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Defined as ( mn )˘ mn 2 and g(m,n)˘(m¯1,m¯n)Findtheformulasforg– f and f –g Note g– f ( m,n )˘)) mn 2¯1 Thus g– f ( m,n )˘ mn¯1 2 Note f– g (m,nWe must show that if f(n) and g(n) are monotonically increasing functions, then so is f(n) g(n) Suppose not Let n1 < n2 and f(n1)g(n1) > f(n2)g(n2) Now, f(n1) ≤ f(n2) and g(n1) ≤ g(n2) f(n1) ≤ f(n2) f(n1)g(n1) ≤ f(n2) g(n1) f(n1)g(n1) ≤ f(n2) g(n2) This contradicts our assumption (b) We must show that if f(n) and g(n) areCalled the inverse of f if g(f(s)) = s for all s ∈ S and f (g(t)) = t for all t ∈ T I proved the following result earlier Theorem Let S and T be sets, and let f S → T be a function f is invertible if and only if f is bijective Example Let S = {a,b,c,d} and T = {1,2,3,Calvin} Define f S → T by f(a) = 1, f(b) = 2, f(c) = 3, f

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(b) Since f and g are integrable on a;b, then f g and f g are integrable Since squares of integrable functions are integrable, then (f g)2 and (f g)2 are integrable Thus, by (a), 4fg is integrable and fg is integrable, as desired Question 5 Consider the function f on 0;1 given byActivity 3 Two kernels of truth Suppose fG→H is a homomorphism, e G and e H the identity elements in G and H respectively Show that the set f1 (e H) is a subgroup of GThis group is called the kernel of f (Hint you know that e G ∈f1 (e H) from beforeUse the definition of a homomorphism and that of a group to check that all the other conditions are satisfied)(a) f g, f g, and fgare continuous at x= a (b) If g(a) 6= 0, then f=gis continuous at x= a Theorem 3 If fis continuous at a, and if gis continuous at f(a), then f g is continuous at a Intermediate Value Theorem (IVT) Suppose fis continuous on a;b If kis any number between f(a) and f(b), then there exists a number c2a;b such that f(c

Find stepbystep Calculus solutions and your answer to the following textbook question Suppose that f and g are continuous on a, b and differentiable on (a, b) Suppose also that f(a)=g(a) and fH is in the right subtree a gdh b fjc ei a g b fjc d ei h Inorder And Postorder • Scan postorder from right to left using inorder to separate left and right subtrees • inorder = g d h b e i a f j c • postorder = g h d i e b j f c aSolutions for Assignment 4 –Math 402 Page 74, problem 6 Assume that φ G→ G′ is a group homomorphism Let H′ = φ(G) We will prove that H′ is a subgroup of G′Let eand e′ denote the identity elements of G and G′, respectivelyWe will use

Ple We can form f g R !A such that g f = idA Solution For each b 2 B such that b = f(a) for some a 2 A, we set g(b) = a This is wellde ned since for each b 2 B there is at most one such a Now pick some element 2 A and for each b 2Proofs involving surjective and injective properties of general functions Let f A !B and g B !C be functions, and let h = g f be the composition of g and f For each of the following statements, either give a formal proof or counterexample (A counterexample means a speci c example

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Theorem 2 If f'(x) = g'(x) for all x in an interval (a, b) of the domain of these functions, then f g is constant or f = g c where c is a constant on (a, b Proof Let F = f − g, then F' = f' − g' = 0 on the interval (a, b), so the above theorem 1 tells that F = f − g is a constant c or f = g c Theorem 3 If F is an antiderivativeDefinedas ( x)˘ 3 p ¯1 andg( x)˘ 3Findthe formulasfor g–fand g– f (x)˘ x¯1;You have certainly dealt with functions before, primarily in calculus, where you studied functions from $\R$ to $\R$ or from $\R^2$ to $\R$ Perhaps you have encountered functions in a more abstract setting as well;

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 GBF Directed by Darren Stein With Michael J Willett, Paul Iacono, Sasha Pieterse, Andrea Bowen What happens after Tanner is outed by his classmates and becomes the title "gay best friend" for three high school queen bees?If g is continuous at x = a, and f is continuous at x = g(a), then the composite function f g given by ( f g)( x) = f ( g(x)) is also continuous at a That is, the composite of two continuous functions is continuous Example Since both f ( x) = x2 1 and g(x) = cos x are continuous on (− ∞, ∞) Therefore, both (f g)( x) = cos 2 x 1F(x)d (a,b) 7!(a,bd) shift up by d f(x)d (a,b) 7!(a,bd) shift down by d cf(x) (a,b) 7!(a,cb) stretch vertically by c 1 cf(x) (a,b) 7!(a, 1 cb) shrink vertically by 1 c f(x) (a,b) 7!(a,b) flip over the xaxis Examples • The graph of f(x)=x2 is a graph that we know how to draw It's drawn on page 59

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Is a neighborhood of f(a), so U = f 1(V) is a neighborhood of a Then B (a) ˆUfor some >0, which implies that f(B (a)) ˆB (f(a)), so fis continuous at a Conversely, if f is continuous at aand V is a neighborhood of f(a), then B (f(a)) ˆV for some >0 By continuity, there exists >0 such that f(B (a)) ˆB (f(a)), so B (a) ˆf 1(V), meaningThen f g(b) = f(g(b)) = f(a) = b, ie f g = idB 119 Show that for an injective function f A !F –g(x)˘ 3 p x3 ¯1 7 Consider the functions f ,g Z£!

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34#)#2 Industry Number of jobs, 15 Share of all advanced industries jobs, 15 Growth, 13Ð15 Management, Scientific, and Technical Consulting Services 17,735 167% 119%Proof This is a straightforward computation left as an exercise For example, suppose that f G 1!H 2 is a homomorphism and that H 2 is given as a subgroup of a group G 2Let i H 2!G 2 be the inclusion, which is a homomorphism by (2) of Example 12F(a) f(b) = f(gcd(a,b)) f(lcm(a,b)) Every completely multiplicative function is a homomorphism of monoids and is completely determined by its restriction to the prime numbers Convolution If f and g are two multiplicative functions, one defines a new multiplicative function f * g, the Dirichlet convolution of f and g, by

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David Fasolo Waxhaw, North Carolina, United States Lawn service owner at DGF LAWN SERVICE 42 connections See David's complete profile on Linkedin and connectTranslates into the following assembly code add t0, g, h add f, g, h add t1, i, j or sub f, f, i sub f, t0, t1 sub f, f, j • Each version may produce a different result because floatingpoint operations are not necessarily associative and commutativeLet f ∶X →Y and g ∶Y →Z be injections, and let a;b ∈X Suppose that g f(a) =g f(b) Since g is injective, we must have f(a) =f(b) Since f is injective we must have a =b Therefore g f is injective Problem 95 Let f and g be bijections Then f and g are both injections, so by problem 94 we know g f

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A,b forwhich f(c) (b a) = ∫ b a f(x)dx This theorem essentially says that if you take the area under f(x) over the interval a,b and 'flatten it out', you get a rectangle whose height is given by the value ofthe function atsome point c in the interval Proof Because f(x) is continuous on a,b, the extreme value theorem tells us it mustLet f A !B A function g B !A is the inverse of f if f g = 1 B and g f = 1 A Theorem 1 Let f A !B be bijective Then f has an inverse Proof Let f A !B be bijective We will de ne a function f 1 B !A as follows Let b 2B Since f is surjective, there exists a 2A such that f(a) = bAnd so Z b a f= Z c a f Z c b f= Z c a f Z b c f by De nition 63 Likewise for the case c a

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Theorem 19 If f A → B and g B → A are two functions such that g f = 1A then f is injective and g is surjective Hence a function with a left inverse must be injective and a function with a right inverse must be surjective Proof g f = 1A is equivalent to g(f(a)) = a for all a ∈ A Showing f is injective Suppose a,a′ ∈ A and fF g A → Ris defined by (f g)(x) = f(x) g(x) Proposition 212 Suppose that f,g A → R and f ≤ g If g is bounded from above then sup A f ≤ sup A g, and if f is bounded from below, then inf A f ≤ inf A g Proof If f ≤ g and g is bounded from above, then for every x ∈ A f(x) ≤ g(x) ≤ sup A gComputer Science 122 Exercise Sheet on Expressions For the following, write the expression given in the other two forms listed 1 infix (A B) * C D / (E F

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C code f = (g h) – (i j);Concavity (new) End Behavior (new) Average Rate of Change (new) Holes (new) Piecewise Functions Continuity (new) Discontinuity (new) Arithmetic & Composition CompositionsG f B (a) (b) b g aB (c) a bB g Bf (d) a b 81 DEFINITION OF THE INTEGRAL 163 Thus in figure b, Z b a f represents the shaded area with the area of the thick box subtracted from it, which is the same as the area of the region marked

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This is our focus46 Bijections and Inverse Functions A function f A → B is bijective (or f is a bijection) if each b ∈ B has exactly one preimage Since "at least one'' "at most one'' = "exactly one'', f is a bijection if and only if it is both an injection and a surjection A bijection is also called a onetoone correspondence For b, I can plug 4 in for i and 3 in for h to give me f= 4 (32) which is then f= 4 1 and then f= 5 The following problems deal with translating from MIPS to C Assume that the variables g, h, i, and j are given and could be considered 32bit integers as declared in a C program

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A,b, then Z b a f(x)dx = F(b)−F(a) where F is any antiderivative of f on a,b Z 3 1 2xdx = x2 3 = 32 −12 = 8 The Second Fundamental Theorem of Calculus Let f be continuous on the closed interval a,b, and define G(x) = Z x a f(t)dt where a ≤ x ≤ b Then G0(x) = d dx "Z x a f(t)dt # = f(x) G(x) = Z x 0 sin2(t)dt G0(x) = sin2(x) HTheorem (713) If g is Riemann integrable on a,b and if f(x) = g(x) except for a finite number of points in a,b, then f is Riemann integrable and Z b a f = Z b a g Theorem (715) Suppose f,g 2 Ra,b Then (a) if k 2 R, kf 2 Ra,b and Z b a kf = k Z b a f (b) f g 2 Ra,b and Z b a (f g) = Z b a f Z b a g (c) If f(x) g(x) 8x 2A is called the inverse function of f if f g = id and g f = id If g is the inverse function of f, then we often rename g as f 1 Examples • Let f R !

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B there is a left inverse g B !R which is defined by f g(y)=f(g(y)) = f(y2) = cosy2 Note that even in this case both g f and f g are defined, they are NOT equal to each other Therefore, the ordering is important when we talk about composition Remark 32 In general, we have g f 6= f g even when both are welldefined 6 D E F f g g o f x>6@r,iob g/&"%1 gp4i"%@b "?vf>6@u"%4` f$&h>6, k à ,z,iob ,&>?1g t e m gsu$& ,&ob $& _8;% =, 1g %,i %$whb$(>?

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Fis onto, because g is in Y but g F (x) for any element x in X Fis onto, because no two elements of X are sent by F to the same element of Y Fis onto, because each element of Y is the image of some element of X Fis not onto, because F (c) = e = F (d) Fis not onto, because F (c) = e = FB is a function A function g B !Definition 111 If f G → H is a homomorphism of groups (or monoids) and e′ is the identity element of H then we define the kernel of f as ker(f) = {g ∈ Gf(g) = e′} The kernel can be used to detect injectivity of homomorphisms as long as we are dealing with groups Theorem 112 (Kernels detect injectivity) Let f G → H be a homo

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2g, and f= f(a;b 1)g, then fis injective but not surjective Problem 5 A function f A!Bis onetoone if, for every a2A, there is only one b2Bsuch that f(a) = b Proof This is false This is the de nition of \wellde ned" Problem 6 If jAj= 4 and jBj= 5, then there cannot be a surjective function from Ato BR be the function defined by f(x)=x 3,andlet ⌘ ⌘ ⇣ ⌘The case that g(a) = g(b) is easy So, assume that g(a) 6= g(b) De ne h(x) = f(x) f(b) f(a) g(b) g(a) g(x) Clearly, h(a) = h(b) Applying Rolle's Theorem we have that there is a c with a < c < b such that h0(c) = 0 = f0(c) f(b) f(a) g(b) g(a) g0(c) For this c we have that (f(b) f(a)) g0(c) = (g(b) g(a)) f0(c) The classical Mean Value

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We also say that f f is a onetoone correspondence Theorem425 The composition of injective functions is injective and the compositions of surjective functions is surjective, thus the composition of bijective functions is bijective That is, let fA → B f A → B and gB → C g B → C

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