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Precalculus Study Guide

Chapter 7: Inverse Functions

Introduction

Algebraically, a function is defined as a set of ordered pairs $\{(x, y)\}$ in which each first element $x$ appears with just a single second element $y$ . There aren't two pairs where the same $x$ is paired with two different $y$ 's.

An invertible function is one-to-one and onto, so that each $x$ is paired with a unique $y$ . There aren't two different $x$ 's paired with the same $y$ .

The inverse function is then the set of ordered pairs $\{(y, x)\}$ in which the order of the members have been reversed.

When these definitions are applied to functions defined by rules operating on sets of real numbers, the characterization of $g (x)$ , the functional inverse for $f (x)$ , is that the compositions of $f (x)$ and $g (x)$ yield the identity function $h (x) = x$ , so that

$f (g (x)) = g (f (x))$ = $x$

The coordinate-interchange property of inverse functions implies that the graphs of a function $f (x)$ and its inverse $g (x)$ are reflections of each other across the line $y = x$ .

In this chapter, we will study the inverses of some of the elementary functions, including $\ln (x)$ , the functional inverse of the exponential function $e^{x}$ .

Chapter Glossary

The following terms in Chapter 7 are linked to the Maple Math Dictionary.

complex number

composition

domain

exponent

exponential

exponentiate

extraneous roots

floating-point

function

identity function

increasing

inverse function

invertible

logarithm

negative

one-to-one

onto

parabola

positive

quadratic equation

quadratic formula

real number

reflection

symmetry

Typical Problems

7.1. Let $f (x) = x^{2} + 2 x + 2$ have as its domain the real numbers $x$ that satisfy $x \geq - 1$ .

(a) Explain the role of the restriction $x \geq - 1$ in the domain of $f (x)$ .

(b) Find $g (x)$ , the functional inverse for $f (x)$ .

(d) Verify that $f (g (x)) = x$ = $g (f (x))$ .

7.2. Verify analytically that $f (x) = \frac{x}{x - 1}$ , $x > 1$ , is its own inverse function.

7.3. Show that for $x$ real, the inverse of $f (x) = e^{x}$ is the function $g (x) = \ln (x)$ with domain $x > 0$ .

7.4. Solve the equation $e^{4 x + 1} = 3$ .

7.5. Solve the equation $\ln (x + 1) + \ln (x + 2) = 3$ .

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 7.1

7.1 - Mathematical Solution

	7.1 (a) - Mathematical Solution
	Restricting the domain of $f (x) = x^{2} + 2 x + 2$ to the set of real numbers $x$ , $x \geq - 1$ , creates a function that is increasing, and hence one-to-one. Such a function is invertible.

7.1 (b) - Mathematical Solution

To find $g (x)$ , the functional inverse of $f (x)$ ,

1) set $y$ equal to $f (x)$ , obtaining $y = x^{2} + 2 x + 2$ ;

2) solve for $x$ , which can be done here with the quadratic formula. In fact, write

$0 = x^{2} + 2 x + 2 - y$

so that

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

Since $x$ must satisfy $x \geq - 1$ , pick the branch that is not always negative. Hence, the first branch (+) is the correct one.

3) switch the letters, obtaining from

$x = \sqrt{y - 1} - 1$

the inverse function

$y = g (x)$ = $\sqrt{x - 1} - 1$

7.1 (c) - Mathematical Solution

Figure 7.1.1 contains a graph of $f (x), g (x)$ , and the line $y = x$ .

Note that the graph of $g (x) = f^{- 1} (x)$ , the inverse of $f (x)$ , is the reflection of the graph of $f (x)$ across the line $y = x$ .

Figure 7.1.1 Graph of $f (x), g (x) = f^{- 1} (x),$ and the line $y = x$

7.1 (d) - Mathematical Solution

To verify that $f (x) = x^{2} + 2 x + 2$ and $g (x) = \sqrt{x - 1} - 1$ are inverse functions, show that $f (g (x)) = x$ and $g (f (x)) = x$ . Hence, we have

$f (g (x))$	$= {(\sqrt{x - 1} - 1)}^{2} + 2 (\sqrt{x - 1} - 1) + 2$
	= $x - 1 - 2 \sqrt{x - 1} + 1 + 2 \sqrt{x - 1} - 2 + 2$
	$= x$

and

$g (f (x))$	$= \sqrt{(x^{2} + 2 x + 2) - 1} - 1$
	$= \sqrt{{(x + 1)}^{2}} - 1$
	$= \|x + 1\| - 1$

However, the domain of $f (x)$ is restricted to $x \geq - 1$ so that $|x + 1| = x + 1$ . Hence, $g (f (x)) = x$ .

7.1 - Maplet Solution

7.1 (a) - Maplet Solution

Problem 7.1 concerns the functional inverse of the function

$f (x) = x^{2} + 2 x + 2$

$x \geq - 1$

The role of the restriction $x \geq - 1$ in the domain of $f (x)$ can be explained with the help of Inverse Function Tutor #1 .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 7.1.2.

The graph of $f (x)$ on the given domain is increasing, and hence one-to-one. This makes $f (x)$ , as defined, invertible.

To launch Inverse Function Tutor #1, click the link: Inverse Function Tutor #1

Figure 7.1.2 Thumbnail image of the Inverse Function Tutor #1

7.1 (b) - Maplet Solution

Problem 7.1 concerns the functional inverse of the function

$f (x) = x^{2} + 2 x + 2$

$x \geq - 1$

The functional inverse of $f (x)$ can be found with Inverse Function Tutor #1 .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 7.1.2 in Part (a).

Clicking the button labeled Inverse executes the three steps

1) Set $y$ equal to $f (x)$

2) Solve for $x$

3) Switch the letters

for the calculation of the rule for $g (x)$ , the functional inverse of $f (x)$ . (The details of these calculations are found, for example, in Part (b) of the Mathematical Solution of Problem 7.1.)

To launch Inverse Function Tutor #1, click the following link: Inverse Function Tutor #1

7.1 (c) - Maplet Solution

Problem 7.1 concerns the functional inverse of the function

$f (x) = x^{2} + 2 x + 2$

$x \geq - 1$

A graph of $f (x)$ , its functional inverse $g (x)$ , and the line $y = x$ is provided by Inverse Function Tutor #1 .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 7.1.2 in Part (a).

To launch Inverse Function Tutor #1, click the following link: Inverse Function Tutor #1

7.1 (d) - Maplet Solution

Problem 7.1 concerns the functional inverse of the function

$f (x) = x^{2} + 2 x + 2, x \geq - 1$

Verification that the rule for $g (x)$ , computed in Part (b), is indeed the functional inverse of $f (x)$ , consists in showing that

$f (g (x)) = x$ = $g (f (x))$

These calculations are performed in Inverse Function Tutor #2 .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 7.1.3.

Clicking the buttons labeled $f (g (x))$ and $g (f (x))$ implements the respective calculation, and shows the result, $x$ , in the appropriate window.

Figure 7.1.3 Thumbnail image of the Inverse Function Tutor #2

To launch Inverse Function Tutor #2, click the link: Inverse Function Tutor #2

7.1 - Interactive Solution

Define $f (x)$ as a function

•	Enter the equation $f (x) = x^{2} + 2 x + 2$

•	Context Panel: Assign Function

Solve $y = f (x)$ for $x = x (y)$

•	Write the equation $y = f (x)$

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

Define the function $g (y) = x (y)$

•	Write $g (y) =$ and complete by Control-dragging the appropriate solution for $x$

•	Context Panel: Assign Function

Obtain Figure 7.1.1

•	Enter $f (x), g (x), x$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 10$

Options: (Range from) $- 1 \leq y \leq 10$

Verifications

•	Enter $f (g (x))$ and press the Enter key.

•	Context Panel: Simplify≻Simplify

Enter $g (f (x))$ and press the Enter key.

Context Panel: Factor

Context Panel: Simplify≻Assuming Positive

7.1 - Programmatic Solution

Invertibility

The rule for the function $f (x)$ is given as a Maple function by entering

>	$f := x \to x^{2} + 2 x + 2 &colon;$ $' f' (x) = f (x)$

but the domain is the set of real numbers $x$ that satisfy $x \geq - 1$ .

Graphing the expression for the rule on a domain larger than the given domain gives the graph in Figure 7.1.4.

$p1 ≔ plot (f (x), x = - 4 .. 2, 0 .. 10, ytickmarks = 4) &colon; p2 ≔ arrow ([- 3, 0], ⟨0, 5⟩, .2, .4, .1, color = blue) &colon; p3 ≔ arrow ([1, 0], ⟨0, 5⟩, .2, .4, .1, color = blue) &colon; p4 ≔ arrow ([- 3, 5], ⟨3, 0⟩, .3, .5, .1, color = green) &colon; p5 ≔ arrow ([1, 5], ⟨- 1, 0⟩, .3, .5, .2, color = green) &colon; display ([p ∥ (1 .. 5)])$

Figure 7.1.4 On $- \infty < x < \infty$ , $f (x) = x^{2} + 2 x + 2$ is not one-to-one, and hence, not invertible

The graph is that of a parabola whose vertex is at the point $(- 1, - 1)$ . In general, the rule $y = x^{2} + 2 x + 2$ maps two $x$ 's to each $y$ . Hence, this rule does not define a one-to-one function and is not invertible. On the restricted domain $x \geq - 1$ , this rule defines an increasing function that is then one-to-one and invertible.

On the prescribed domain $x \geq - 1$ , the rule $y = x^{2} + 2 x + 2$ defines the increasing function $f (x)$ that is then one-to-one and hence invertible on that domain.

Obtaining the Inverse

To obtain the inverse function, apply the following three steps.

•	Set $y$ equal to $f (x)$

•	Solve for $x$

•	Switch the letters

The first and second steps are implemented together, yielding

>	$s1 ≔ solve (y = f (x), x)$

Since $x$ must satisfy $x \geq - 1$ , we must pick the solution that is not always negative. This is the first solution, selected via

>	$X ≔ s1 [1]$

Switching the letters leads to

>	$Y ≔ eval (X, y = x)$

Thus, the rule for the inverse function is

$y = \sqrt{x - 1} - 1$

which we express as the Maple function

>	$g ≔ unapply (Y, x) &colon;' g' (x) = g (x)$

Figure 7.1.5 shows a graph of $f (x)$ , the inverse function, and the line $y = x$ .

>	$p6 ≔ plot (f (x), x = - 1 .. 2, color = black) &colon; p7 ≔ plot (x, x = - 1 .. 10, color = green) &colon; p8 ≔ plot (g (x), x = - 1 .. 10, color = red) &colon; display ([p ∥ (6 .. 8)], scaling = constrained)$

Figure 7.1.5 Graph of $f (x)$ (black), $g (x) = f^{- 1} (x)$ (red), and the line $y = x$ (green)

If the graph of $f (x)$ is reflected across the line $y = x$ , the reflection will coincide with the graph of the inverse of $f (x)$ . This can be seen if Maple's reflect command from the plottools package is used. This command reflects a graph across a line, with the line being specified as a list of two points on that line.

>	$p9 ≔ reflect (p6, [[0, 0], [1, 1]]) &colon; p10 ≔ plot (g (x), x = - 1 .. 10, color = red, thickness = 3) &colon; display ([p10, p9], scaling = constrained)$

Figure 7.1.6 Graph of $g (x) = f^{- 1} (x)$ (red) and (in black) the reflection of $f (x)$ across the line $y = x$

The graphs of the reflection of $f (x)$ and the inverse of $f (x)$ coincide.

Verifying the Inverse

To verify analytically that the correct inverse has been found, show that the compositions $f (g (x))$ and $g (f (x))$ both reduce to the identity function. Thus, one must show that

$f (g (x)) = x$

and

$g (f (x)) = x$

both hold. The first of these is shown in Maple via

>	$simplify (f (g (x)))$

whereas the second is shown via

>	$simplify (g (f (x))) assuming x \geq - 1$

Problem 7.2

	7.2 - Mathematical Solution
	To show that $f (x) = \frac{x}{x - 1}$ is its own inverse, write $g (x) = \frac{x}{x - 1}$ , and show $f (g (x)) = g (f (x))$ = $\frac{(\frac{x}{x - 1})}{(\frac{x}{x - 1} - 1)}$ = $\frac{x}{x - (x - 1)} = \frac{x}{1}$ = $x$

7.2 - Maplet Solution

To verify that the function

$f (x) = \frac{x}{x - 1}$ , $x > 1$

is its own inverse, we must show that $f (f (x)) = x$ . This verification can be provided with Inverse Function Tutor #2 .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 7.2.1.

Enter the same rule for both $f (x)$ and $g (x)$ , then click either of the buttons labeled $f (g (x))$ or $g (f (x))$ to obtain the validating $x$ .

To launch Inverse Function Tutor #2, click the link: Inverse Function Tutor #2

Figure 7.2.1 Thumbnail image of the Inverse Function Tutor #2

7.2 - Interactive Solution

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

•	Enter the equation $f (x) = \frac{x}{x - 1}$

•	Context Panel: Assign Function

•	Enter $f (f (x))$ and press the Enter key.

•	Context Panel: Simplify≻Simplify

Graph $f (x)$

•	Enter $f (x)$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$0 \leq x \leq 10$

Options: (Range from) $0 \leq y \leq 10$
Constrained Scaling

•	Control-drag $x$ onto the graph

7.2 - Programmatic Solution

The rules for the function $f (x)$ and its inverse $g (x) = f^{- 1} (x)$ (which is the same rule as for $f (x)$ ), are written as Maple functions via the notation

>	$f ≔ x \to x / (x - 1) &colon; g ≔ x \to x / (x - 1) &colon;' f' (x) = f (x) &semi;' g' (x) = g (x)$

Verification that $f (g (x)) = g (f (x))$ = $x$ , consists of the following two calculations.

>	$simplify (f (g (x))) &semi; simplify (g (f (x)))$

Hence, $g (x) = f (x)$ is the inverse of $f (x)$ .

Figure 7.2.2 shows the strong symmetry exhibited by the function $f (x)$ so that it can be its own inverse.

>	$p11 ≔ plot (f (x), x = 1 .. 10, y = 0 .. 10, color = black) &colon; p12 ≔ plot (x, x = 0 .. 10, color = green) &colon; display ([p ∥ (11 .. 12)], scaling = constrained)$

Figure 7.2.2 Graph of $f (x) = \frac{x}{x - 1}$ and the line $y = x$

Problem 7.3

7.3 - Mathematical Solution

If the (real) logarithm function has been defined as the inverse of the exponential function, then the compositions $e^{\ln (x)} = x$ = $\ln (e^{x})$ are identities. If the logarithm function is defined by some other process, then the validity of these identities must be demonstrated. However, prior to a course in integral calculus, the logarithm is generally defined as the inverse of the exponential function, and these compositions reduce to $x$ by definition.

Thus, given the exponential statement $a = e^{b}$ , the logarithm of $a$ with base $b$ is defined as the exponent that must be placed on $e$ to obtain $a$ . Clearly, that exponent must be $b$ , and we write $\log_{&ExponentialE;} (a) = b$ or more simply, $\ln (a) = b$ . Consequently, $e$ raised to the exponent that produces $x$ must evaluate to $x$ . That is the content of the first composition $e^{\ln (x)} = x$ .

The second composition can be "established" by remembering one of the properties of the log function for real quantities. Write

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

by using the "multiplier" rule

$\log (A^{B}) = B \log (A)$

for real logarithms. Then, since $\ln (e) = 1$ because the exponent that must be placed on $e$ to get $e$ must be 1, we finally get

$\ln (e^{x}) = x \ln (e)$ = $x (1) = x$

7.3 - Maplet Solution

To show that for $x$ real, the inverse of $f (x) = e^{x}$ is the function $g (x) = \ln (x)$ with domain $x > 0$ , use Inverse Function Tutor #2 .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 7.3.1.

After entering the rules $f (x) = e^{x}$ and $g (x) = \ln (x)$ , click the buttons labeled $f (g (x))$ and $g (f (x))$ to obtain the validating $x$ in each case.

Clicking the button labeled Graph produces a graph of $f (x)$ in black, $g (x)$ in red, the line $y = x$ in green, and the reflection of $f (x)$ across the line $y = x$ in blue. This reflection coincides with the graph of $g (x)$ , thereby giving additional visual evidence that $g (x)$ is the functional inverse of $f (x)$ .

Figure 7.3.1 Thumbnail image of the Inverse Function Tutor #2

To launch Inverse Function Tutor #2, click the link: Inverse Function Tutor #2

7.3 - Interactive Solution

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

Note: How to Enter $e^{x}$

In Maple, the formal name of the exponential function $e^{x}$ is exp(x).

When working in Math mode, this function can be entered with its formal name, or as ${&ExponentialE;}^{x}$ , where the letter " $&ExponentialE; "$ must be entered in one of the following ways.

•	In Math mode, type $e$ and use Command Completion to select the exponential "e" from the pop-up. Command Completion is available in the Tools menu as "Complete Command" Command Completion from the keyboard: CTRL Spacebar (PC) with a Mac equivalent.

•	Select $&ExponentialE;$ from the Constants and Symbols palette.

•	Select e from the Common Symbols palette.

•	Write $f (x) = {&ExponentialE;}^{x}$

•	Context Panel: Assign Function

•	Write $g (x) = \ln (x)$

•	Context Panel: Assign Function

•	Write $f (g (x))$ and press the Enter key.

•	Write $g (f (x))$ and press the Enter key.

•	Context Panel: Simplify≻Assuming Positive

7.3 - Programmatic Solution

That $g (x) = \ln (x)$ , $x > 0$ , is the inverse of $f (x) = e^{x}$ , $x$ real, is shown by verifying the compositions $f (g (x))$ and $g (f (x))$ both simplify to $x$ . To do this in Maple, enter both functions as

>	$f ≔ \exp &colon; g ≔ \ln &colon;' f' (x) = f (x) &semi;' g' (x) = g (x)$

The composition $f (g (x)) = e^{\ln (x)}$ is obtained in Maple with

>	$f (g (x))$

Note that the simplification to $x$ is immediate.

The composition $g (f (x)) = \ln (e^{x})$ is obtained in Maple with

>	$g (f (x))$

This composition is not automatically simplified to $x$ because that simplification is valid only for real $x$ , and Maple is designed to be correct for complex numbers. To signal to Maple that $x$ is real, use

>	$g (f (x)) assuming x ∷ real$

Had we anticipated the need to work with just real quantities, we could have installed the RealDomain package. Doing that now with

>	$with (RealDomain) &colon;$

tells Maple that new quantities should be treated as real. Thus, if we enter

>	$g (f (x))$

the functions $f$ and $g$ are still treated in full generality because they were defined before the RealDomain package was initiated. However, if we now enter

>	$\ln (\exp (x)) &semi; restart$

we see that the simplification to $x$ is now immediate.

Problem 7.4

7.4 - Mathematical Solution

The equation

$e^{4 x + 1} = 3$

is solved by taking the natural logarithm of both sides. This leads to the equation

$\ln (e^{4 x + 1}) = \ln (3)$

For logarithms of positive quantities, the multiplier rule

$\ln (A^{B}) = B \ln (A)$

can be applied to give

$(4 x + 1) \ln (e) = \ln (3)$

Since $\ln (e) = 1$ , we immediately have

$4 x + 1 = \ln (3)$

from which it follows that

$4 x = \ln (3) - 1$

and

$x = \frac{\ln (3) - 1}{4}$

7.4 - Maplet Solution

A solution of the equation

$e^{4 x + 1} = 3$

can be obtained with Inverse Function Tutor #3 .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 7.4.1.

After entering the equation, clicking the button labeled Graph produces a graph of the left-hand side of the equation in black, and the right-hand side, in red. The intersection of these two curves provides the desired solution.

To obtain an analytic solution of the equation, make the appropriate choice between the buttons labeled either Exponentiate both sides or Natural log of both sides.

Figure 7.4.1 Thumbnail image of the Inverse Function Tutor #3

(In Problem 7.4, the correct choice is the latter.)

The button labeled Simplify will simplify the logarithm on the left-hand side of the equation produced by the previous button. The button labeled Solve will yield the exact value of $x$ that satisfies the given equation, whereas the button labeled Convert to floating-point will give the solution as a decimal number.

To launch Inverse Function Tutor #3, click the following link: Inverse Function Tutor #3

7.4 - Interactive Solution

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

Note: How to Enter $e^{x}$

In Maple, the formal name of the exponential function $e^{x}$ is exp(x).

When working in Math mode, this function can be entered with its formal name, or as ${&ExponentialE;}^{x}$ , where the letter " $&ExponentialE; "$ must be entered in one of the following ways.

•	In Math mode, type $e$ and use Command Completion to select the exponential "e" from the pop-up. Command Completion is available in the Tools menu as "Complete Command" Command Completion from the keyboard: CTRL Spacebar (PC) with a Mac equivalent.

•	Select $&ExponentialE;$ from the Constants and Symbols palette.

•	Select e from the Common Symbols palette.

Exact Solution

•	Enter $e^{4 x + 1} = 3$ and press the Enter key.

•	Context Panel: Solve≻Solve

•	Context Panel: Select Element≻1

•	Context Panel: Approximate≻10

Numeric Solution

•	Enter $e^{4 x + 1} = 3$ and press the Enter key.

•	Context Panel: Solve≻Numerically Solve

Stepwise Solution

•	Enter $e^{4 x + 1} = 3$

•	Context Panel: Manipulate Equation Apply $\ln$ to both sides Return Steps

•	Context Panel: Simplify≻Assuming Positive

•	Context Panel: Manipulate Equation Add $- 1$ to both sides Multiply both sides by $1 / 4$ Return Steps

7.4 - Programmatic Solution

Enter the equation $e^{4 x + 1} = 3$

>	$q ≔ \exp (4 x + 1) = 3$

Solve the equation exactly.

>	$X ≔ solve (q, x)$

Floating-point form of the exact solution:

>	$evalf (X)$

Numeric solution:

>	$fsolve (q, x)$

Solve the equation stepwise

Compute the natural logarithm of both sides of the equation.

>	$q1 ≔ map (\ln, q)$

Move the exponent on $e$ in front of the logarithm.

>	$q2 ≔ simplify (q1) assuming x ∷ real$

Subtract 1 from each side.

>	$q3 ≔ q2 - (1 = 1)$

Divide both sides by 4.

$q3 / 4$

Problem 7.5

7.5 - Mathematical Solution

The equation

$\ln (x + 1) + \ln (x + 2) = 3$

is solved by first combining the two logarithms on the left, using the rule

$\ln (A) + \ln (B) = \ln (A B)$

valid as long as all the logarithms remain real. This gives

$\ln ((x + 1) (x + 2)) = 3$

from which the logarithm on the left is removed by exponentiating both sides. Thus, we have

$e^{\ln ((x + 1) (x + 2))} = e^{3}$

Since exponential and log functions are inverse, the left side simplifies to $(x + 1) (x + 2)$ by definition, and the equation becomes

$(x + 1) (x + 2)$ = $e^{3}$

Expanding the product on the left gives

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

or the quadratic equation

$x^{2} + 3 x + 2 - e^{3} = 0$

Application of the quadratic formula gives the two solutions

$x_{1} = \frac{- 3 + \sqrt{{(- 3)}^{2} - 4 (1) (2 - e^{3})}}{2}$ = $\frac{- 3 + \sqrt{1 + 4 e^{3}}}{2}$ = 3.00949

$x_{2} = \frac{- 3 - \sqrt{{(- 3)}^{2} - 4 (1) (2 - e^{3})}}{2}$ = $\frac{- 3 - \sqrt{1 + 4 e^{3}}}{2}$ = -6.00949

where the approximations are provided by an appropriate computing device such as a calculator.

The second solution of the quadratic, namely, the negative number, is not a solution of the original equation. If that negative number were substituted for $x$ in the original equation, the logarithm of a negative number would arise. The simplification induced by use of the multiplication rule is not valid for such logarithms, and hence, the second solution of the quadratic equation is not a solution of the original equation.

7.5 - Maplet Solution

A solution of the equation

$\ln (x + 1) + \ln (x + 2) = 3$

can be obtained with Inverse Function Tutor #3 .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 7.5.1.

To obtain an analytic solution of the equation, make the appropriate choice between the buttons labeled either Exponentiate both sides or Natural log of both sides.

Figure 7.5.1 Thumbnail image of the Inverse Function Tutor #3

(In Problem 7.5, the correct choice is the former.)

The button labeled Simplify will simplify the exponential on the left-hand side of the equation produced by the previous button. The button labeled Solve will yield the exact value of $x$ that satisfies the given equation, whereas the button labeled Convert to floating-point will give the solution as a decimal number.

To launch Inverse Function Tutor #3, click the following link: Inverse Function Tutor #3

7.5 - Interactive Solution

Exact Solution

•	Enter the equation $\ln (x + 1) + \ln (x + 2) = 3$ and press the Enter key.

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

•	Context Panel: Approximate≻10

Direct Numeric Solution

•	Enter the equation $\ln (x + 1) + \ln (x + 2) = 3$ and press the Enter key.

•	Context Panel: Solve≻Numerically Solve

Obtain the graph in Figure 7.5.1

•	Enter $\ln (x + 1) + \ln (x + 2)$ , the left side of the equation.

•	Context Panel: Plots≻Plot Builder

•	Control-drag the number 3 onto the graph.

Stepwise Solution

•	Enter the equation $\ln (x + 1) + \ln (x + 2) = 3$ and press the Enter key.

•

Context Panel: Manipulate Equation
        Apply exp to both sides
        Apply simplify to left side
        Apply expand to left side
        Add $- 2$ to both sides
        Add ${(3 / 2)}^{2} = 9 / 4$ to both sides
        Apply factor to left side
        Take square root of both sides
        Choose root with $+ \sqrt{}$
        Add $- 3 / 2$ to both sides
        Return Steps

•	Context Panel: Approximate≻10

7.5 - Programmatic Solution

Enter the equation.

>	$q ≔ \ln (x + 1) + \ln (x + 2) = 3$

Draw Figure 7.5.2, a graph of the left-hand side (in black) and the right-hand side (in red). A solution to the equation occurs at any intersection of the two curves. There is at least one solution to the equation, approximately $x = 3$ , found by clicking the cursor on the intersection of the red and black curves, and reading the coordinate in the upper-left corner of the screen.

>	$plot ([lhs (q), rhs (q)], x = - 3 .. 10, color = [black, red], legend = [Left Side of Equation, Right Side of Equation]) &semi;$

Figure 7.5.2 Graph of $y = \ln (x + 1) + \ln (x + 2)$ and $y = 3$

Obtain an exact solution.

>	$X ≔ solve (q, x)$

Convert this exact solution to floating-point form.

>	$evalf (X)$

Find the numeric solution directly.

>	$fsolve (q, x)$

Stepwise Solution

Combine the two logs on the left with the law

$\ln (A) + \ln (B) = \ln (A B)$

valid for real logarithms.

>	$q1 ≔ combine (q) assuming x > - 1$

Exponentiate both sides of the resulting equation.

>	$q2 ≔ map (\exp, q1)$

Multiply the factors on the left.

>	$q3 ≔ expand (q2)$

Bring all terms to the left.

>	$q4 ≔ q3 - \exp (3)$

Solve the resulting quadratic exactly.

>	$QX ≔ solve (q4, x)$

Express these two exact solutions in floating-point form.

>	$evalf ([QX])$

Note that there are two solutions to the quadratic equation, but only one solution satisfies the original equation. The negative number is not a solution of the original equation because substitution of that value into either $\ln (x + 1)$ or $\ln (x + 2)$ will result in the log of a negative number. Since the logarithm of a negative number is a complex number, the negative solution of the quadratic equation cannot be a solution of the original equation.

This extraneous solution was introduced when the logarithms were combined with the rule

$\ln (A) + \ln (B) = \ln (A B)$

which is valid only as long as the logarithms remain real.

Exercises - Chapter 7

For each function $f (x)$ given in Exercises 7.1 - 7.10,

(a) find $g (x)$ , the functional inverse for $f (x)$ ;

(b) graph $f (x), g (x)$ , and the line $y = x$ on the same set of axes;

If no restriction is imposed on the rule for $f (x)$ , then the domain is the largest set of real numbers for which the rule is defined.

7.1. $f (x) = 3 x - 5$

7.2. $f (x) = 2 - 7 x$

7.3. $f (x) = x^{2} + 6 x + 45, x > - \frac{1}{3}$

7.4. $f (x) = 6 - x - x^{2}$ , $x > - \frac{1}{2}$

7.5. $f (x) = \sqrt{3 x + 2}$

7.6. $f (x) = \frac{10}{x^{2} + 10 x + 25}, x > - 5$

7.7. $f (x) = \frac{1}{x^{2} + 1}, x > 0$

7.10. $f (x) = x^{2 / 3}, x > 0$

Solve the equations in Exercises 7.11 - 7.17. Be sure to verify that the answers obtained are indeed solutions of the original equations, and not extraneous solutions generated by the solution process.

7.11. $4 e^{3 x - 1} = 5$

7.12. $5 e^{2 - 3 x} = e^{3 - 2 x}$

7.13. $3 e^{2 x^{2}} = 2 e^{3 x}$

7.14. $\ln (5 x - 4) + \ln (6 x + 1) = \ln (7 x + 3)$

7.15. $\ln (2 x^{2} + 1) = 1$

7.16. $\ln (3 x + 2) - 2 \ln (4 x - 7) = 1$

7.17. $2 \ln (5 x - 4) = 2 \ln (6 x + 1) - 1$

7.18. It is proposed to solve the equation

$5 e^{- 4 x} = 7 e^{6 x^{2}} + 3$

by the steps listed below. Determine if the calculation is correct. If it is incorrect, find and explain the error.

$\ln (5) - 4 x = \ln (7) + 6 x^{2} + \ln (3)$

$6 x^{2} + 4 x + \ln (\frac{21}{5}) = 0$

$x = \frac{- 4 \pm \sqrt{16 - 4 (6) \ln (\frac{21}{5})}}{12}$

Go to Chapter 1 2 3 4 5 6 8 9 10 11

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