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Mathematics Assigment COMPOSITION FUNCTION and INVERSE FUNCTIONBy :Candra Putrie R. (06) Farah Ratna A. (12)XI A7 : 

Mathematics Assigment COMPOSITION FUNCTION and INVERSE FUNCTIONBy :Candra Putrie R. (06) Farah Ratna A. (12)XI A7

COMPOSITION FUNCTION and INVERSE FUNCTION : 

COMPOSITION FUNCTION and INVERSE FUNCTION

Composition Function and Inverse Function : 

Composition Function and Inverse Function Definition of Relation and Function/Mapping. The Kinds of Function. a. Constant Function b. Identity Function c. Absolute Val. Function d. Even Function and Odd Function e. Linear Function - The Straight Line Equation f. Step Function g. Quadratic Function Characteristic of Function (sifat-sifat fungsi). a. Injective Function b. Into Function c. Surjective Function d. Bijective Function Domain and Range of Function. -Algebra Function Composition of Function Inverse Function -Formula Inverse Function The Way To Find Inverse Function Of y = ax2 + bx +c ; a ? 0 Inverse Function of Composition Function - Conclusion

Definition of Relation and Function / Mapping : 

Definition of Relation and Function / Mapping Relation of set A to set B is the way which make a pair element of set A to element of set B, so that is gotten order pair (x.y) with x element of set A (x?A) and y element of set (y?B).

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Function / Mapping is relation which make a pair every in element of set A to exactly one element in set B A B A is called Domain/daerah asal = D B is called Codomain/daerah lawan/kawan = C Set of pairs of Codomain is called Range = R R ? B (codomain) Example : Which is the function ? A B A B A B * * * * * * * * * * * * * * * * * * * * * Function not function not function x f(x)= y Menu

The Kinds of Function : 

The Kinds of Function Constant Function if f : c ? x can be written f(x) = c, so it is called constant function example : Draw the function below : a. f(x) = -3 b. f(x) = 2½

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Answer Constant Function y x 2.5 3 f(x) = -3 f(x) = 2 ½ Menu

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2.Identity Function If f : x ? x can be written f(x) = x, so it is called identity function example : Draw the function of f(x) = x

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Identity Function y x -1 -1 1 1 f(x) = x ? y = x Menu

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3. Absolute Value Function If f(x) : x ? |x| can be written f(x) = |x| so it is called absolute value function example : Draw the function below : a. f(x) = |x| b. f(x) = -|x| c. f(x) = -|x+1|+1

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Absolute value function, f(x) = |x| x -2 -1 0 1 2 y 2 1 0 1 2 y x 1 -2 1 2 1 2 f(x) = |x|

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Absolute Value function, f(x) = -|x| x -2 -1 0 1 2 y -2 -1 0 -1 -2 y x 1 2 -1 -2 1 2 f(x) = -|x|

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Absolute value function, f(x) = -|x+1|+1 x -2 -1 0 1 2 y 0 1 0 -1 -2 y x 1 2 -1 -2 -1 -2 1 f(x) = -|x+1|+1 Menu

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4. Even Function and Odd Function f(x) is called odd function if it has characteristic f(-x) = -f(x) f(x) is called even function if it has characteristic f(-x) = f(x) Example : Prove that function below include even function or odd function! f(x) = x3-2x g. f(x) = sec 2x f(x) = -x4+3x2+1 h. f(x) = -cos3x f(x) = 2x5-x2-2 i. f(x) = 2x3 + x f(x) = sin x x+1 f(x) = 2 cotan x f(x) = -x4 + 3x2 x2-2

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Even and Odd function f(x) = x3-2x f(-x) = -(x)3-2(-x) = -x3 + 2x -f(x) = -x3 + 2 f(-x) = -f(x) ? f(x) = odd function f(x) = -x4+3x2+1 f(-x) = -(-x)4 + 3(-x)2+1 = -x4 + 3x2 +1 f(x) = f(-x) ? f(x) = even function f(x) = 2x5-x2-2 f(-x) = 2(-x)5-(-x)2-2 = -2x5-x2 -2 -f(x) = -2x5+x2+2 f(x) not even or odd function

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d. f(x) = sin x f(-x)= sin(-x) = sin x f(x) = f(-x) ? f(x) = even function e. f(x) = 2 cotan x f(-x) = 2 cotan (-x) ?kw iv = -2 cotan x f(-x) = -f(x) ? f(x) = odd function f. f(x) = -x4 + 3x2 x2-2 f(-x) = x4 + 3(-x)2 (-x)2-2 = x4 + 3x2 x2-2 -f(x) = x4 - 3x2 -x2+2 f(x) ? not even or odd function

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g). f(x) = sec 2x f(x) = sec (-2x) = sec 2x kw IV f(x) = f(-x) ? f(x) even function h). f(x) = - cosec 3x f(x) = - cosec -3x = - cosec 3x f (-x) = f(x) ? f(x) odd function

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i). f(x) = 2x3 + x x+1 f(-x) = 2(-x)3+(-x) -x+1 = -2x3-x -x+1 -f(x) = -2x3 + x -x-1 f(x) not odd / not even function

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Conclusion 1. If the exponents is even number and in kw IV is positive (+) so it is include even function. 2. If the exponents is odd number and in kw IV is negative (-) so it is include odd function. 3. If the exponents is both (even&odd) number so neither. Menu

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5. Linear Function If f : x ? ax + b can be written f(x) = ax + b, so it is called linear function. y g y2 (x2.y2) y2 – y1 y1 (x1,y1) a x1 x2 X g is linear function g(x) = ax + b that form a with x axis

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If g(x) = ax+b g(x2) = ax2+b g(x1) = ax1+b g(x2) – g(x1) = a(x2-x1) y2–y1 = a (x2–x1) a = y2-y1 x2-x1 So, a = m, tan a = y2-y1, is called slope x2-x1 Conclusion If y = ax+b then its slope is a If A(x1.y1) and B(x2.y2) then its slope is mAB = y2-y1 x2-x1 If g line form a with x axis then its slope is m = tan a Menu

The Straight Line Equation : 

The Straight Line Equation Through the point (0.0) with slope m?y=mx Through the point (0.c) with slope is m?y=mx+c Through the point (x1.y1) with slope is m?y-y1 = m(x-x1) Through 2 points (x1.y1) & (x2.y2) slope = y-y1 x-x1 y2-y1 x2-x1 If intersect x axes in (a.0)&(0.b), m ? bx+ay = ab If parallel ? m1=m2 (same slope) If perpendicular ? m2.m1 = -1 y l k a1 = ? (x axes.l) a2 = ? (x axes.k) ß = a1 - a2 ß tan ß = tan (a1 - a2 ) = tan a1 - tan a2 1 + tan a1 tan a2 x = ml – mk 1 + ml mk a1 a2 =

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Examples Find slope of Point A (-2.1) B (3.-4) l line form angle 60° with x axes positive Find the straight line equation Through point (-1.4) and slope 3 Through point (2.1) and parallel 2x-y+3=1 Through point (-3.2) and perpendicular x+2 = 2y-1 3 Parallel x+y = -10 and through intersection point of 2x+y = -1 and x+y = -7

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Solution 1.a) m = y2-y1 = -4-1 x2-x1 3+2 = -5 5 = -1 b) m = tan a = tan 60 = 2.a) y-y1 = m(x-1) y-4 = 3(x+1) y-4 = 3x+3 y = 3x+7 2.b) y = 2x+3 m = 2 y-y1 = m (x-x1) y-1 = 2 (x-2) y-1 = 2x-4 y = 2x-3

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2x+y = -1 x+y = -7 x = 6 y = -1 x+y = -10 y = -x-10 m = m2=-1 y-y1 = m (x-x1) y+13 = -1(x-6) y+13 = -x-6 y = -x-7 c.) y-y1 = m (x-x1) y-2 = -6(x+3) y-2 = -6x-18 y = -6x-16 Menu -

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Step Function. is a function consist of two or more function Example : 1. Draw the graph of function below : a). f(x) = -2 for -1= x =3 1 for 3< x =6 2 for 6< x =9 b). f(x) = -1 for x = 0 x+1 for x > 0

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Solution : a. y x 1 2 2 -1 3 6 9 º º f(x) = -2 for -1= x =3 1 for 3< x =6 2 for 6< x =9

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b. 5 4 3 2 1 1 2 3 4 5 -7 -6 -5 -4 -3 -2 -1 -1 f(x) = x+1 f(x) = -1 ? f(x) = -1 for x = 0 x+1 for x > 0 Menu y x

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7. Quadratic Function. If f: x ? ax2+bx+c, a?0 and a,b and c ? R so it is called Quadratic Function Examples : Draw the graph of absolute quadratic function. a). f(x) = |x2-2x-3| b). f(x) = - |2x2-5x-3| c). f(x) = |-2x2+4x| d). f(x) = - |-x2+7x-12| e). f(x) = |sin x| f). f(x) = -|cos x| g). f(x) = -|sin 2x| h). f(x) = |cos 2x| + 1

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Solution f(x) = |x2-2x-3| D = b2 - 4ac = (-2)2 - 4.1(-3) = 16 (i) x=0 ? y= -3 (0.-3) y=0 ? x2-2x-3 = 0 (x-3)(x+1) = 0 x=3 v x=-1 (3.0) (-1.0) (ii) x = -b = 2 = 1 2a 2 y = D = 16 =-4 -4a -4 (1.-4) -1 3 4 3 2 1 -1 -2 -3 -4 1 f(x) = |x2-2x-3| y x

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b. f(x) = - |2x2-5x-3| D = (-5)2 - 4.2(-3) = 25+24 = 49 (i) x = 0 ? y= -3 (0.-3) y = 0 ? 2x2-5x-3 = 0 (2x+1) (x-3)=0 x=-½ v x=3 (-½.0) (3.0) (ii) x = -b = 5 = 1¼ 2a 4 y = D = 49 = -6? -4a -8 (1¼. -6?) 1 1 2 3 -1 -2 -3 -4 -5 -6 -7 ? ? f(x)=-|2x2-5x-3| y x

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c. f(x) = |-2x2 +4x| D = (4)2-4(-2)0 = 16 (i) x=0 ? y = 0 (0.0) y=0 ? -2x2 +4x x (-2x + 4) x=0 V x=2 (0.0) (2.0) (ii) x = -b = -4 = 1 2a -4 y = D = 16 = 2 -4a 8 ? (1.2) x -1 2 3 Y -6 0 -6 helping point 1 0 1 2 3 6 4 2 -2 -4 -6 f(x) = |-2x2+4x| y x

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d. f(x) = -|-x2+7x-12| D = 49-4(-1)(-12) = 49-48 = 1 (i) x = 0 ? y = -12 (0.-12) y = 0 ? -x2+7x-12 (-x+4) (x-3) x = 4 V x = 3 (4.0) (3.0) (ii) x = -b = -7 = 3½ 2a -2 y = D = 1 -4a 4 ? (3½ . ¼) 1 -2 -4 -6 -8 -10 -12 1 2 3 4 f(x) = - |-x2+7x-12| y x

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f(x) = ? sin x ? f(x) = ? sin x ?

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f(x) = ? cos x ? 90 180 270 360 1 -1 f(x) = ? cos x ?

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1 -1 90 180 270 360 f(x) = -|sin 2x| f(x) = sin x f(x) = -|sin 2x|

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2 1 -1 45 90 135 180 225 270 315 360 f(x) = Cos 2x f(x) = |cos 2x| + 1 f(x) = |cos 2x| + 1 Menu

Characteristic of Function : 

Characteristic of Function Injective Function (fungsi satu-satu) If X1,X2 element set A and X1?X2 so map f(X1)?f(X2) is called Injective Function (fungsi yang petanya berbeda-beda). Examples : f : A?B if (x) = |x| 3. f : B?R if (x) = x+1 4. f : B?C if f(x) = x2 * * * * * * * * * Injective * * * * * * * * Not * * * * * * * * Injective 1 2 1 2 3 4 . . -2 1 2 3 4 . . Inj . . -1 0 1 2 . -½ 0 1 2 3 . . Inj -2 -1 0 1 2 . . 0 1 2 3 4 . . Not

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Into Function (fungsi ke dalam) If each x element A, so there is which has same map in B (ada peta yang sama) Examples : 1 * * * * * * * into * * * * * * * * * * * * not into 2. f : B?C if f(x) = |x| 3. f : R?R : f(x) = X3 -2 -1 0 1 2 . . 0 1 2 3 . . into -½ -1 0 v3 2 ? -1 0 3 v3 8 not Menu

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Surjective Function(fungsi onto) If X1,X2 element A, so map of that function same with codomain (fungsi yang kodomainnya habis) Examples : 1 * * * * * * * * * * surjective * * * * * * * * * * surjective * * * * * * * * not 2. f : B?C : f(x) = x2 -2 -1 0 1 2 0 1 2 3 4 3. f : C?C ; f = |x| 0 1 2 3 4 0 1 2 3 4 not surjective Menu

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Bijective Function (korespondensi satu-satu) is the function which have characteristic function of injective and surjective. Examples : 1 * * * * * * * * * not * * * * * * * * * * bijective 2. f : R?R ; f(x) = x +1 -½ -1 0 1 2 ½ 0 1 2 3 bijective 3. f : B ? R ; f(x) = x2 -2 -1 0 1 2 -1 0 1 2 3 4 not Menu

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Observe the function below, injective, surjective, into or bijective! a. f : A?R ; f(x) = 2x-1 b. f : R?R ; f(x) = x2 c. f : R?R ; f(x) = x3 Solution : a. 1 2 3 4 5 . . . . . -1 0 1 2 3 4 5 . . b. . . -2 -1 0 1 2 . . . . -2 -1 0 1 2 3 4 . . injective into

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. . -2 -1 0 1 2 . . . . -8 . . -1 0 1 . . 8 . . c. bijective Menu

Domain and Range of Function : 

Domain and Range of Function Domain can be formed by determine of the solution set which make possible f(x) is real number. Notes (i) f(x) : polynomials ? Df = {x|x?R} (ii) f(x) : ? Df = {M = 0} (iii) f(x) : ? Df = {M > 0} (iv) f(x) : g(x) ? Df = {h(x) ? 0} h(x) Range (Rf) is subset of codomain can be found by enter Df to f(x) which is known. Menu

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Example : Determine Df and Rf of the function: f(x) = 2x-5 f(x) = x2-4x+3 f(x) = -x3+2x-1 f(x) = f(x) = f(x) = f(x) =

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Solution : Df = {x|x?R} Df = {x|x?R} Df = {x|x?R} f(x) = v2x-x2 2x-x2 = 0 x (2-x) = 0 x = 0 v x = 2 Df = {x|0 = x = 2, x?R} ? ? + - - 0 2

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e. f(x) = x2-3x-4 >0 (x-4) (x+1) >0 x = 4 v x = -1 Df = {x|x<-1vx>4, x?R} º º -1 4 + - + f. f(x) = 2x+1 = 0 x+2 2x = 1 v x+2 = 0 x = -½ v x = -2 Df = {x|x<-2 v x=-½, x?R} g. f(x) = 2 5-3x 5-3x ? 0 -3x ? -5 x ? 5/3 Df = {x|x ? 5/3} º ? + - + -2 -½

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Range Rf = {y|y?R} Rf = {y|y?R} Rf = {y|y?R} f(x) = 0 = x = 2 f(0) = = 0 f(1) = = 1 f(2) = = = 0 f (2/3) = = = Rf = {y|y = 0, y?R} e. f(x) = x<-1 or x>4 f(-3/2) = = f(-2) = f(-3) = Rf = {y|y > 0, y?R}

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f. f(x) = x < -2 v x = -½ f(-3) = f(-½) = = 0 f(-4) = f(0) = f(1) = = 1 Rf = {y|y = 0, y?R} g. Rf = {y|y?R} Menu

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ALGEBRA OF FUNCTION If f and g are two functions which is defined in real numbers, so, can be determined by operations of that function among others: (f + g) (x) = f (x) + g (x) (f – g) (x) = f (x) – g (x) (f x g) (x) = f (x) . g (x) (f/g) (x) = f (x) : g (x), g (x) is not equal to 0 Domain of the functions above: D (f + g) = D (f – g) = D (f x g) Df ? Dg D (f/g) = Df ? Dg, Dg ? 0 Back

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Example: If f (x) = 2x – 3 and g (x) = 2x2 + x – 5 find: (f + g) (x) (g – f) (x) (f.g) (x) (g/f) (X) Answer: f(x) + g(x) = 2x2 + 3x – 8 (2x2 + x – 5) – (2x – 3) = 2x2 – x – 2 (2x2 + x – 5) (2x – 3) = 4x3 + 2x2 – 10x – 6x2 – 3x + 15 = 4x3 – 4x2 – 13x – 15

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d. 2x2 + x – 5/2x – 3 x=2/3 2 1 -5 3 6 2 4 1 H(X) = (2x + 4) : 2 = X + 2 So, (x + 2) + 1/(2x – 3) 2. Find domain of no.1!! a, b, c = {x|x ? R} d (g/f) = {x|x ? 3/2, x ? R}

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3. F = {(-2, -9), (-1, 7), (0,-5), (1, -3), (2, -1), (3, 1)} g = {(-2, 8), (-1, 3), (0, 6), (1, 9), (2, 12)} Find: a. f + g b. f – g c. f.g d. f/g Answer: Df=Dg f+g f-g f.g f/g -2 -9+8=1 -17 -72 -9/8 -1 7+3=10 -4 21 7/3 0 -5+6=1 -11 -30 -5/6 1 -3+9=6 -12 -27 -3/9 2 -1+12=11 -13 -12 -1/12

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f+g = {(-2, -1) (-1, 10) (0, 1) (1, -6) (2,-11)} f-g = {(-2,-17) (-1,-4) (0,-11) (1, -6) (2,-13)} f.g = {(-2,72) (-1,21) (0,-30) (1,-27) (2, -12)} f/g = {(-2,-9/8) (-1,7/3) (),-5/6) (1,3/9) (2,-1/12)}

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COMPOSITION FUNCTION Composition Function of Two Functions g f (fog) g : x ? y ? g (x) = y f : t ? z ? f (y) = z f {g(x)} = z h : x ? z ? h (x) = z (fog) (x) = {f(x)} X? ?Y ?Z

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(fog) (x) = f {g(x)} (gof) (x) = g {f(x)} Condition of (fog) Rg ? Df ? ? D (fog) c Dg R (fog) c Rf Back

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Example: F = {(-1,4) (1,6) (2,3) (8,5)} g = {(3,8) (4,1) (5,-1) (6,2} Find: a. (f o g) and (g o f) b. (f o g) (3) and (g o f) (2) 2. f = {(0,1) (2,4) (5,-1) (4,5)} g = {(2,0) (1,2) (5,3) (6,7)} Find: a. (f o g) and (g o f) b. (f o g) (0) and (g o f) (1)

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3. g = {(3,2) (7,-6) (-5,4) (2,8) (8,-1)} (g o f) = {(0,2) (1,-6) (2,4) (4,8) (3,-1)} Find f function!! 4. f(x) and g(x) is defined in R f(x) = x + 5 ; g(x) = 2x2 Answer: 1. a g Rg=Df f 3 -1 4 4 1 6 5 2 3 6 3 5 so, (fog) = {(3,5) (4,6) (5,4) (6,4)} (f o g)

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f Rf=Dg g -1 4 8 1 6 1 2 5 -1 8 3 2 (g o f) b. (fog) (3) = 5 (gof) (2) = 8

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2. a g Rg=Df f 2 9 1 1 2 4 5 5 -1 6 3 5 4 7 (fog) (fog) = {(2,1) (1,4)}

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f Rf=Dg g 0 2 0 2 1 2 5 4 3 4 5 7 6 -1 (gof) (gof) = {(0,2) (4,3)} b. (fog) (0) = ø (gof) (1) = ø

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3. f g 0 3 2 1 7 6 2 -5 4 4 2 8 3 8 1 (gof) f = {(0,3) (1,7) (2,-5) (4,2) (3,8)}

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4. f(x) = x + 5; g(x) = 2x2 a. (fog) (x) = f(g(x)) = f(2x2) = 2x2 + 5 b. (gof) (x) = g(f(x)) = g(x + 5) = 2 (x + 5)2 = 2(x2 + 10x + 25) = 2x2 + 20x + 50 c. (fog) (-1) = 2x2 + 5 = 2(-1)2 + 5 = 7 d. (gof) (3) = 2x2 + 20x + 50 = 2(3)2 + 20(3) + 50 = 18 + 110 = 128

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5. f(x) = x2 – x + 1 g(x) = x – 5 Find: a. (fog) (x) (fog) (x) = f(g(x)) = f(x – 5) = (x – 5)2 – (x – 5) + 1 = x2 -10x + 25 – x + 5 + 1 = x2 – 11x + 31 b. (gof) (x) (gof) (x) = g(f(x)) = g (x2 – x + 11) = x2 – x + 1 – 5 = x2 –x – 4

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c. (fog) (x) = 57 x2 – 11x + 31 – 57 = 0 x2 – 11x – 26 = 0 (x + 2) (x – 13) = 0 x = -2 or x = 13 d. (gof) (p) = 2; find p!! p2 – p - 4 = 2 p2 – p – 6 = 0 (p – 3) (p + 2) = 0 p = 3 or p = -2

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INVERSE FUNCTION f f -1 f : x ? y ? f(x) = y f-1 : y ? x ? f-1 (y) =x From f(x) function can be found inverse of f (x) it has characteristic bijective. Inverse of f(x) can be written f-1 (x) x? ?y

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Example: Find f-1 (x) of the function below: f(x) = 3x-4 f(x) = v(2x-7); x ? 7/2 Answer: f(x) = 3x – 4 y = 3x – 4 x = y + 4 3 y ? x = x + 4 3 So, f-1 (x) = x + 4 3

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b. f(x) = v(2x-7); x ? 7/2 y = v(2x-7) y2 = 2x – 7 2x = y2 – 7 x = y2 – 7 2 So, f-1 (x) = x2 – 7 2

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Another Examples: f(x) = x (2x – 1); x ? ½ 2. f(x) = 5 (4x + 1); x ? -1/4 3. f(x) = (2 – 3x) (x + 5); x ? -5 f(x) = x2/3 + 10 f(x) = x2; x = 0 f(x) = 4x2, x = 0 f(x) = v2 3x+1 f(x) = 34x – 1 + 5 f(x) = ½ log (2x – 5) f(x) = 6 log 3x + 2

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Answer: y = x (2x – 1) 2xy – y = x 2xy – x = y x(2y – 1) = y x = y (2y – 1) f-1 (x) = x (2x – 1) y = 5 (4x + 1) 4xy + y = 5 x = (5 – y) 4y f-1 (x) = (5 – x) 4x y = (2 – 3x) (x + 5) xy + 5y = 2 – 3x xy – 3x = 2 – 5y

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x(y + 3) = 2 – 5y x = (2 – 5y) (y + 3) f-1 (x) = (2 – 5x) (x +3); x ? -3 4. y = x2/3 + 10 x2/3 = y – 10 x = 3v (y – 10)2 f-1 (x) = 3v (y – 10)2 5. y = x2 x = ± vy f-1 (x) = - vx 6. y = 4x2 x = ± vy 4 f-1 (x) = - vx 4

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y = v2 3x+1 3x + 1 = v2 log y 3x = v2 log y – 1 x = (v2 log y – 1) 3 f-1 (x) = (v2 log x – 1) 3 y = 34x – 1 + 5 34x – 1 = y – 5 4x + 1 = 3 log (y – 5) x = {3 log (y – 5) + 1} 4 f-1 (x) = {3 log (x – 5) + 1} 4 y = ½ log (2x – 5) 2x – 5 = ½ y 2x = ½ y + 5 x = (½ y + 5) 2 f-1 (x) = (½ x + 5) 2 10. y = 6 log 3x + 2 f-1 (x) = 6x-2 3

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Formulas of Inverse Function f(x) = ax ? f-1(x) = aLogX f(x) = aLogX ? f-1(x) = ax f(x) = ax + b ? f-1(x) = x – b a f(x) = ax + b cx + d So, f-1 (x) = b – dx cx - a Notes: (f-1)-1 (x) = f(x) Back

Slide 74: 

THE WAY TO FIND INVERSE FUNCTION OF y = ax2 + bx +c ; a ? 0 Its characteristic isn’t bijective, ‘coz, it has the same mapping. We can change that function becomes bijective by limiting the domain function by looking at the symmetrical equation, That is : x = -b/2a Df = {x|x = -b/2a, x ? R} Df = {x|x = -b/2a, x ? R} Then, the inverse function can be gotten by completing square

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Example: y = -x2 + 6x – 8 x = -b/2a = -6/-2 = 3 Df = {x|x = 3, x ? R} Df = {x|x = 3, x ? R} (x – 3)2 = -y -8 + 9 (x – 3)2 = -y + 1 x – 3 = ± v-y + 1 x = 3 ± v-y + 1 f-1(x) = 3 + v-y + 1 ; x = 1 for x = 3 f-1(x) = 3 - v-y + 1 ; x = 1 for x = 3 Back

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INVERSE FUNCTION OF COMPOSITION FUNCTION fog g f g’ f-1 ( g-1 o f -1) = (fog) -1 (f o g) : x ? z ? (f o g) (x) = z (f o g) -1 : z ? x ? (f o g)-1 (x) = x x? ?y ?z

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Example: f and g function are defined in R, f(x) = 10 – 2x; and g(x) = -2/x; x ? 0. Please find: f-1 (x) and g-1 (x) f-1 (3) and g-1(2) (f o g) (x) (f o g)-1 (x) (g o f) (x) (g o f)-1 (x)

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Solution: f-1 (x) = x – 10 -2 g-1 (x) = -2 x f-1 (3) = 3 – 10 -2 = -7/-2 = 7/2 = 3.5 g-1 (2) = -2/2 = -1

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c. (f o g) (x) = f(g(x)) = f (-2/x) = 10 + 4 x = 10x + 4 ; x ? 0 x d. (f o g)-1 (x) = 4 x – 10 ; x ? 10 e. (g o f) (x) = g(f(x)) = g(10 – 2x) = -2 10 – 2x ; x ? 5 = -1 5 - x

Slide 80: 

f. (g o f)-1 (x) = -2 – 10x -2x = 1 + 5x x

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2. f(x) = 2 + x; g(x) = 4x; h(x) = 5x – 2 Determine: (f o g o h) (x) = f(g(h(x))) = f(g(5x – 2)) = f(4(5x – 2)) = f(20x – 8) = 2 + (20x – 8) = 20x – 6 b (f o g o h) -1 (x) = (x + 6) 20

Slide 82: 

f-1 (x) and g-1 (x) and h-1 (x) f(x) = 2 + x f-1 (x) = x – 2 g-1(x) = x 4 h-1 (x) = (x + 2) 5

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d. (g o h o f) (x) = g(h(f(x))) = g(h(2 + x)) = g(5(2 + x) – 2) = g(8 + x) = 4(8 + x) = 32 + 4x e. (f-1 o h-1 o g-1) (x) = f-1 (h-1 (g-1 (x))) = f-1 (h-1 (x/4)) = f-1 (x/4 + 2) 5 - 2 = f-1 (x + 8) 20 = x – 32 20

Slide 84: 

f. (f o g-1 o h) (x) = f(g-1 (h(x))) = f(g-1 (5x – 2)) = f(5x – 2) 4 = 5x – 2 + 2 4 = 5x + 6 4 g. (g-1 o f-1 o h-1) (2) = g-1 (f-1 (h(2))) = g-1 (f-1 (4/5) = g-1(4/5 – 2) = g-1 (-6/5) = -6/5 4 = -6 20 = -3 10

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h. (f-1 o g o h-1) -1 (1) = (h o g-1 o f) (1) = h(g-1 (f(1))) = h(g-1 (3)) = h(3/4) = 15/4 – 2 = 7/4 = 1 3/4

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Conclusion: (f o g) -1 (x) = (g-1 o f-1) (x) (g o f) -1 (x) = (f-1 o g-1) (x) (f o g o h) -1 (x) = (h-1 o g-1 o f-1) (x) (f o g -1 o h -1 ) -1 (x) = (h o g o f -1 ) (x) Back

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