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9 1-Functions

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36 views11 pages

9 1-Functions

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minhhangdao1107
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1 introduction to Functions Relations and Functions The elements of the set X = (1, 2, 3, 4) are associated with the elements of the set ¥ = (3, 5, 7, 9) as depicted by the arrow diagram shown below. This association between X and ¥ is called a retetion from X to ¥. ‘The relation between the element 1 of X and the element . ae 3 of Vis depicted by 1 +> 3. This indicates that Lis the 1) —» 3 starting element and 3 is the ending element and thus we {2-15 may say that I is mapped to 3. Similarly, we have ( 3p 7 2 + 5,3 b> 7 and 4 +9 9, The relation satisfies the Niger te Ne / ¥ ¥ following condition: each element x of X is mapped to exactly one element y of ¥ This relation is called a function or a mapping. A funetion from X to ¥ is a relation which maps each, element x € X toa unique element y € Y. ‘The diagram below shows anothet relation from X to ¥. ‘We have: Sones TLIIT ‘The element 1 is mapped to two elements of ¥. Is this relation a function? 196 Functions 15. Variables x and y are known to be related by an equation of the form axy ~ b = a(x! + bs), where a and b are constants. Observed values of the two variables are shown in the following table: [equ |e aa . Sil ert | 250 | 2.67 U lt Plot xy — 37 against x and use the graph to estimate (a) the value of y when x = 1.5, (b) the values of a and b. 16, The following values of x and y arc believed to obey a law of the form x"y* = 200, ‘where m and n ate constants 7 eal etal eos fe ta y eres ere | aca [nae z Eland Rewrite the given law in a form suitable for drawing a straight line graph. Draw the graph and hence estimate (a) the value-of m and of n, (b) the value of x when y = 10. Linear Law 195 Which of the following relations is not a function? State your reason fa) o) 5 3 (a) The relation is a function, (b) The relation is not a function (c) The relation is @ function, (@ The relation is a function. Functions ate generally denoted by the small letters f, 8, hy If denotes the function from X = {1, 2, 3, 4} to ¥ = (3, 5, 7, 9) defined by: ited. F205, 1237 .and F499 “The element 3 is called the image of 1. Similarly, 5, 7 and 9 are images of 2, 3 and 4 respectively. Fauhermore, any element x of X is related to an element y of Y by the rule y= 2x +L ‘Therefore, the function can be simply defined by using the rule as fixty, wherey=2x+! or 4 fixe ee. When we write ff) = 2x + 1 the image of 1 is {(1) = 3 whereas f(x) isthe image of Pov ahs function, each element of X = (1, 2, 3, 4) must be a stating element and the orig called the domain off. The set of all images R = (3, 5. 7, 9) is ealled the range or image set of f notion f is defined by fs x > 2x7 + 1 with domain fiple2 A X= (-1, 2, 3). Find the image set of f. Ci fo) = 20 +1 The images of -1, 2 and 3 are: f(D) = 2-1 +1 =3 12) =20) +1 £3) =2G¥ +1 The image set is R= (3.9. 12% Functions 197 GBM, 198 Functions ‘The diagram shows part of the mapping fs xb» mx +c. Find (a) the value of m and of ¢, (b) the image of 5 under f, (©) the clement whose image is 3. (a) f@) = mre f:2—7 = f) ice 2m + ¢ fi4n-41 = 4) ie 4m +6 ==1 Solving (1) and (2), we get: m (b) f@) = 4x 415 The image of 5 is 5) =-4x 5 +15 = (©) Let x be the element whose image is 3. Then f(@)=3 = -4r415=3 5 x The diagram shows part of the mapping f. Find (a) the values of a, and c, £ ov three (B) the possible values of k’ | 3 wines a — 1 ot Cay = ae (@ f:le8 = (I) =8 eee fom os HO fe (2) f:-2H-l > f(2)= cee eet Solving (1), (2) and (3), we get: a=2,b=Sandc=1 (b) From (a), f@) = 2 + 5x41 fikts 2k => fk) =2k, ie. 2h + Sk + 2k Qe + 3k + Ck + Ie 0 0 or~1 2 : Hence the possible values of k are -4 and ~1. ‘A function fis defined by f x 9 2° + 5x~ 5 for x > 0. Find the value of x which is unchanged by the mapping Since the image of is, fx) ie, 2 4Sr-5 = v4dr-5 w+ 5- DD x=-Sorl Since -5 is not in the domain, the possible value is 1 +1 are not defined when x = 2. So the functions Expressions such as e+ 2 xB and gixh? can be defined for all values of x except x = 2 and thus we write aetl xed fixe x#2 and gir Hence the element 2 has no images under the functions f and . GEE Aiic 6 A tunetion Fin detined by fx > 3+ 5, x #0. Calculate ee (a) the image of 5 under f, = (b) the possible values of x whose image is 8, axed £0) (a) The image of 5 is (5) = 3x5 + 2a 16 ©) fo) = 8 art i=8 345 =8r 3x - 8r+5=0 Gr- Se 1) =0 5 slot 3 ‘The possible values of x are $ and 1. hire: In this example, te elements $ and 1 have the same image 8 Functions 199 bee ee a aph of a Function A useful representation of the function f : x 1 2x + straight line y = 2+ 1 : 7 are shown ‘The mappings f: 1 +9 3. and £2 3.19 sh with the corresponding points (1, 3) and (3, 7) in the diagram. An immediate observation is that each point (x, y) on the line corresponds to a mapping fixey which maps a number x on the x-axis to a number y on the y-axis For a function g : x +» 2x +1 with domain (ei 1 to), dome x Comeapensins to the range 1 < f(x) < 13 (b) Given thet giz (r+ DY, @ 0

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