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Oct 01, 09 · Suppose f and g are differentiable functions with the values shown in the table below For each of the following functions h, find h '(3) x f(x) g(x) f '(x) g '(x) 3 3 7 6 6 (a) h(x) = f(x) g(x) h '(3)= 0 (b) h(x) = f(x) g(x) h '(3)= (c) h(x) = f(x) / g(x) h '(3)= i was able to get the answer to a by guessing and adding things together but i really dont know how to solve this. = f!(x)g(x)−f(x)g!(x) g(x)2 • d dx!. What ‰ C‰ ¾(from‹¯‹® üa½o½o½o «€€ Ð ˜Š.

F(x) = G(x) C (3) What is thatconstant?Evaluating equations (2)and(3)atx = a gives us F(a) = G(a) C = ∫ a a f(t)dt = 0 which leads to G(a) C = 0 C = G(a) Similarly, evaluating the equations atx = b gives F(b) = G(b) C = ∫ b a f(t)dt or ∫b a f(t)dt = G(b) C = G(b) G(a) andwe are done 5. Simple and best practice solution for g=(xc)/x equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand,. 1 day ago · J z¼¤ðx–vŒ©¨ ×í •‘ eŒ¢h,,“®`©9 6G7L ˜% Ù¢åY KÀÐ_ÍÇ u "x± pP¥ s £èE†×E cËœf\\•°ò ¥r Õ »©¨ Ç&€ ¿hñb3¼þîãY—ö²ô J¿Ý¡ûÓf ÅÖ¯aü ÷‚õ ³iç÷µýkÍ¿þ´íÚææ3–˵ML×¾¼u ©hªI&‹Óªv—™Ê ›ýùkÀ¬ øSGè ¤d!y‹ ¼ E.

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There exists x 2A such that (g f)(x) = g(f(x)) = z Therefore if we let y = f(x) 2B, then g(y) = z Thus g is surjective Problem 338 In each part of the exercise, give examples of sets A;B;C and functions f A !B and g B !C satisfying the indicated properties (a) g is not injective but g f is injective. &úÐ Æ 6¡j˜b3TÈ dàÙ™é ºQ0y ¢˜¿)piŠŠ & )Ãtp‡qTWwbT6DS®e ÌðQ¬Ö8“È ÁfHäL ý š@ o R Ëj °1G í „% ZÇjH>¿B"* 4 ÌXÕçe©uW»1fæî¸cÓ¿ù ¿É* &kD‡ñŒ ’b\¥&Fx—³›Çâ b"X )´%”Eö WH1š ÿû’dÉ ÕhHAé B5A & U§ ‡°gHá ädŒ`ñ€dH¹ÖÍ †Ã {uQé´sÛ. From Subject wu saturday Google Docs Date Wed, 22 Jun 11 1020 0400 MIMEVersion 10 ContentType multipart/related;.

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