Chapter 3: Applications of Differentiation
Section 3.7: What Derivatives Reveal about Graphs
Graph fx=6−5 x+3 x2−x33−4 x+x2 for x∈−5,10, indicating any asymptotes this rational function might have. Then, use the tools of the calculus to analyze the features of this graph.
Tools≻Load Package: Student Calculus 1
Context Panel: Assign Function
fx=6−5 x+3 x2−x33−4 x+x2→assign as functionf
Rational Function Tutor
Figure 3.7.4(a) contains an image of the
tutor applied to fx.
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
Enter a rational function PxQx:
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∈−5,10 and y∈−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.
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
limx→1−fx = ∞
limx→1+fx = −∞
Expression palette: Limit template
Limits from the left and right at x=3
limx→3−fx = ∞
limx→3+fx = −∞
On the basis of these limits, both x=1 and x=3 are deemed to be the equations of vertical 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→±∞ is constant.
Expression palette: Limit template
Limits as x→±∞
limx→−∞fx = ∞
limx→∞fx = −∞
Since the limits as x→±∞ are not constant, no line of the form y=constant is a horizontal 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, fx≐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
denomfx = x2−4⁢x+3→assign to a named
Use the numer command to extract the numerator.
Context Panel: Assign to a Name≻n
numerfx = −x3+3⁢x2−5⁢x+6→assign to a namen
Apply the quo command for long division of the numerator by the denominator.
Assign the remainder to the name 'r'.
quon,d,x,'r' = −x−1
View the remainder.
r = −6⁢x+9
By carrying out the division, the rule for the function becomes
fx= −x−1+9−6 xx2−4 x+3= −x−1−9/2x−3−3/2x−1
In this form, it is clear that for large x, fx≐ −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 fx=6−5 x+3 x2−x33−4 x+x2 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 fx
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:
The local minima occur at:
The function is increasing on the intervals:
The function is decreasing on the intervals:
The function is concave up on the intervals:
The function is concave down on the intervals:
The points of inflection occur at:
The zeros occur at x=:
Table 3.7.6 Data generated by the Curve Analysis tutor for fx=6−5 x+3 x2−x33−4 x+x2,x∈−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
→assign to a name
For the second derivative:
Expression palette: Differentiation template
Apply to the first derivative.
Context Panel: Assign to a Name≻Fpp
ⅆⅆ x Fp
Figures 3.7.4(d) and 3.7.4(e) contains graphs of f′ and f″, respectively.
Figure 3.7.4(d) Graph of f′
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 c1 and c2 by solving the equation f′x=0
Set f′x equal to zero and press the Enter key.
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 c1 and c2.)
Figure 3.7.4(f) Roots dialog
Apply the Second-Derivative test
Since f″c1 is positive, the point c1,fc1 is a relative minimum, a conclusion that is consistent with Figure 3.7.4(c).
f″c1 = 1.038901641
fc1 = 1.785177942
Since f″c2 is negative, the point c2,fc2 is a relative maximum, a conclusion that is consistent with Figure 3.7.4(c).
f″c2 = −0.866074053
fc2 = −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
Zeros of the Function
Find the x-intercepts by solving the equation fx=0
Write fx=0 and press the Enter key.
Context Panel: Solve≻Solve
Since I=−1 in Maple, there is only one real x-intercept, namely, x=2.
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 c1,fc1 = −0.5318343933,1.785177943 is a relative minimum. The point c2,fc2 = 5.216882010,−8.602473055 is a relative maximum.
From Figure 3.7.4(c), the endpoint −5,f−5 = −5,7716 is a relative maximum, whereas the endpoint 10,f10 = 10,−24821 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 c1,1, 1,3, and 3,c2.
From Table 3.7.4(b) and Figure 3.7.4(c), the function decreases on the intervals −5,c1 = −5,−0.5318343933,
and c2,10 = 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,fp = 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 fx is unbounded on an unrestricted domain.
Some Useful Commands
CriticalPointsfx,x=−5..10,numeric = −0.5318343933,5.216882010
ExtremePointsfx,x=−5..10,numeric = −5.,−0.5318343933,5.216882010,10.
InflectionPointsfx,x=−5..10,numeric = 1.818917126
Rootsfx,x=−5..10 = 2
Asymptotesfx,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, c1 = −0.5318343933 and c2 = 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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