) = \sin (x)$

•	Context Panel: Assign Function

•	Type $f (x), 3 f (2 x - π / 4) + 5$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder $- 2 π \leq x \leq 2 π$ (Use 2*Pi for $2 π$ ) Options: Global Defaults & Settings Axes≻Advanced Settings≻Tickmarks≻ $x$ change "default" to "spacing(Pi,0)" Apply

9.1 - Programmatic Solution

Entering $f (x) = \sin (x)$ as a Maple function with

>	$f ≔ x \to \sin (x) &colon;' f' (x) = f (x)$

we obtain the required graph as Figure 9.1.3.

>	$plot ([f (x), 3 f (2 x - π / 4) + 5], x = - 2 π .. 2 π, color = [red, black], scaling = constrained, tickmarks = [spacing (π), default])$

Figure 9.1.3 Graphs of $\sin (x)$ (red) and $3 \sin (2 x - \frac{π}{4}) + 5$ (black)

The amplitude of $\sin (x)$ is 1, whereas the amplitude of

$g (x) = a f (b x + c) + d$

= $3 \sin (2 x - \frac{π}{4}) + 5$

is 3. The angular frequency of $\sin (x)$ is 1, but the angular frequency of $g (x)$ is 2. The phase angle for $\sin (x)$ is 0, but for $g (x)$ , it is $\frac{π}{4}$ . The center-line for $\sin (x)$ is $y = 0$ , but for $g (x)$ , it is $y = 5$ .

Problem 9.2

	9.2 - Mathematical Solution
	The domain of $\cos (x)$ that is used to define the principal branch of the arccosine function is the interval $[0, π]$ .

9.2 - Maplet Solution

The domain on which $\cos (x)$ is inverted to the principal branch of the arccosine function is the interval $[0, π]$ .

This can be deduced from the graph generated by the built-in Inverse Tutor whose home is the Student Calculus 1 package.

The inverse function, namely, $\arccos (x)$ , is entered into the tutor, and the Display button yields a plot showing the graph of this function in red, and its reflection across $y = x$ in blue. The reflection is a graph of the functional inverse of $\arccos (x)$ , that is, of $\cos (x)$ . From this blue graph, the domain of that part of $\cos (x)$ defining a one-to-one invertible function can be deduced.

The Inverse Tutor from the Calculus1 subpackage of the Student package can be accessed from the Tools menu. Follow the path

Figure 9.2.1 Thumbnail image of the built-in Inverse Tutor (Student Calculus 1)

Tools≻Tutors≻Calculus: Single Variable≻Function Inverse

Alternatively, it can be launched in a Maple worksheet with the command

Student[Calculus1][InverseTutor]();

However, the easiest way to launch most of the built-in tutors is to load the appropriate package from the Tools menu, following a path such as Tools≻Load Package: Student Calculus 1. Then, from the Context Panel of the item to be brought into the Tutor, select Tutors and the appropriate Tutor.

It can also be launched by clicking on the link: Inverse Tutor

9.2 - Interactive Solution

•	Type $[\arccos (x), x]$

•	Context Panel: Plots≻Plot Builder $- 4 \leq x \leq 4$ Options: Axes≻Advanced Settings for $x$ -axis, change Tickmarks from "default" to "spacing(Pi,0)" and change Pi to Pi/4

9.2 - Programmatic Solution

To determine the domain on which $\cos (x)$ is inverted to the principal branch of $\arccos (x)$ , reflect the graph of $\arccos (x)$ across the line $y = x$ . The resulting graph in Figure 9.2.2 will contain the portion of the graph of $\cos (x)$ that we seek. The reflection of the inverse shows the original function, but only that part belonging to the inverse function.

>	$p_{1} ≔ plot (\arccos (x), x = - 4 .. 4) &colon; display (reflect (p_{1}, [[0, 0], [1, 1]]), scaling = constrained, tickmarks = [spacing (π / 2), default], view = [0 .. π, - 1 .. 1], labels = [x, y])$

Figure 9.2.2 The functional inverse of $\cos (x)$ reflected across the line $y = x$

From Figure 9.2.2, we can infer that the domain of $\cos (x)$ used to define the principal branch of the arccosine function is the interval $[0, π]$ .

Problem 9.3

	9.3 - Mathematical Solution
	The identity $\cosh^{2} x - \sinh^{2} x = 1$ is proven by converting each hyperbolic function to its exponential equivalent. Thus, write ${(\frac{e^{x} + e^{- x}}{2})}^{2} - {(\frac{e^{x} - e^{- x}}{2})}^{2} = \frac{e^{2 x} + 2 + e^{- 2 x} - (e^{2 x} - 2 + e^{- 2 x})}{4}$ = $\frac{4}{4} = 1$

9.3 - Maplet Solution

A verification of the identity

$\cosh^{2} x - \sinh^{2} x = 1$

can be obtained with the Hyperbolic Function Identities Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 9.3.1.

To enter the identity, note that the Maple notation for expressions such as $\cosh^{2} x$ is cosh(x)^2.

To replace each hyperbolic function with its exponential equivalent, press the button labeled Convert to exponential form.

Figure 9.3.1 Thumbnail image of the Hyperbolic Function Identities Tutor

To simplify the resulting expression, press the button labeled Simplify.

To determine if the two sides of the resulting equation are the same, press the button labeled Verify. Of course, for this identity, the simplification step results in the identity $1 = 1$ , so that a visual inspection would suffice.

To launch the Hyperbolic Function Identities Tutor, click the link: Hyperbolic Function Identities Tutor

9.3 - Interactive Solution

Type the left-hand side as $\cosh^{2} (x) - \sinh^{2} (x)$

Context Panel: Conversions≻Exponential

9.3 - Programmatic Solution

The identity

$\cosh^{2} x - \sinh^{2} x = 1$

is proven by converting each hyperbolic function to its exponential equivalent. In Maple, these calculations begin with the left-hand side.

Enter the left-hand side. (The parentheses are essential.)

>	$q ≔ \cosh^{2} (x) - \sinh^{2} (x)$

Convert to exponential form.

>	$convert (q, \exp)$

Problem 9.4

9.4 - Mathematical Solution

To derive the formula

$arcsinh (x) = \ln (x + \sqrt{x^{2} + 1})$

apply the three steps for calculating the inverse of the function $f (x) = \sinh (x)$ . Begin with the equation

$y = \sinh (x)$

In order to solve for $x$ , convert the hyperbolic function to its exponential equivalent, obtaining

$y = \frac{e^{x} - e^{- x}}{2}$

Recognizing that $e^{- x} = \frac{1}{e^{x}}$ , make the substitution $z = e^{x}$ , obtaining

$y = \frac{z}{2} - \frac{1}{2 z}$

$0 = z^{2} - 2 y z - 1$

Use the quadratic formula to solve for $z = e^{x}$ , obtaining

$z = \frac{- (- 2 y) \pm \sqrt{{(- 2 y)}^{2} - 4 (- 1)})}{2}$ = $y \pm \sqrt{y^{2} + 1}$

Now, $y = \sinh (x)$ means that when $x = 0$ , $y = \sinh (0)$ = 0. If, therefore, $y = 0$ , then the solution with the positive square root, namely,

$e^{x} = y + \sqrt{y^{2} + 1}$

becomes

${&ExponentialE;}^{0} = 1$ = 0 + 1 = 1

Hence, we select this solution, and write

$e^{x} = y + \sqrt{y^{2} + 1}$

Finally, isolating $x$ by taking the natural logarithm of both sides, we have

$x = \ln (y + {\sqrt{y^{2} + 1}}^{})$

If we switch the letters, we get the desired result, namely,

$y = \ln (x + \sqrt{x^{2} + 1})$

9.4 - Maplet Solution

The Inverse Hyperbolic Function Tutor will guide a user through the derivation of the logarithmic form of an inverse hyperbolic function such as $arcsinh (x)$ .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 9.4.1.

Select $arcsinh (x)$ from the drop-down list in the tutor, noting that it appears in the form of an equation $y = arcsinh (x)$ .

In succession, press the buttons labeled as on the left in Table 9.4.1. The results are as on the right.

The choice of the correct solution of the equation quadratic in z requires the most thought. For the function $arcsinh (x)$ , the two solutions will be

Figure 9.4.1 Thumbnail image of the Hyperbolic Function Identities Tutor

$z = x \pm \sqrt{x^{2} + 1}$

Since $x$ is in the domain of $arcsinh (x)$ , it can be both positive and negative. However, $z = e^{y}$ which must always be positive. The only way to guarantee that $z$ remains positive is to choose the solution with the plus sign before the root, else since the radical will always be slightly larger than $x$ , the expression could become negative if the solution with the minus sign is chosen.

To launch the Inverse Hyperbolic Function Tutor, click the link: Inverse Hyperbolic Function Tutor

Button	Result
Solve for x	$x = \sinh (y)$
Convert to exponentials	$\sinh (y)$ replaced with $\frac{e^{y} - e^{- y}}{2}$
Set exp(y) = z	$e^{- y} = \frac{1}{z}$ and an equation quadratic in $z$ results
Solve for z	There are two solutions for $z$ . Pick the correct one
Replace z with exp(y)	Restore $e^{y}$
Solve for y	Isolate $y$
Correct Expression for y	Gives Maple's logarithmic expression for initial function
Table 9.4.1 Actions provided by Inverse Hyperbolic Function Tutor

9.4 - Interactive Solution

Stepwise Solution

•	Type the equation $y = \sinh (x)$ and press the Enter key.

•	Context Panel: Conversions≻Exponential

•	Control-drag the exponential form of the equation.

•	Change $e^{x}$ to $z$ , and $e^{- x}$ to $\frac{1}{z}$ . Press the Enter key.

•	Context Panel: Solve≻Obtain Solutions for≻ $z$

•	Type $e^{x} =$ and complete by Control-dragging the appropriate solution. Press the Enter key.

•	Context Panel: Solve≻Isolate Expression for≻ $x$

•	Control-drag, and change $x$ to $y$ and $y$ to $x$ .

•	Press the Enter key.

Conversion via Maple

•	Type arc $\sinh (x)$

•	Context Panel: Apply a Command convert ln

9.4 - Programmatic Solution

To derive the formula

$arcsinh (x) = \ln (x + \sqrt{x^{2} + 1})$

apply the three steps for calculating the inverse of the function $f (x) = \sinh (x)$ .

Enter the equation $y = \sinh (x)$ .

>	$q ≔ y = \sinh (x)$

Convert the hyperbolic function to exponential form.

>	$q1 ≔ simplify (convert (q, \exp))$

Make the substitutions $z = e^{x}$ and $\frac{1}{z} = e^{- x}$ .

>	$q2 ≔ eval (q1, \{\exp (x) = z, \exp (- x) = 1 / z\})$

Solve for $z = e^{x}$ .

>	$q3 ≔ solve (q2, z)$

Select the solution that satisfies $(x, y) = (0, 0)$ .

>	$q4 ≔ \exp (x) = q3 [1]$

Solve explicitly for $x$ .

>	$q5 ≔ isolate (q4, x)$

Switch the letters $x$ and $y$ .

>	$y = eval (rhs (q5), y = x)$

Compare to Maple's built-in conversion to log form.

>	$convert (arcsinh (x), \ln)$

Problem 9.5

9.5 - Mathematical Solution

To solve the equation

$2 e^{3 x} = 5$

first isolate the exponential term on the left, that is, divide both sides by 2, to obtain

$e^{3 x} = \frac{5}{2}$

Take the natural logarithm of both sides to obtain

$\ln (e^{3 x}) = \ln (\frac{5}{2})$

For $x$ real, we have

$\ln (e^{3 x}) = 3 x \ln (e)$ = $3 x$

so we actually have

$3 x = \ln (\frac{5}{2})$

Division by 3 completes the solution so that

$x = \frac{1}{3}$ $\ln (\frac{5}{2})$

9.5 - Maplet Solution

The equation $2 e^{3 x} = 5$ can be solved with the Transcendental Equation Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 9.5.1.

The equation is entered in the form $f (x) = 0$ , and the button labeled Graph $f (x)$ provides a plot of $f (x)$ , the expression obtained when the equation is rewritten to have zero on its right-hand side.

The button labeled Solve analytically will yield an exact solution where possible.

The button labeled Float the results will convert the exact solution to floating-point (decimal) form.

Figure 9.5.1 Thumbnail image of the Transcendental Equation Tutor

The button labeled Solve numerically will provide a numeric solution of the equation.

To launch the Transcendental Equation Tutor, click the following link: Transcendental Equation Tutor

9.5 - Interactive Solution

Immediate Maple Solution

•	Type $2 e^{3 x} = 5$ and press the Enter key. Be sure to use the exponential e, not the letter $e$ .

•	Context Panel: Solve≻Solve

Stepwise Solution

•	Type $2 e^{3 x} = 5$ and press the Enter key. Be sure to use the exponential e, not the letter $e$ .

•	Context Panel: Manipulate Equation Multiply both sides by $1 / 2$ Apply ln to both sides Press Return Steps

•	Context Panel: Simplify≻Assuming Real

•	Context Panel: Manipulate Equation Multiply both sides by $1 / 3$ Press Return Steps

9.5 - Programmatic Solution

The equation

>	$q ≔ 2 \exp (3 x) = 5$

is immediately solved for $x$ in Maple via

>	$solve (q, x)$

Proceeding stepwise, first isolate the exponential term on the left, that is, divide both sides by 2, to obtain

>	$q1 ≔ q / 2$

Take the natural logarithm of both sides to obtain

>	$map (\ln, q1)$

Although $\ln (x)$ and $e^{x}$ are inverse functions for $x$ real, the natural logarithm of a complex number is a more complicated concept. Maple will not make the simplification

$\ln (e^{3 x}) = 3 x \ln (e)$ = $3 x$

unless it knows that $x$ is real. Hence, the proper way to obtain the desired simplification is

>	$q2 ≔ map (\ln, q1) assuming x ∷ real$

Division by 3 completes the solution so that

$q2 / 3$

is obtained.

Problem 9.6

9.6 - Mathematical Solution

To solve the equation

$\ln (2 x + 5) = 3$

first exponentiate both sides, obtaining

$e^{\ln (2 x + 5)} = e^{3}$

$2 x + 5 = e^{3}$

since the simplification

$e^{\ln (f (x))} = f (x)$

is valid for any $x$ .

All that remains is to solve for $x$ , by first subtracting 5 from each side of the equation to get

$2 x = e^{3} - 5$

then dividing the resulting equation by 2, to obtain

$x = \frac{e^{3} - 5}{2}$

9.6 - Maplet Solution

The equation $\ln (2 x + 5) = 3$ can be solved with the Transcendental Equation Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 9.6.1.

The button labeled Solve analytically will yield an exact solution where possible.

The button labeled Float the results will convert the exact solution to floating-point (decimal) form.

Figure 9.6.1 Thumbnail image of the Transcendental Equation Tutor

The button labeled Solve numerically will provide a numeric solution of the equation.

To launch the Transcendental Equation Tutor, click the following link: Transcendental Equation Tutor

9.6 - Interactive Solution

Immediate Maple Solution

•	Type $\ln (2 x + 5) = 3$ and press the Enter key.

•	Context Panel: Solve≻Solve

Stepwise Solution

•	Type $\ln (2 x + 5) = 3$ and press the Enter key.

•	Context Panel: Manipulate Equation Apply exp to both sides Add $- 5$ to both sides Multiply both sides by $1 / 2$ Press Return Steps

9.6 - Programmatic Solution

The equation

>	$q ≔ \ln (2 x + 5) = 3$

is immediately solved for $x$ in Maple via

>	$solve (q, x)$

Proceeding stepwise, first exponentiate both sides, obtaining

>	$q1 ≔ map (\exp, q)$

Notice that Maple implements the simplification

$e^{\ln (f (x))} = f (x)$

immediately.

All that remains is to solve for $x$ , done stepwise by first subtracting 5 from each side of the equation via

>	$q2 ≔ q1 - (5 = 5)$

then dividing the resulting equation by 2, to obtain

$q2 / 2$

Problem 9.7

9.7 - Mathematical Solution

The equation

$\cos (x) = 2 - x$

has a real root in the interval $[2, 3]$ , as can be seen from Figure 9.7.1 where $\cos (x)$ and $2 - x$ are separately graphed in black and red, respectively. If the equation is rearranged to read

$x = 2 - \cos (x)$

it will be in the form

$x = g (x)$

where

$g (x) = 2 - \cos (x)$

If an initial guess at the root $r$ is given by $x_{0}$ , and a new approximation to the root is given by

$x_{1} = g (x_{0})$

Figure 9.7.1 Left and right hand sides of the equation $\cos (x) = 2 - x$

the first step has been taken in an iterative process that might converge to the root $r$ .

In general, under the right conditions on $g (x)$ , the sequence

$x_{k + 1} = g (x_{k}), k = 0, 1, \dots$

converges to a root $r$ .

Implementing this strategy with $x_{0} = 2$ , we get the sequence in Table 9.7.1. The sequence converges to the root $r$ that lies in the interval $[2, 3]$ .

$x_{1} = 2.416146836$

$x_{2} = 2.748203710$

$x_{3} = 2.923615314$

$x_{4} = 2.976336857$

$x_{5} = 2.986376308$

$x_{6} = 2.987978108$

$x_{7} = 2.988224469$

$x_{8} = 2.988262135$

$x_{9} = 2.988267888$

$x_{10} = 2.988268767$

Table 9.7.1 Iterates for $x = g (x)$

9.7 - Maplet Solution

The equation $\cos (x) = 2 - x$ can be solved with the Transcendental Equation Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 9.7.2.

The button labeled Solve analytically will yield an exact solution where possible.

The button labeled Float the results will convert the exact solution to floating-point (decimal) form.

Figure 9.7.2 Thumbnail image of the Transcendental Equation Tutor

For this problem, there is no analytic solution available. Hence, Maple simply rewrites the equation using $_Z$ in place of $x$ . However, floating this result gives $x = 2.988268926$ , essentially the same value found by a strictly numeric solution.

The button labeled Solve numerically provides a numeric solution of the equation.

To launch the Transcendental Equation Tutor, click the following link: Transcendental Equation Tutor

9.7 - Interactive Solution

Immediate Maple Solution

•	Type $\cos (x) = 2 - x$

•	Context Panel: Solve≻Numerically Solve

Iterative Solution

•	Type $g (x) = 2.0 - \cos (x)$

•	Context Panel: Assign Function

•	Type $g (3.0)$ and press the Enter key.

•	Type $g ()$ and Control-drag the value of $x_{k}$ between the parentheses. Press the Enter key.

•	Repeat until the digits in $x_{k}$ stop changing.

9.7 - Programmatic Solution

The real root of the equation

>	$q ≔ \cos (x) = 2 - x$

is not a rational number. In fact, Maple's solve command gives the construction

>	$r ≔ solve (q, x)$

for the root. However, if this root is subjected to the evalf command, the following numeric approximation for the root appears.

>	$evalf (r)$

Not every equation can be solved exactly, so Maple's solve command need not return a solution. When it does return a solution, it need not be in a recognizable form.

However, there is at least one real root that can be computed numerically, and for this, we use Maple's fsolve command, implemented via

>	$fsolve (q, x)$

That there is only one real root can be determined by writing the equation in the form $f (x) = 0$ , where $f (x)$ is the function whose rule is

>	$f ≔ lhs (q) - rhs (q)$

Then, the roots of the original equation are the zeros of the function $f (x)$ , which is graphed in Figure 9.7.3.

>	$plot (f, x = 2 .. 4)$

Figure 9.7.3 Graph of $f (x) = \cos (x) - 2 + x$

There are a number of different algorithms that can be used to obtain a numeric solution of the equation $f (x) = 0$ . One of the simplest to understand is the bisection method. If the signs of $f (a)$ and $f (b)$ are different, then a zero of $f (x)$ is known to lie in the interval $[a, b]$ . An approximation of this zero is the midpoint of the interval, given by $c = \frac{a + b}{2}$ . The zero will lie in that part of the interval $[a, b]$ for which the sign of $f (c)$ differs from the sign at the endpoint of the interval. Repeated application of this strategy narrows down the bound on the zero until it is at an acceptable level.

The first step in this construction is illustrated in Figure 9.7.4.

$m := (1 + 4) / 2 &colon;$
$p_{1} := plot (f, x = 1 .. 4, color = black) &colon;$
$p_{2} := arrow ([m, 0], [m, eval (f, x = m)], 0.01, 0.05, 0.2, color = black) &colon;$
$p_{3} := textplot ([2.5, 0.1, typeset (` m`)]) &colon;$ $p_{4} := textplot (\{[1.1, - 0.6, typeset (` f` (a) < 0)], [2.2, - 0.5, typeset (` f` (` m`) < 0)], [3.4, 1.3, typeset (` f` (b) > 0)]\}, align = RIGHT) &colon;$ $display ([p_{1}, p_{2}, p_{3}, p_{4}], labels = [x, f], labelfont = [TIMES, ITALIC, 12], tickmarks = [5, 4])$

Figure 9.7.4 Illustration of the bisection method for root-finding

Table 9.7.2 lists the steps of the bisection algorithm.

Step	Comment
1. Test that $f (a) f (b) < 0$ .	This checks for a root in the interval $[a, b]$ .
2. Obtain $m = \frac{b + a}{2}$ .	This is the midpoint of the interval $[a, b]$ .
3. If $f (a) f (m) < 0$ then $b = m$ . Return to step 2.	Zero of $f (x)$ lies between left endpoint and midpoint.
4. If $f (a) f (m) > 0$ then $a = m$ . Return to step 2.	Zero of $f (x)$ lies between right endpoint and midpoint.
5. If $f (a) f (m) = 0$ then stop, since $m$ is a root of $f (x) = 0$ .	Zero of $f (x)$ is at the midpoint.
Table 9.7.2 Steps in the bisection method for root-finding

An unsophisticated procedure called bis, implements the bisection algorithm, and is given below. It contains a test to be sure there is a sign-change in the initial interval. Then, a prescribed number of interval-halvings are executed.

$bis ≔ proc (f, a, b, n) local F, A, B, C, k, L, H &semi; A ≔ a &semi; B ≔ b &semi; C ≔ evalf ((A + B) / 2) &semi; if type (f, procedure) then F ≔ f else F ≔ unapply (evalf (f), x) &semi; fi &semi; L ≔ [0, [A, B], C, F (C), B - A] &semi; if evalf (F (a) &ast; F (b)) > 0 then ERROR (`root not bracketed`) &semi; fi &semi; for k from 1 to n do if evalf (F (A) &ast; F (C)) < 0 then B ≔ C else A ≔ C &semi; fi &semi; C ≔ evalf ((A + B) / 2) &semi; L ≔ L, [k, [A, B], C, F (C), B - A] &semi; od &colon; H ≔ [`k `, `` (alpha, beta), `m `, `f(m)`, abs (beta - alpha)] &colon; Matrix ([H, L]) &semi; end &colon;$

The following command guarantees that all of the results obtained by the bis command will be displayed.

>	$interface (rtablesize = infinity) &colon;$

The bis command is invoked via

>	$bis (f, 2, 4, 20)$

The first column records the number of interval-halvings, the second gives the interval in which the root is located, the third is the midpoint of the interval in the second column, the fourth is the function value, or residual, at the midpoint, and the fifth is the length of the interval in the second column.

The iterative method used earlier can likewise be implemented in Maple. Define the function

>	$g ≔ x \to 2 - \cos (x) &colon;' g' (x) = g (x)$

and execute the iteration

>	$x [0] ≔ 2.0 &semi; for k from 1 to 10 do x [k] ≔ g (x [k - 1]) &semi; od &semi; unassign (' x') &semi;$

Exercises - Chapter 9

In Exercises 9.1 - 9.16, graph the given pair of functions on the same set of axes.

9.1. $\sin (x)$ and $4 \sin (3 x - \frac{π}{3})$

9.2. $\sin (x)$ and $- 2 \sin (\frac{x}{2} + \frac{π}{6})$

9.3. $\sin (x)$ and $3 \sin (\frac{π}{2} - x)$

9.4. $\sin (x)$ and $- 2 \sin (\frac{x}{3} - \frac{π}{4})$

9.5. $\cos (x)$ and $3 \cos (2 x + \frac{π}{4})$

9.6. $\cos (x)$ and $- 3 \cos (\frac{x}{4} + \frac{π}{3})$

9.7. $\cos (x)$ and $- \cos (\frac{π}{6} - 2 x)$

9.8. $\cos (x)$ and $4 \cos (2 x - \frac{π}{6})$

9.9. $e^{x}$ and $e^{2 x}$

9.10. $e^{x}$ and $e^{- x / 2}$

9.11. $e^{x}$ and $e^{x + 1}$

9.12. $e^{x}$ and $e^{2 - 3 x}$

9.13. $\sinh (x)$ and $\sinh (2 x)$

9.14. $\sinh (x)$ and $\sinh (x - 1)$

9.15. $\cosh (x)$ and $\cosh (3 x)$

9.16. $\cosh (x)$ and $- 2 \cosh (1 - x)$

9.17. State the domain on which $\tan (x)$ is inverted to the principal branch of the arctangent function.

In Exercises 9.18 - 9.21, prove the given identities. Hint: Use the exponential equivalents of the hyperbolic functions.

9.18. $\sinh (x + y) = \sinh (x) \cosh (y) + \cosh (x) \sinh (y)$

9.19. $\sinh (x - y) = \sinh (x) \cosh (y) - \cosh (x) \sinh (y)$

9.20. $\cosh (x + y) = \cosh (x) \cosh (y) + \sinh (x) \sinh (y)$

9.21. $\cosh (x - y) = \cosh (x) \cosh (y) - \sinh (x) \sinh (y)$

In Exercises 9.22 - 9.27, prove the given identities. Hint: Use the identities in Exercises 9.18 - 9.21 and proceed as for the equivalent trigonometric identities, or simply use the exponential equivalents of the hyperbolic functions.

9.22. $\sinh (2 x) = 2 \sinh (x) \cosh (x)$

9.23. $\cosh (2 x) = \cosh^{2} x + \sinh^{2} x$

9.24. $\cosh (2 x) = 2 \cosh^{2} x - 1$

9.25. $\cosh (2 x) = 1 + 2 \sinh^{2} x$

9.26. $\sinh (x) = \sqrt{\frac{\cosh (2 x) - 1}{2}}$

9.27. $\cosh (x) = \sqrt{\frac{\cosh (2 x) + 1}{2}}$

In Exercises 9.28 - 9.33, solve the given equation for $x$ .

9.28. $4 e^{- 3 x} = 7$

9.29. $3 e^{2 - 5 x} = 21$

9.30. $5 e^{2 / x} = 19$

9.31. $3 \ln (4 x - 5) = 8$

9.32. $7 \ln (4 - x) = 5$

9.33. $2 \ln (3 x + 9) = 13$

Go to Chapter 1 2 3 4 5 6 7 8 10 11

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