In Example 4, explain how one can compute f(−5)f(−5). What is f(−5)f(−5)? Double-check that your answer makes sense by looking at the graph of f(x)f(x) in Figure 6.9.
In Example 4, explain how one can compute f(−5)f(−5). What is f(−5)f(−5)? Double-check that your answer makes sense by looking at the graph of f(x)f(x) in Figure 6.9.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
In Example 4, explain how one can compute f(−5)f(−5). What is f(−5)f(−5)? Double-check that your answer makes sense by looking at the graph of f(x)f(x) in Figure 6.9.
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Example 4
Figure 6.7 is the graph of the derivative f'(x) of a function f (x). It is given that f(0) = 100. Sketch
the graph of f (x), showing all critical points and inflection points of f and giving their coordinates
20+
f'(x) 0
10
MIS
THS
200.
000S
/30
1ed a T 10 20
blueo ainT
-10
Figure 6.7: Graph of derivative
Transcribed Image Text:RetractE
PIL OT
322
Chapter Six CONSTRUCTING ANTIDERIVATIVES A G
: 0, x = 20, and x = 30, where f" (x) = 0. The inflection pet
0, 10, 20, 25, 30. Using the
Solution
The critical points of f occur at x =
of the critical points and inflection points of f., we evaluate f(x) for x =
the definite integrals using the areas of triangular regions under the graph of f'(x), rememberi
that areas below the x-axis are subtracted. (See Figure 6.8.)
Critical point
(20, 300)
Inflection point
(25, 275)
20+
300 -
dhw eno andanut
Shaded area
1 रे
Colgms
10
10+
2001
Critical point
(30, 250)
Inflection point
(10, 200)
Critical point (0, 100)
100
Mgn e 10
20
30
-10+
10
20
30
Figure 6.8: Finding f (10) = f(0) + So" f'(x) dæ
Figure 6.9: Graph of f(x)
Since f(0) = 100, the Fundamental Theorem gives us the following values of f, which are marked
in Figure 6.9.
noituloe
f'(x) dx = 100 + (Shaded area in Figure 6.8) = 100+(10)(20) = 200,
ol arigerg Inore f (10) = f (0) +
2
0.0 sTugi
r20
f (20) = f(10) +| f'(x) dx = 200 +(10)(20) = 300,
10asomf ai
1
alaoq ai nortW e
25
f (25) = f(20) +
f'(x) dx = 300 -
20
%3D
(5)(10) = 275,
30
f(30) = f(25) + f'(x) dæ = 275 –-÷(5)(10) = 250.
oussf25 inioe ad
Example 5
Suppose F(t)
een
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