1 (iv) f(x) on [-}, 1]. x* + x + 1 x + 1 (v) f(x) on [-1, }). x? + 1 (vi) f(x) on [0, 5]. x? - 1
1 (iv) f(x) on [-}, 1]. x* + x + 1 x + 1 (v) f(x) on [-1, }). x? + 1 (vi) f(x) on [0, 5]. x? - 1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Solve question 1 with subparts (iv), (v),(vi) parts
![8:37 AM Ø
VOLTE E 4 28%
(Once again, the reader is invited to supply the details of this part of the
argument.) I
PROBLEMS
1. For each of the following functions, find the maximum and minimum
values on the indicated intervals, by finding the points in the interval
where the derivative is 0, and comparing the values at these points with
the values at the end points.
(i) f(x) = x³ – x² - 8x + 1 on [-2, 2].
(ii) f(x) = x³ + x + 1
(iii) f(x) = 3x* – 8x³ + 6x²
on [-1, 1].
on (-}, }).
%3D
1
(iv) f(x)
on [-, 1].
xi + x + 1
x + 1
(v) f(x)
on [-1, }).
=
x? + 1
(vi) f(x)
on [0, 5].
х? —
1
11. Significance of the Derivative 191
2. Now sketch the graph of each of the functions in Problem 1, and find all
local maximum and minimum points.
3. Sketch the graphs of the following functions.
(i), f(x) = x +
1
3
(ii) ƒ(x) = x +
x²
x2
(iii) f(x)
x? – 1
1
(iv) f(x)
1 + x?
4. (a) If a1 < · · ·
< an, find the minimum value of f(x) = } (x – a;)².
i=1
*(b) Now find the minimum value of f(x) = ) x – a;|. This is a prob-
lem where calculus won't help at all: on the intervals between the
a's the function f is linear, so that the minimum clearly occurs at one
of the ai, and these are precisely the points where f is not differ-
entiable. However, the answer is easy to find if you consider how
f(x) changes as you pass from one such interval to another.
*(c) Let a > 0. Show that the maximum value of
1
1
f(x)
1 + |x|
1 + |x - a
is (2 + a)/(1 + a). (The derivative can be found on each of the
intervals (- o, 0), (0, a), and (a, ∞) separately.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49e60f1d-a03d-4abd-b02e-d7681ad9c98e%2Fb1056f01-814e-4fef-8ade-63ea45a25ae3%2Fdr1ckq5_processed.png&w=3840&q=75)
Transcribed Image Text:8:37 AM Ø
VOLTE E 4 28%
(Once again, the reader is invited to supply the details of this part of the
argument.) I
PROBLEMS
1. For each of the following functions, find the maximum and minimum
values on the indicated intervals, by finding the points in the interval
where the derivative is 0, and comparing the values at these points with
the values at the end points.
(i) f(x) = x³ – x² - 8x + 1 on [-2, 2].
(ii) f(x) = x³ + x + 1
(iii) f(x) = 3x* – 8x³ + 6x²
on [-1, 1].
on (-}, }).
%3D
1
(iv) f(x)
on [-, 1].
xi + x + 1
x + 1
(v) f(x)
on [-1, }).
=
x? + 1
(vi) f(x)
on [0, 5].
х? —
1
11. Significance of the Derivative 191
2. Now sketch the graph of each of the functions in Problem 1, and find all
local maximum and minimum points.
3. Sketch the graphs of the following functions.
(i), f(x) = x +
1
3
(ii) ƒ(x) = x +
x²
x2
(iii) f(x)
x? – 1
1
(iv) f(x)
1 + x?
4. (a) If a1 < · · ·
< an, find the minimum value of f(x) = } (x – a;)².
i=1
*(b) Now find the minimum value of f(x) = ) x – a;|. This is a prob-
lem where calculus won't help at all: on the intervals between the
a's the function f is linear, so that the minimum clearly occurs at one
of the ai, and these are precisely the points where f is not differ-
entiable. However, the answer is easy to find if you consider how
f(x) changes as you pass from one such interval to another.
*(c) Let a > 0. Show that the maximum value of
1
1
f(x)
1 + |x|
1 + |x - a
is (2 + a)/(1 + a). (The derivative can be found on each of the
intervals (- o, 0), (0, a), and (a, ∞) separately.)
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