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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
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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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
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.
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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