2j) Let us consider the function f(x) = e-ª - 1. Since the image of the exponential function e is (0, +∞o), also the image of e- is (0, +∞o). Therefore J = (-1, +∞o). Hence J is bounded from below and is unbounded from above, inf J = -1 and there is no min J.
2j) Let us consider the function f(x) = e-ª - 1. Since the image of the exponential function e is (0, +∞o), also the image of e- is (0, +∞o). Therefore J = (-1, +∞o). Hence J is bounded from below and is unbounded from above, inf J = -1 and there is no min J.
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
The question asks to find the maximum, minimum, Supremum and infimum. I attached the solution the part I don’t understand is 2J)
Why is there no minimum ?
![d) D = {reR:
sin x ≥ 1}.
e) E= {xER: e* - 1 ≤0}.
f) F = {²-\x: xER}.
g) G= {x|x|-x-2: TER}.
h) H = {√²- |x|: xER}.
i) I = {| sin r|-1: TER}.
j) J = {e-1: TER}.
I](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2F1db51a44-0da5-49be-9263-001f8996f67c%2F7uxw9w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:d) D = {reR:
sin x ≥ 1}.
e) E= {xER: e* - 1 ≤0}.
f) F = {²-\x: xER}.
g) G= {x|x|-x-2: TER}.
h) H = {√²- |x|: xER}.
i) I = {| sin r|-1: TER}.
j) J = {e-1: TER}.
I
![2j) Let us consider the function f(x) = e- - 1. Since the image of the exponential function e
is (0, +∞o), also the image of e- is (0, +∞o). Therefore J = (-1, +∞). Hence J is bounded
from below and is unbounded from above, inf J = − 1 and there is no min J.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2F1db51a44-0da5-49be-9263-001f8996f67c%2Fdji7a75_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2j) Let us consider the function f(x) = e- - 1. Since the image of the exponential function e
is (0, +∞o), also the image of e- is (0, +∞o). Therefore J = (-1, +∞). Hence J is bounded
from below and is unbounded from above, inf J = − 1 and there is no min J.
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