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The Substitution Rule If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then Z f(g(x))g0(x)dx = f(u)du Example Calculate R tanx dx Solution Z tanx dx = sinx cosx dx This suggests substitution u = cosx, since then du = −sinx dx and so, sinx dx = −du Z. ¢, µ ½ ª Á Ä, x o ³ ê Ä ¢ È ¢ Æ ¢ ¤ Ó ¡ Å u s ® v Ý ¼ c Æ ¢ í ê é Ì Å é B » ê ä ¦, Ý ¼ ù vL ÍL E=L1 \L2 Æ È é B. Then g f A !C is de ned by (g f)(1) = 1 This map is a bijection from A = f1gto C = f1g, so is injective and surjective However, g is not injective, since g(1) = g(2) = 1, and f is not surjective, since 2 62f(A) = f1g Problem 339 De ne functions f and g from Z to Z such that f is not surjective and yet g f is surjective Solution Let f.

HOMEWORK #7 SOLUTIONS TO SELECTED PROBLEMS 3 Corollary Assume charK = p If L=K is a flnite extension and fi 2 L, there exists e ‚ 0 such that fipe is separable over K Proof Let f 2 Kt be the minimal polynomial of fi, and write f(t) = h(tpe) for e ‚ 0 and h 2 Kt irreducible and separableThen 0 = f(fi) = h(fipe), so that fipe is a root of the separable irreducible polynomial. Feb 01, 18 · A ab cef B c d e g and C b c f g and U a bc d e f g h Draw venn diagrams to represent the following sets i A union B intersection C ii B C union C B Mathematics. Oct 17, 16 · If a ∥ b and e ∥ f, what is the value of y?.

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H = f(g(x 0)∆g)−f(g(x 0)) = f(g ∆g)−f(g) Thus we apply the fundamental lemma of differentiation, h = f0(g)η(∆g)∆g, 1 f0(g)η(∆g) ∆g h Note that f0(g(x)) > 0 for all x ∈ (a,b) and η(∆g) → 0 as h → 0, thus, lim h→0 ∆g/h = lim h→0 1 f0(g)η(∆g) 1 f0(g(x)) Thus g0(x) = 1 f0(g(x)), g 0(f(x)) = 1 f0(x) 3 Suppose g is a real function on R1, with bounded. } %T8p _ Z Z# F 4Ä x K r M. Title HBTeachNonExamindd Author christopherdinardo Created Date 11/19/ AM.

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