Chapter 3: Applications of Differentiation
Section 3.7: What Derivatives Reveal about Graphs
Example 3.7.1
For the function fx=x cosx, x∈0,2 π use first principles to obtain the data in Table 3.7.1.
Solution
Define the Function
Define the function f
Control-drag (or type) fx=…
Context Panel: Assign Function
fx=x cosx→assign as functionf
Graph the Function and its Derivatives
Graph f,f′,f″
Figure 3.7.1(a) Graph of fx
Figure 3.7.1(b) Graph of f′x
Figure 3.7.1(c) Graph of f″x
Obtain the Critical Numbers
Obtain the critical numbers c1 and c2 by solving the equation f′x=0
Tools≻Load Package: Student Calculus 1
Loading Student:-Calculus1
Write the equation f′x=0; press the Enter key.
Context Panel: Student Calculus1≻Solve≻Find Roots Complete the Roots dialog as per Figure 3.7.1(d) Press OK.
Context Panel: Assign to a Name≻c
f′x=0
cos⁡x−x⁢sin⁡x=0
→roots
0.8603335890,3.425618459
→assign to a name
c
Figure 3.7.1(d) Roots dialog
The two solutions of f′x=0 in the interval 0,2 π are in the list whose name is c.
The individual critical numbers can now be referenced by typing c1 and c2.
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 3.7.1(a).
f″c1 = −2.077216671
fc1 = 0.5610963382
Since f″c2 is positive, the point c2,fc2 is a relative minimum, a conclusion that is consistent with Figure 3.7.1(a).
f″c2 = 3.848816238
fc2 = −3.288371396
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: Student Calculus1≻Solve≻Find Roots See Figure 3.7.1(d); bound roots in 1,6. Press OK.
Context Panel: Assign to a Name≻p
f″x=0
−2⁢sin⁡x−x⁢cos⁡x=0
2.288929728,5.086985094
p
The individual candidates can now be referenced by typing p1 and p2.
Zeros of the Function
Find the x-intercepts by solving the equation fx=0
Write fx=0 and press the Enter key.
Context Panel: Student Calculus1≻Solve≻Find Roots See Figure 3.7.1(d); bound roots in 0,2 π. Uncheck "calculate numerically".
fx=0
x⁢cos⁡x=0
0,12⁢π,32⁢π
Conclusions
The point c1,fc1 = 0.8603335890,0.5610963382 is a relative maximum. The point 2 π,f2 π = 2⁢π,2⁢π is also a relative maximum. From Figure 3.7.1(a), it is also the absolute maximum
The point 0,0 is a relative minimum. The point c2,fc2 = 3.425618459,−3.288371396 is also a relative minimum. From Figure 3.7.1(a), this point is also the absolute minimum.
From Figures 3.7.1(a) and 3.7.1(b), the function increases on the intervals 0,c1 = 0,0.8603335890 and c2,2 π = 3.425618459,2⁢π. The function decreases on the interval c1,c2 = 0.8603335890,3.425618459.
From Figures 3.7.1(a) and 3.7.1(c), the function is concave upward on the interval p1,p2 = 2.288929728,5.086985094, and concave downward on the intervals 0,p1 = 0,2.288929728, and p2,2 π = 5.086985094,2⁢π.
From Figures 3.7.1(a) and 3.7.1(c), the points p1,fp1 = 2.288929728,−1.506070450, and p2,fp2 = 5.086985094,1.861310957 are inflection points.
Some Useful Commands
Applicable Commands
CriticalPointsfx,x=0..2 π,numeric = 0.8603335890,3.425618459
ExtremePointsfx,x=0..2 π,numeric = 0.,0.8603335890,3.425618459,6.283185308
InflectionPointsfx,x=0..2 π,numeric = 2.288929728,5.086985094
Rootsfx,x=0..2 π,numeric = 0.,1.570796327,4.712388980
Rootsfx,x=0..2 π = 0,12⁢π,32⁢π
The numeric option was necessary in the first three commands because of the complexity of the equations being solved. For functions whose "special points" can be given analytically, the option can be omitted. For the given function, the zeros can be determined analytically, as shown by the second call to the Roots command.
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