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Answer to If f(x) = g(u) and u = u(x) then a) f '(x) = g '(u) b) f '(x) = g '(u) u '(x) c) f '(x) = u '(x) d) None of the above By signing up, When you drive to the store, you're probablyF(c) = lim c œ x f(c) = f(x) The bottom line is that F (x) = lim h œ 0 F(xh) F(x) h = f(x) 5The second fundamental theorem of integral calculus We are nowin a position toprove the final andmost important theorem in this sequence ofresults Theorem Let f(x) be a continuous function on the interval a,b Let G(x) be any function withthe/ 0 ' ( 1 0 2

To get that conclusion, we need to know that f(x) g(y) for all x;y2nd use Proposition 2(c) Suppose that f S → T and g T → U are bijections Let f−1 T → S and g−1 U → T be their respective inverses I'll prove that f−1 ·g−1 is the inverse of g ·f f−1 g−1 ·(g ·f) (s) = f−1 g−1(g(f(s))) Definition of composite = f−1(f(s)) g−1(g(junk)) = junk = s f−1(f (junk)) = junk (g f)f−1 ·g2 < k _ j h k k b c k d Z h e b f i b Z ^ Z b f _ g b F k l b k e Z \ Z D _ e ^ u r Z i b g n h j f Z l b d _ I h _ ^ b l _ e b 6 I j b a _ j u 4

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The form z= f(xat) g(x−at) is a solution of the wave equation ∂2z ∂t2 = a2 ∂2z ∂x2 Solution Let u= xatand v= x−at Then z= f(u) g(v) and the chain rule gives ∂z ∂x = df du ∂u ∂x dg dv ∂v ∂x = df du dg dv, ∂z ∂t = df du ∂u ∂t dg dv ∂v ∂t = a df du −a dg dv Thus ∂2z ∂x2 = ∂ ∂x ∂z ∂xW e _ f _ g l ^ b Z n j Z f u K l j m g Z b a i j h \ h e h d b H Q < b g l k Z f h g Z j _ a Z x s b c K r b \ d Z ^ e y f Z l h \ ,5(' 0$7 b 3UR5R 0 13 2 1 11 40 12 8 R5 F Z l u i j h r b \ g u _ WIRED MAT 50, 80, 105;Õ / Ü 8 >

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9 6 = 5 > C < ;Find (f g)(x) for f and g below f(x) = 3x 4 (6) g(x) = x2 1 x (7) When composing functions we always read from right to left So, rst, we will plug x into g (which is already done) and then g into f What this means, is that wherever we see an x in f we will plug in g That is, g acts as our new variable and we have f(g(x))Title Qualtrics Survey Software Author rmarriott Created Date AM

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Otherwise, if f gand gis bounded from above, then f(x) g(x) sup A g for every x2A Thus, fis bounded from above by sup A g, so sup A f sup A g Similarly, f g implies that sup A( f) sup A ( g), so inf f inf g Note that f gdoes not imply that sup Af inf g;3 / 3 (c) Yes Suppose (u, v) is a particular but arbitrarily chose element in the codomainThen, u and v are both real numbers Let x = 1v and y = (u1)/3Then, x and y are also both real numbers So, (x, y) is in the domainBy the definition of F, we have F( , )=F(1− , 1 3)=(3∙ 1 3 −1,1−(1− ))=( , ) (d) Yes, because F is both injective and surjective0 1 4 ;

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Math 432 Real Analysis II Solutions to Homework due March 11 Question 1 Let f(x) = k be a constant function for k 2R 1 Show that f is integrable over any a;b by using Cauchy's " P condition for integrability< = 4 ;Integration by Substitution Let u = g(x) and F(x) be the antiderivative of f(x) Then du = g0(x)dx and Z f g(x) g0(x)dx = Z f(u)du = F(u)C Also, Z b a f g(x) g0(x)dx = Z g(b) g(a) f(u)du = F g(b) −F g(a) Ex u = g(x) = x2, du = g0(x)dx = 2xdx Z 5 2 2xex2dx = Z 52=25 22=4 eu du = eu 25 4 = e25 −e4 Integration Rules Examples Z kdu = kuC Z

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