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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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98. 8.1 As:- Giden :- If f If of and Junctions at that Bristof To Grove :- lin (f+g) x NC lin f(x) = f(c) 2-C of rand, 8- g • for any czo, 7 i T = Point 2 in live g(x) = g(e) f(c) + gle). such that I f(x) - f(c) | < 1/2 €, when o/x-c/ <S₁ 1 g(x) = g(e) | < 1/₁₂ c, when, 0</x-c/<5₂. 8 = men (S₁+5₂), then, for 0</x- c/ <S₁ =) If +9)2 = (f (c) + g(e)) < e =) lin (f+g) x = f(c) + g(e). М-С care | f(x) = f(c) < 1/₂) = ₂ [g(n)- g (c) | < / €. -|(f+g) x = (f(c) + g(c)) | = | f(x) = f(c) + g(x) = g()) </f(u)-f(c) 1 + g(x)-g()) Fince, lin f(x) = f(c), lin g(n) = g(e) NIC n +Ve numbers St. Sz two when 0</2-c/<s Continuous Suche
Transcribed Image Text:98. 8.1 As:- Giden :- If f If of and Junctions at that Bristof To Grove :- lin (f+g) x NC lin f(x) = f(c) 2-C of rand, 8- g • for any czo, 7 i T = Point 2 in live g(x) = g(e) f(c) + gle). such that I f(x) - f(c) | < 1/2 €, when o/x-c/ <S₁ 1 g(x) = g(e) | < 1/₁₂ c, when, 0</x-c/<5₂. 8 = men (S₁+5₂), then, for 0</x- c/ <S₁ =) If +9)2 = (f (c) + g(e)) < e =) lin (f+g) x = f(c) + g(e). М-С care | f(x) = f(c) < 1/₂) = ₂ [g(n)- g (c) | < / €. -|(f+g) x = (f(c) + g(c)) | = | f(x) = f(c) + g(x) = g()) </f(u)-f(c) 1 + g(x)-g()) Fince, lin f(x) = f(c), lin g(n) = g(e) NIC n +Ve numbers St. Sz two when 0</2-c/<s Continuous Suche
Hence, This is Junction is 802.7 in the at Pewaf $i- False, at is so f R Point a P. Let N = C The Junction f(x) = 1x1 but is 0. treue 8.37 True, every function which is Point 0 continuous. Every that is not derivable necessarily continuous function of be derivable Hence, f'(c) = lin f(n) - fle). NIC N-C Taking limits tody Sum continuous function. differentiable at f of Continmous continuous not differentiable at f(x) - fle) = f(x) = f(e) (n-e), (n+c) - (x-c) is as x-e, we have, lin {f(n) - fle)} = lin {_f(x) = f(e) (n.c) } NC NC N-C continuous at مدا exists. - lin { f(x) = f(c) 2. kein (re) ひう = f'(@) · 0 at lin f(n) = f(e), and NIC 20 2 nz@. therefore,
Transcribed Image Text:Hence, This is Junction is 802.7 in the at Pewaf $i- False, at is so f R Point a P. Let N = C The Junction f(x) = 1x1 but is 0. treue 8.37 True, every function which is Point 0 continuous. Every that is not derivable necessarily continuous function of be derivable Hence, f'(c) = lin f(n) - fle). NIC N-C Taking limits tody Sum continuous function. differentiable at f of Continmous continuous not differentiable at f(x) - fle) = f(x) = f(e) (n-e), (n+c) - (x-c) is as x-e, we have, lin {f(n) - fle)} = lin {_f(x) = f(e) (n.c) } NC NC N-C continuous at مدا exists. - lin { f(x) = f(c) 2. kein (re) ひう = f'(@) · 0 at lin f(n) = f(e), and NIC 20 2 nz@. therefore,
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