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Chapter 3: Applications of Differentiation

Section 3.7: What Derivatives Reveal about Graphs

Example 3.7.4

Graph $f (x) = \frac{6 - 5 x + 3 x^{2} - x^{3}}{3 - 4 x + x^{2}}$ for $x \in [- 5, 10]$ , indicating any asymptotes this rational function might have. Then, use the tools of the calculus to analyze the features of this graph.

Solution

Initialize

•	Tools≻Load Package: Student Calculus 1

Loading Student:-Calculus1

•	Control-drag $f (x) = \dots$

•	Context Panel: Assign Function

$f (x) = \frac{6 - 5 x + 3 x^{2} - x^{3}}{3 - 4 x + x^{2}}$ $\overset{assign as function}{\to}$ $f$

Rational Function Tutor

•	Figure 3.7.4(a) contains an image of the tutor applied to $f (x)$ .

•	If this tutor is launched from the Tools menu, the numerator and denominator of the rational function must be entered separately.

•	The tutor selects the size of the graphing window, but this can be modified via the "Plot Options" button.

•	Information about asymptotes is lost when the tutor is closed, although the graph will be written to the worksheet when the "Close" button is pressed.

•	Table 3.7.4(a) uses a Task Template to access the tutor, and provides for the retention of information about asymptotes.

Figure 3.7.4(a) Rational Function tutor applied to $f$

Alternate access to the Rational Function tutor is through the task template in Table 3.7.4(a).

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Table 3.7.4(a) Rational Function tutor accessed through task template

After entering the function into the appropriate pane in the task template, press the "Asymptotes" button. The equations for any horizontal, vertical, or oblique (slant) asymptotes will be written to the respective windows. Press the tutor button to launch the tutor. Press the "Close" button in the tutor to write the graph to the Plot window in the task template.

The graph shown in Table 3.7.4(a) was obtained by using the "Plot Options" button to modify the graphing window by setting $x \in [- 5, 10]$ and $y \in [- 40, 40]$ . In that figure, the graph of $f$ is drawn in red, the vertical asymptotes in dashed green, and the oblique asymptote in dashed blue. There is no horizontal asymptote for this function.

Asymptotes

Vertical Asymptotes

•

The Rational Function tutor (Figure 3.7.4(a)) and the related task template in Table 3.7.4(a) indicate that $x = 1$ and $x = 3$ are the equations of vertical asymptotes. A vertical asymptote can occur where the denominator of the rational function vanishes, but is confirmed only by taking limits on either side of such a zero.

•	Expression palette: Limit template Limits from the left and right at $x = 1$

$lim_{x \to 1 -} f (x)$ = $\infty$

$lim_{x \to 1 +} f (x)$ = $- \infty$

•	Expression palette: Limit template Limits from the left and right at $x = 3$

$lim_{x \to 3 -} f (x)$ = $\infty$

$lim_{x \to 3 +} f (x)$ = $- \infty$

On the basis of these limits, both $x = 1$ and $x = 3$ are deemed to be the equations of vertical asymptotes.

Horizontal Asymptotes

•	The Rational Function tutor (Figure 3.7.4(a)) and the related task template in Table 3.7.4(a) give no indication of the existence of horizontal asymptotes. Such asymptotes occur when either limit as $x \to \pm \infty$ is constant.

•	Expression palette: Limit template Limits as $x \to \pm \infty$

$lim_{x \to - \infty} f (x)$ = $\infty$

$lim_{x \to \infty} f (x)$ = $- \infty$

Since the limits as $x \to \pm \infty$ are not constant, no line of the form $y = constant$ is a horizontal asymptote.

Oblique Asymptote

•

The Rational Function tutor (Figure 3.7.4(a)) and the related task template in Table 3.7.4(a) indicate that $y = - x - 1$ is the equation of an oblique asymptote. That such an asymptote might exist is suggested by the difference in the degrees of the numerator and denominator of $f$ . Since the degree of the numerator is 3 and that of the denominator is 2, the large- $x$ behavior of $f$ is linear, that is, for large $x$ , $f (x) ≐ a x + b$ .

•	The equation of an oblique asymptote can be found by synthetic (or long) division of the numerator by the denominator. This is done in Maple with the quo (i.e., quotient) command.

•	Use the denom command to extract the denominator.

•	Context Panel: Assign to a Name≻ $d$

$denom (f (x))$ = $x^{2} - 4 x + 3$ $\overset{assign to a name}{\to}$ $d$

•	Use the numer command to extract the numerator.

•	Context Panel: Assign to a Name≻ $n$

$numer (f (x))$ = $- x^{3} + 3 x^{2} - 5 x + 6$ $\overset{assign to a name}{\to}$ $n$

•	Apply the quo command for long division of the numerator by the denominator. Assign the remainder to the name $' r'$ .

$quo (n, d, x,' r')$ = $- x - 1$

•	View the remainder.

$r$ = $- 6 x + 9$

By carrying out the division, the rule for the function becomes

$f (x) = - x - 1 + \frac{9 - 6 x}{x^{2} - 4 x + 3} = - x - 1 - \frac{9 / 2}{x - 3} - \frac{3 / 2}{x - 1}$

In this form, it is clear that for large $x$ , $f (x) ≐ - x - 1$ .

The fraction $r / d$ is split into two simpler fractions by the algebraic process called "partial fraction decomposition." The conversion of a fraction to this form can be done through the Conversions≻Partial Fractions option in the Context Panel. A stepwise (tutorial) approach is available via a task template.

Curve Analysis Tutor

•	Figure 3.7.4(b), an image of the tutor, illustrates the features of the graph of $f (x) = \frac{6 - 5 x + 3 x^{2} - x^{3}}{3 - 4 x + x^{2}}$ that can be determined from $f$ itself, and from the derivatives $f'$ and $f ″$ .

•	Where $f$ is increasing or decreasing, its graph is drawn in red or black, respectively,

•	Intervals where the graph of $f$ is concave up or down are shaded in gray or yellow, respectively.

•	Relative extrema and inflection points are shown in green.

•	Selecting one of the eight radio-buttons and clicking the "Calculate" button yields the information listed in Table 3.7.4(b).

•	Figure 3.7.4(c) uses the FunctionChart (a.k.a. FunctionPlot) command to draw the graph contained in Figure 3.7.4(b). The command provides slightly more control over the features of the graph. The symbols for the seven green points can be made larger, and arrows are used to indicate concavity.

•	The $x$ -intercepts are marked with circles; the inflection points, with crosses; and the extreme points with diamonds. These distinctions are not visible in the tutor.

Figure 3.7.19 Curve Analysis tutor applied to $f (x)$

>	Student:-SetColors(red,black,green,gray,yellow): Student:-Calculus1:-FunctionChart((-x^3+3x^2-5x+6)/(x^2-4*x+3),x=-5..10,pointoptions=[symbolsize=20],caption=[],concavity=[filled(gray,yellow)]);

Figure 3.7.20 Graph by FunctionChart

The graph in Figure 3.7.4(c) provides a bit more insight than the graph in the Curve Analysis tutor (Figure 3.7.4(b)). However, the Curve Analysis tutor does provide useful calculations. Table 3.7.4(b) displays the information that would be provided by the "Calculate" button in the tutor.

The local maxima occur at:

[-5., 4.81]

[5.22, -8.60]

The local minima occur at:

[-.532, 1.79]

[10., -11.8]

The function is increasing on the intervals:

[-.532, 1.00]

[1.00, 3.00]

[3.00, 5.22]

The function is decreasing on the intervals:

[-5., -.532]

[5.22, 10.]

The function is concave up on the intervals:

[-5., 1.00]

[1.82, 3.00]

The function is concave down on the intervals:

[1.00, 1.82]

[3.00, 10.]

The points of inflection occur at:

[1.00, Float(-infinity)]

[1.82, -.835]

[3.00, Float(-infinity)]

The zeros occur at $x =$ :

Table 3.7.6 Data generated by the Curve Analysis tutor for $f (x) = \frac{6 - 5 x + 3 x^{2} - x^{3}}{3 - 4 x + x^{2}}, x \in [- 5, 10]$

The function is undefined at $x = 1$ and $x = 3$ because these are zeros of the denominator. While it is technically true that the concavity of the function changes across these two asymptotes, no standard calculus texts would call these "points" inflection points because they are not points on the graph of $f$ .

First and Second Derivatives

Obtain and graph both the first and second derivatives of $f$ .

For the first derivative:

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

•	Context Panel: Simplify≻Simplify

•	Context Panel: Assign to a Name≻Fp

$f' (x)$

$\frac{- 3 x^{2} + 6 x - 5}{x^{2} - 4 x + 3} - \frac{(- x^{3} + 3 x^{2} - 5 x + 6) (2 x - 4)}{{(x^{2} - 4 x + 3)}^{2}}$

$\overset{simplify}{=}$

$- \frac{x^{4} - 8 x^{3} + 16 x^{2} - 6 x - 9}{{(x^{2} - 4 x + 3)}^{2}}$

$\overset{assign to a name}{\to}$

$Fp$

For the second derivative:

•	Expression palette: Differentiation template Apply to the first derivative.

•	Context Panel: Simplify≻Simplify

•	Context Panel: Assign to a Name≻Fpp

$\frac{&DifferentialD;}{&DifferentialD; x} Fp$

$- \frac{4 x^{3} - 24 x^{2} + 32 x - 6}{{(x^{2} - 4 x + 3)}^{2}} + \frac{2 (x^{4} - 8 x^{3} + 16 x^{2} - 6 x - 9) (2 x - 4)}{{(x^{2} - 4 x + 3)}^{3}}$

$\overset{simplify}{=}$

$- \frac{6 (2 x^{3} - 9 x^{2} + 18 x - 15)}{{(x^{2} - 4 x + 3)}^{3}}$

$\overset{assign to a name}{\to}$

$Fpp$

Figures 3.7.4(d) and 3.7.4(e) contains graphs of $f'$ and $f ″$ , respectively.

>	F:=(-x^3+3x^2-5x+6)/(x^2-4*x+3): plot(simplify(diff(F,x)),x=-5..10,-5..15,color=red);

Figure 3.7.4(d) Graph of $f'$

>	F:=(-x^3+3x^2-5x+6)/(x^2-4*x+3): plot(simplify(diff(F,x,x)),x=-3..7,-10..10,color=green,discont=true);

Figure 3.7.4(e) Graph of $f ″$

The equation $f' (x) = 0$ has two real solutions, but the equation $f ″ (x) = 0$ has only one. Since $f$ is not defined at $x = 1$ and $x = 3$ , both points where $f'$ also does not exist, the values $x = 1$ and $x = 3$ are not critical numbers. Moreover, for the same reason, the neither $x = 1$ nor $x = 3$ are candidates for inflection points.

Obtain the Critical Numbers

Obtain the critical numbers $c_{1}$ and $c_{2}$ by solving the equation $f' (x) = 0$

•	Set $f' (x)$ equal to zero and press the Enter key.

•	Context Panel: Student Calculus1≻Solve≻Find Roots Configure Roots dialog as per Figure 3.7.4(f)

•	Context Panel: Assign to a Name≻ $c$ (Reference roots as $c_{1}$ and $c_{2}$ .)

Figure 3.7.4(f) Roots dialog

$Fp = 0$

$- \frac{x^{4} - 8 x^{3} + 16 x^{2} - 6 x - 9}{{(x^{2} - 4 x + 3)}^{2}} = 0$

$\overset{roots}{\to}$

$[- 0.5318343934, 5.216882010]$

$\overset{assign to a name}{\to}$

$c$

Second-Derivative Test

Apply the Second-Derivative test

•	Since $f ″ (c_{1})$ is positive, the point $(c_{1}, f (c_{1}))$ is a relative minimum, a conclusion that is consistent with Figure 3.7.4(c).

$f ″ (c_{1})$ = $1.038901641$

$f (c_{1})$ = $1.785177942$

•	Since $f ″ (c_{2})$ is negative, the point $(c_{2}, f (c_{2}))$ is a relative maximum, a conclusion that is consistent with Figure 3.7.4(c).

$f ″ (c_{2})$ = $- 0.866074053$

$f (c_{2})$ = $- 8.602473055$

Candidates for Inflection

Obtain candidates for inflection points by solving the equation $f ″ (x) = 0$

•	Set $f ″ (x)$ equal to zero and press the Enter key.

•	Context Panel: Solve≻Numerically Solve

•	Context Panel: Assign Name≻ $p$

$Fpp = 0$

$- \frac{6 (2 x^{3} - 9 x^{2} + 18 x - 15)}{{(x^{2} - 4 x + 3)}^{3}} = 0$

$\overset{solve}{\to}$

$1.818917126$

$\overset{assign to a name}{\to}$

$p$

Zeros of the Function

Find the $x$ -intercepts by solving the equation $f (x) = 0$

•	Write $f (x) = 0$ and press the Enter key.

•	Context Panel: Solve≻Solve

$f (x) = 0$

$\frac{- x^{3} + 3 x^{2} - 5 x + 6}{x^{2} - 4 x + 3} = 0$

$\overset{solve}{\to}$

$\{x = 2\}, \{x = \frac{1}{2} - \frac{1}{2} I \sqrt{11}\}, \{x = \frac{1}{2} + \frac{1}{2} I \sqrt{11}\}$

Since $I = \sqrt{- 1}$ in Maple, there is only one real $x$ -intercept, namely, $x = 2$ .

Conclusions

Figures 3.7.4(d) and 3.7.4(e), graphs of $f'$ and $f ″$ , respectively, are useful for clarifying intervals of increase/decrease, and concavity, and for determining if the candidates for inflection are indeed inflection points. Wherever the red curve in Figure 3.7.4(d) is below the $x$ -axis, the function $f$ is decreasing; above the $x$ -axis, increasing. Wherever the green curve in Figure 3.7.4(e) is below the $x$ -axis, the function $f$ is concave downward; above the $x$ -axis, concave upward.

•	The point $[c_{1}, f (c_{1})]$ = $[- 0.5318343933, 1.785177943]$ is a relative minimum. The point $[c_{2}, f (c_{2})]$ = $[5.216882010, - 8.602473055]$ is a relative maximum.

•	From Figure 3.7.4(c), the endpoint $[- 5, f (- 5)]$ = $[- 5, \frac{77}{16}]$ is a relative maximum, whereas the endpoint $[10, f (10)]$ = $[10, - \frac{248}{21}]$ is a relative minimum.

•	Because of the vertical asymptotes, this function has no absolute extrema.

•	From Table 3.7.4(b) and Figure 3.7.4(c), the function increases on the intervals $[c_{1}, 1)$ , $(1, 3)$ , and $(3, c_{2}]$ .

•	From Table 3.7.4(b) and Figure 3.7.4(c), the function decreases on the intervals $[- 5, c_{1}]$ = $[- 5, - 0.5318343933]$ , and $[c_{2}, 10]$ = $[5.216882010, 10]$ .

•	From Table 3.7.4(b) and Figure 3.7.4(c), the function is concave upward on the intervals $[- 5, 1]$ , and $[p, 3)$ ; it is concave downward on the intervals $(1, p]$ and $(3, 10]$ .

•	From Table 3.7.4(b) and Figure 3.7.4(c), the point $[p, f (p)]$ = $[1.818917126, - 0.8405414371]$ is an inflection point.

•

The endpoints of a finite domain for the function have to be considered when searching for extrema. If the domain is unrestricted, that is, if it is the full set of real numbers for which the rule of the function is defined, then for this function, there would not be a global maximum or minimum because $|f (x)|$ is unbounded on an unrestricted domain.

Some Useful Commands

Applicable Commands
$CriticalPoints (f (x), x = - 5 .. 10, numeric)$ = $[- 0.5318343933, 5.216882010]$
$ExtremePoints (f (x), x = - 5 .. 10, numeric)$ = $[- 5., - 0.5318343933, 5.216882010, 10.]$
$InflectionPoints (f (x), x = - 5 .. 10, numeric)$ = $[1.818917126]$
$Roots (f (x), x = - 5 .. 10)$ = $[2]$
$Asymptotes (f (x), x = - 5 .. 10)$ = $[y = - x - 1, x = 1, x = 3]$

•	The ExtremePoints command returns the endpoints of the interval of investigation as extreme points.

•	Consequently, there are just two critical numbers, namely, $c_{1}$ = $- 0.5318343933$ and $c_{2}$ = $5.216882010$ ; and but one inflection point at $p$ = $1.818917126$ .

•	Note the additional Asymptotes command, which returned equations of the two vertical asymptotes and the one oblique asymptote.

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