Chapter 3: Applications of Differentiation
Section 3.7: What Derivatives Reveal about Graphs
Example 3.7.2
Graph fx=x3+5 x2−17 x−9, x∈−9,5; 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 fx=…
Context Panel: Assign Function
fx=x3+5 x2−17 x−9→assign as functionf
Curve Analysis Tutor
Figure 3.7.2(a), an image of the tutor, illustrates the features of the graph of fx=x3+5 x2−17 x−9 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 3.7.2(a).
Figure 3.7.2(b) uses the FunctionChart (a.k.a. FunctionPlot) command to draw the graph contained in Figure 3.7.2(a). 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 3.7.2(a) Curve Analysis tutor applied to fx=x3+5 x2−17 x−9
Student:-SetColors(red,black,green,gray,yellow): Student:-Calculus1:-FunctionChart(x^3+5*x^2-17*x-9,x=-9..5,pointoptions=[symbolsize=20],caption=[],concavity=[filled(gray,yellow)]);
Figure 3.7.2(b) Graph via the FunctionChart command
Table 3.7.2(a) displays the information that would be provided by the "Calculate" button in the tutor.
The local maxima occur at:
[-4.57, 77.7]
[5., 156.]
The local minima occur at:
[-9., -180.]
[1.24, -20.5]
The function is increasing on the intervals:
[-9., -4.57]
[1.24, 5.]
The function is decreasing on the interval:
[-4.57, 1.24]
The function is concave up on the interval:
[-1.67, 5.]
The function is concave down on the interval:
[-9., -1.67]
The points of inflection occur at:
[-1.67, 28.7]
The zeros occur at x=:
-7.19
-.470
2.66
Table 3.7.2(a) Data generated by the Curve Analysis tutor for fx=x3+5 x2−17 x−9,x∈−9,5
Graph the Function and its Derivatives
Graph f,f′,f″
Figure 3.7.2(c) Graph of fx
Figure 3.7.2(d) Graph of f′x
Figure 3.7.2(e) Graph of f″x
Obtain the Critical Numbers
Obtain the critical numbers c1 and c2 by solving the equation f′x=0
Write f′x=0 and press the Enter key.
Context Panel: Solve≻Numerically Solve
Context Panel: Conversions≻To List
Context Panel: Assign Name≻c
f′x=0
3x2+10x−17=0
→solve
−4.572599296,1.239265962
→to list
→assign to a name
c
The equation f′x=0 is quadratic, so there is an exact solution for the critical numbers. However, these numbers are binomial surds (radicals) that are hard to estimate without conversion to floating-point form.
Second-Derivative Test
Apply the Second-Derivative test
Since f″c1 is negative, the point c1,fc1 is a relative maximum, a conclusion that is consistent with Figure 3.7.2(c).
f″c1 = −17.43559578
fc1 = 77.67056587
Since f″c2 is positive, the point c2,fc2 is a relative minimum, a conclusion that is consistent with Figure 3.7.2(c).
f″c2 = 17.43559577
fc2 = −20.48538069
Candidates for Inflection
Obtain candidates for inflection points by solving the equation f″x=0
Write f″x=0 and press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻x
Context Panel: Assign Name≻p
f″x=0
6x+10=0
→solutions for x
−53
p
Zeros of the Function
Find the x-intercepts by solving the equation fx=0
Write fx=0 and press the Enter key.
fx=0
x3+5x2−17x−9=0
−7.190234786,−0.4704431317,2.660677918
Maple can provide the exact solution of a cubic equation. Unfortunately, it is the rare cubic for which the roots are simple expressions. Here, even after drastically simplifying the exact solutions, the expressions are unwieldy. Hence, a numeric solution is obtained instead.
Conclusions
The point c1,fc1 = −4.572599296,77.67056587 is a relative maximum. The point 5,f5 = 5,156 is also a relative maximum. From Figure 3.7.2(c), it is the absolute maximum.
The point c2,fc2 = 1.239265962,−20.48538069 is a relative minimum. The point −9,f−9 = −9,−180 is also a relative minimum. From Figure 3.7.2(c), it is the absolute minimum.
From Figures 3.7.9 and 3.7.10, the function increases on the intervals −9,c1 = −9,−4.572599296 and c2,5 = 1.239265962,5. The function decreases on the interval c1,c2 = −4.572599296,1.239265962.
From Figures 3.7.2(c) and 3.7.2(e), the function is concave upward on the interval p,5 = −53,5, and concave downward on the interval −9,p = −9,−53 .
From Figures 3.7.2(c) and 3.7.2(e), the point p,fp = −53,77227≐−1.6667,28.593 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 as x→±∞.
Some Useful Commands
Applicable Commands
CriticalPointsfx,x=−9..5,numeric = −4.572599296,1.239265962
ExtremePointsfx,x=−9..5,numeric = −9.,−4.572599296,1.239265962,5.
InflectionPointsfx,x=−9..5 = −53
Rootsfx,x=−9..5,numeric = −7.190234786,−0.4704431317,2.660677918
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