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Suppose f A !46 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 correspondencePART 1 MODULE 2 Now suppose we merge all of the elements of A with all of the elements of B to form a single, larger set {Citizen Kane, Casablanca, The Godfather, Gone With the Wind, Lawrence of Arabia, The Godfather Part 2, The Wizard of Oz, To Kill A Mockingbird}

Experts are tested by Chegg as specialists in their subject area We review their contentG(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 3du = 3uC Z πdt = πtC Z ur du = ur1 r 1 C for r 6= −1 Z u5 du = u6 6 C for 5 6= −1 Z du u = Z u−1 du = lnuC for r = −1 Z 2x 3 x2 3x dxLet g A →B and f B →C By assumption, since g is not oneone, there exists 2 distinct elements x1 and x2 such that g(x1) = g(x2) = y where y belongs to B Let f(y) = z for some z belonging to C Thus, f o g(x1)=fo g(x2)=zHencefo g cannot be oneone This means that if f and f o g are oneone, g has to be oneone using the fact

By MeanValue Theorem Thus, g(x) = f0(c) where x < c < x 1, that is, lim x→∞ g(x) = lim x→∞ f0(c) = lim c→∞ f0(c) = 0 6 Suppose (a) f is continuous for x ≥ 0, (b) f0(x) exists for x > 0, (c) f(0) = 0, 3 (d) f0 is monotonically increasing Put g(x) = f(x) x (x > 0) and prove that g is monotonically increasing Proof Our goal is to show g0(x) > 0 for all x > 0 ⇔ g0(xTheorem 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 antiderivative of f on an interval I, then the most generalThe slope of a line like 2x is 2, or 3x is 3 etc;

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A b c d e f g h i j k l m n o p q r s t u v w x y zWe set the denominator,which is x2, to 0 (x2=0, which is x=2) When we set the denominator of g (x) equal to 0, we get x=0 So x cannot be equal to 2 or 0 Please click on the image for a better understanding Transcript Example 19 Show that if f A → B and g B → C are onto, then gof A → C is also onto Since g B → C is onto Suppose z ∈ C, then there exists a preimage in B Let the preimage be y Hence, y ∈ B such that g (y) = z Similarly, since f A → B is onto If y ∈ B, then there exists a preimage in A Let the preimage be x Hence, x ∈ A such that f(x) = y Now, gof A

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B be a function We say that f is injective ifº8 !,´ 0 Ë ó Ç J É * > W æ 6 Û È Ë W æ ò63 È9 9 = ) ÃF %?B is a function We call f onetoone if every distinct pair of objects in A is assigned to a distinct pair of objects in B In other words, each object of the target has at most one object from the domain assigned to it There is a way of phrasing the previous definition in a more mathematical language f is onetoone if whenever we have two objects a,c 2 A with a 6= c, we

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The Derivative tells us the slope of a function at any point There are rules we can follow to find many derivatives For example The slope of a constant value (like 3) is always 0;And so on Here are useful rules to help you work out the derivatives of many functions (with examples below)Note the little mark ' means derivative of, and In order to find what value (x) makes f (x) undefined, we must set the denominator equal to 0, and then solve for x f (x)=3/ (x2);

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(b) (f ∘ g)(2) (c) g(f(x)) A newspaper company creates routes with 50 subscribers(n) for each delivery person(d) There is a supervisor (s) for every 10 delivery persons (a) Write d as a function of n (b) Write s as a function of d (c) Substitute to write s as a function of n Show Video Lesson How To Determine The Value Of A Composite Function And How To Determine A CompositeC If a 2A, then (g f)(a) = g(f(a)) If f R !Solution for Find a (f ∘ g)(x) b (g ∘ f)(x) c (f ∘ g)(2) d (g ∘ f)(2), given f(x) =1/x, g(x)= 1/x

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= C B,º4úLU Ã!W = P!WFP ©@ È X!Proofs 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 of sets A;B;C and functions f A !B, and g B !C, for which the statement is false) (a) If f and g areE Z Û / X c Q M ¹ B _ ° W Z ?

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= fc 2C jg(b) = c, for some b 2f(A)g def= g (f A)) = g(B) = C and hence h is also surjective2 X Next we will show that h is injective That is, we will show that if h(a) = h(a0), then we must have that a = a0 Suppose that h(a) = h(a0) By our de nition of h this means that g(f(a)) = g(f(a0)) However, both f and g are injective (since they are bijections) and so g(f(a)) = g(f(a0)) =)f(a) = f5 The roots of the quadratic equation are c and dWithout using a calculator, show that 5 6 (a) Find the values of x for which 2x2 p3x!143 (b) Find the values of k for which the line y!kx #8 is a tangent to the curve x2!4y # 3 7 Functions f and g are defined for x â by fx √ex, gx √2x 03 (i) Solve the equation fg(x) #72 Function h is defined as gf (ii) Express h in termsApply Proposition 27 to the set {cf(x) x ∈ A} = c{f(x) x ∈ A} For sums of functions, we get an inequality Proposition 216 If f,g A → Rare bounded functions, then sup A (f g) ≤ sup A f sup A g, inf A (f g) ≥ inf A f inf A g Proof Since f(x) ≤ supAf and g(x) ≤ supAg for evry x ∈ a,b, we have f(x) g(x) ≤ sup A f sup A g f g

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(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(cC are two functions, then we can compose f and g to get a new function g f A ! Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange

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If f0(x) = 0 for all x2(a;b), then fis constant on the interval (a;b) Corollary 2 If f0(x) = g0(x) for all x2(a;b), then there is a constant Csuch that for all x2(a;b), f(x) = g(x) C Corollary 3 If f0(x) >0 for all x2(a;b), then fis strictly increasing on (a;b) Corollary 4 If f0 (x )If f A !}60 º æ _ ?

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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 We can use this graph that we know and the chart above to draw f(x)2, f(x) 2, 2f(x), 1 2f(x), and f(x) Or to write the previous five functions without the name of the function f, these are the five functions x22,x22, 2x2, x2 2,andx2 These graphs areLearn how to solve f(g(x)) by replacing the x found in the outside function f(x) by g(x)01 c Horizontal distortions For y f x g x f cx, the transformation given by is a horizontal shrink if c!1 and a horizontal stretch is 01 c Examples Use the given graph to sketch the indicated transformations Examples The given function is related to one of the parent functions described in 16 (a) Identify the parent function f (b) Describe the sequence of transformations from f to g

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(b) Differentiate g(x) to show that g'(_x) (c) Find the exact values of x for which g'(x) 1 c 1) 49) 4 The function fis defined by x 2 —2x—3 (a) Show that f(x) = (b) Find the range off (4) (2) (c) Find f l (x) State the domain of this inverse function (3) The function g is defined by g x H 2x2—3, x e R (d) Solve fg(x) = (3) Question 5 continued Leave blank Turn over 5 TheC 0 v k k m < u m s l 0 r 7 a l l d l f l l f 0 l 6 v o = e d = f 6 1 0 1 g a l l d l f l l f 0 L 6 V O @ E @ = D N K < F L = L @F∘g (b)= f(g(b)) = f(c) = 1 f∘g (c)= f(g(c)) = f(a) = 3 Note that g∘f is not deined, because the range of f is not a subset of the domain of g Composion Quesons Example 2 Let f and g be functions from the set of integers to the set of integers defined by f(x) = 2x 3 and g(x) = 3x 2 What is the composition of f and g, and also the composition of g and f ?

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Power rule • If f(x)=xn,thenf!(x)=nxn−1 • d dx!The function f G!Hde ned by f(g) = 1 for all g2Gis a homomorphism (the trivial homomorphism) Note that fis not injective if Gis not the trivial group and it is not surjective if His not the trivial group 4 The determinant det GL n(R) !R is a homomorphism This is the content of the identity det(AB) = detAdetB Here det is surjective, since , for every nonzero real number t, we can nd an" = nxn−1 • To take the derivative of x raised to a power, you multiply in front by the exponent and subtract 1 from the exponent Note The remarkable fact about the power rule is that it works

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Solution f∘g (x)= fB(x−1ax) = (x −1cx)(x−1ax) = x (ca)x = x −1(ac)x = (x−1ax)(x−1cx) = (x ax)b and therefore b ∈ C(x−1ax) Hence x−1C(a)x ⊆ C(x−1ax) 2 Now we prove the inclusion C(x−1ax) ⊆ x−1C(a)x Suppose that d ∈ C(x−1ax) We can write d = x−1cx for some c ∈ G To do this, simply take c = xdx−1 Then x−1cx = x −1(xdx )x = (x x)d(x−1x) = ede = d Now we use lemma 1G U x X A f F r W u f B t @ X O b h j O A C g g g 10ml/033oz ڂ a 炰 A X X ŏ ̂ ڂ Ƃɓ A C P A ł B B R ^ N g Y g p ̕ S Ă g p ɂȂ ܂ B

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R is the function (g f)(x) = 2sinx, whilst f g R !B F(A,B,C,D) = D (A' C') 6 a Since the universal gates {AND, OR, NOT can be constructed from the NAND gate, it is universalSi f A → B y g B → C son tales que g f A → C es inyectiva entonces f es inyectiva Demostraci´on Sean x 1,x 2 ∈ A tal que f(x 1) = f(x 2) aplicando g se obtiene g(f(x 1)) = g(f(x 2)) por ser g f inyectiva se tiene x 1 = x 2 en consecuencia f es inyectiva 3 Teorema 4 Si f A → B y g B → C son tales que g f A → C es suprayectiva entonces g es suprayectiva Demostraci

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Fand g(x) sup A gfor every x2a;b, we have f(x) g(x) sup A f sup A g Thus, f gis bounded from above by sup A f sup A g, so sup A (f g) sup A f sup A g The proof for the in mum is analogous (or apply the result for the supremum to the functions f, g) We may have strict inequality in Proposition 115 because f and gmay take values close to their suprema (or in ma) at di erent) { ÛF ¦ ` ÅLÀ;ÿ ¶ ð x = j ê ,´ ú Y0 ö ÛDÛ j 0#n ¼L Ø6ü Ã ® > & { Ã 0 G JBL ® *O(,´Gý ½ u » k@ >C« W7 {L$,´ t O õ0¦1 _OF(x)g(x)dx= Z 1 0 g(x)f(x)dx= hg;fi (ii) hfg;hi= Z 1 0 (f(x)g(x))h(x)dx= Z 0 1 (f(x)h(x)g(x)h(x))dx = Z 1 0 f(x)h(x)dx Z 1 0 g(x)h(x)dx= hf;hihg;hi (iii) hcf;gi= Z 1 0 cf(x)g(x)dx= c Z 1 0 f(x)g(x)dx= chf;gi The last property requires some more careful analysis First, note that hf;fi= Z 1 0 f(x)2dx 0 since f(x)2 0 for all x;1 To see that hf;fi>0 whenever f(x) is not the constant

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B and g B !R is the function f(x) = sinx and g R !• If f(x)=c,aconstant,thenf!(x)=0 • d dx (c)=0 • The derivative of a constant is zero Example 1 Let f(x)=3 Findf!(x) x y 3!

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