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. CasEkample sove withntegral using aresums.imeExample 4Figure 6.7 is the graph of the derivative f'(x) of a function f (x). It is given that f(0) = 100. Sketchthe graph of f (x), showing all critical points and inflection points of f and giving their coordinates20+f'(x) 010MISTHS200.000S/301ed a T 10 20blueo ainT-10Figure 6.7: Graph of derivative RetractEPIL OT322Chapter Six CONSTRUCTING ANTIDERIVATIVES A G: 0, x = 20, and x = 30, where f" (x) = 0. The inflection pet0, 10, 20, 25, 30. Using theSolutionThe 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), rememberithat areas below the x-axis are subtracted. (See Figure 6.8.)Critical point(20, 300)Inflection point(25, 275)20+300 -dhw eno andanutShaded area1 रेColgms1010+2001Critical point(30, 250)Inflection point(10, 200)Critical point (0, 100)100Mgn e 102030-10+102030Figure 6.8: Finding f (10) = f(0) + So" f'(x) dæFigure 6.9: Graph of f(x)Since f(0) = 100, the Fundamental Theorem gives us the following values of f, which are markedin Figure 6.9.noituloef'(x) dx = 100 + (Shaded area in Figure 6.8) = 100+(10)(20) = 200,ol arigerg Inore f (10) = f (0) +20.0 sTugir20f (20) = f(10) +| f'(x) dx = 200 +(10)(20) = 300,10asomf ai1alaoq ai nortW e25f (25) = f(20) +f'(x) dx = 300 -20%3D(5)(10) = 275,30f(30) = f(25) + f'(x) dæ = 275 –-÷(5)(10) = 250.oussf25 inioe adExample 5Suppose F(t)een

Answered: Image /qna-images/answer/76aa01a5-e3df-4ec2-9535-fed8aad2e6aa.jpg

Answered: In Example 4, explain how one can…

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.

Cas
Ekample s
ove with
ntegral using are
sums.
ime
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

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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