ir,( )0, rj ﺠـ .ﻁﺭﻴﻘﺔ ﻟﻺﻨﺸﺎﺀ : ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻟﻴﻜﻥ aﻋﺩﺩﺍ ﺤﻘﻴﻘﻴﺎﻹﻨﺸﺎﺀ ﺍﻟﻤﺴﺘﻘﻴﻡ ) (dﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ ) y = axﺃﻱ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ ,( x → axﻴﻜﻔﻲ ﺃﻥ ﻨﻌﻴﻥﻨﻘﻁﺘﻴﻥ ﻤﺘﻤﺎﻴﺯﺘﻴﻥ ﻤﻥ ﻫﺫﺍ ﺍﻟﻤﺴﺘﻘﻴﻡ ,ﺃﺤﺩﻫﻤﺎ 0ﻭﺍﻷﺨﺭﻯ ﺘﺨﺘﺎﺭ ﻤﺜﻼ ﻟﻤﺎ y = a , x = 1ﻓﻨﻌﻴﻥrjﻫﻭ ,ﺍirﻟ,ﻤﺴ0ﺘﻘﻴﻡ )( )(OAﺍﻟﻨﻘﻁﺔ ) A(1, aﻭﺍﻟﻤﺴﺘﻘﻴﻡ ) (d ﻤﺜﻼ :ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡﻟﻨﻨﺸﺊ ) (d3 ), (d 2 ), (d1ﺍﻟﺘﻤﺜﻴﻼﺕ ﺍﻟﺒﻴﺎﻨﻴﺔ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻟﻠﺩﻭﺍل h, g, fﺤﻴﺙ : f : x → f (x) = 2x g : x → g(x) = −3x h : x → h(x) = 0ﻜل ﻤﻥ fﻭ gﻭ hﺩﺍﻟﺔ ﺨﻁﻴﺔ ﻤﻨﻪ ﻜل ﻤﻥ ) (d3 ), (d 2 ), (d1ﻤﺴﺘﻘﻴﻡ ﻴﺸﻤل ﺍﻟﻤﺒﺩﺃ 0ﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯ ﻟﻤﺎ: x = 1 f (x) = 2ﻤﻨﻪ ﺍﻟﻨﻘﻁﺔ ) A(1,2ﺘﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻡ ) (d1 g(x) = −3ﻤﻨﻪ ﺍﻟﻨﻘﻁﺔ ) B(1,−3ﺘﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻡ) (d2 h(x) = 0ﻤﻨﻪ ﺍﻟﻨﻘﻁﺔ ) C(1,0ﺘﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻡ ) (d3 ﻭﻋﻠﻴﻪ ) (d3 ) = (OC), (OB) = (d2 ), (OA) = (d1
(d 2 ) (d 1 ) :ﻤﻨﻪ ﺍﻟﺸﻜل rj • A (d 3 ) 0 ir C •B
• ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ ﺕ f : 1ﺩﺍﻟﺔ ﺨﻁﻴﺔ ﺒﺤﻴﺙ f (13) = 2 ﺃﺤﺴﺏ )f (−5), f (−2), f (1 ﺕ k, h, g, f : 2ﺩﻭﺍل ﺨﻁﻴﺔ ﺒﺤﻴﺙ: f (3) = 2ﻭ g(2) = −3ﻭ h(5) = 5ﻭ k(−1) = 2 ﻋﺭﻑ ﻜل ﻤﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺨﻁﻴﺔ k, h, g, fﺒﺩﺴﺘﻭﺭ( ) ( )0,ir, rj ﺍﻟﻤﺘﻌﺎﻤﺩ ﻭﺍﻟﻤﺘﺠﺎﻨﺱ ﺍﻟﻤﻌﻠﻡ ﻨﻔﺱ ﺇﻟﻰ ﺇﻟfﻰ,ﻤgﻌﻠ,ﻡir,krj, hﺒﺎ,ﻟﻨ0ﺴﺒﺔ ﺒﻴﺎﻨﻴﺎ ﺍﻟﺩﻭﺍل ﺍﻟﺨﻁﻴﺔ ﻤﺜل :ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺕ3ﻓﻲ ﺍﻟﺸﻜل ﺍﻟﻤﻭﺍﻟﻲ ) (d3 ), (d 2 ), (d1ﻫﻲ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﺍﻟﺘﻤﺜﻴﻼﺕ ﺍﻟﺒﻴﺎﻨﻴﺔ ﻟﻠﺩﻭﺍل . h, g, f) (d 2 ) (d1 ) (d 3 rj ir 0 (1ﺃﺜﺒﺕ ﺃﻥ ﻜﻼ ﻤﻥ ﺍﻟﺩﻭﺍل h, g, fﻫﻲ ﺩﺍﻟﺔ ﺨﻁﻴﺔ. (2ﻋﻴﻥ ﻜﻼ ﻤﻥ ﺍﻷﻋﺩﺍﺩ )h(−2), g(1), f (2 (3ﺃﻋﻁ ﻋﺒﺎﺭﺓ ﻜل ﻤﻥ ) g(x), f (xﻭ ) h(xﺒﺩﻻﻟﺔ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ) xﺇﻋﻁ ﻋﺒﺎﺭﺓ ) f (xﺒﺩﻻﻟﺔ xﻤﻌﻨﺎﻩ ﺇﻋﻁﺎﺀ ﺍﻟﺩﺴﺘﻭﺭ ﺍﻟﺫﻱ ﻴﻌﺭﻑ ﺍﻟﺩﺍﻟﺔ ( f
ﺕ: 4ﺃﻋﻁ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﻜل ﻤﻥ ﺍﻟﺩﻭﺍل k, h, g, fﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺎﺘﻴﺭ: )k(x = )x, h(x = 1 x, )g(x = −x 3, f (x) = 5x 2ﺕ f : 5ﺩﺍﻟﺔ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻔﻬﺎ Rﻭﻓﻲ ﺍﻟﺠﺩﻭل ﺍﻟﻤﻭﺍﻟﻲ ﺒﻌﺽ ﺍﻟﻘﻴﻡ ﻟﻠﻌﺩﺩ ) f (xﺤﺴﺏ ﻗﻴﻡ ﺍﻟﻌﺩﺩ x x 23 45 6 7 89)f (x 1 1,5 2 2,5 3 3,5 3,5 3,5 ﻫل ﺍﻟﺩﺍﻟﺔ fﺨﻁﻴﺔ ؟
• ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ: ﺕ f :1ﺩﺍﻟﺔ ﺨﻁﻴﺔ ﺒﺤﻴﺙ f (13) = 2 ﺒﻤﺎ ﺃﻥ fﺨﻁﻴﺔ ﻓﺈﻥ f (x) = ax :ﺤﻴﺙ aﺜﺎﺒﺕ ﺤﻘﻴﻘﻲ ﻭﻋﻠﻴﻪa = 6 : 1 a = 2 ﻓﺈﻥ: ﻭﺒﻤﺎ ﺃﻥ f (13) = 2 3ﻭﻤﻨﻪ f (x) = 6x :ﻭﺒﺎﻟﺘﺎﻟﻲ f (1) = 6 :ﻭ f (−2) = −12ﻭ f (−5) = −30 ﺕ : 2ﻴﻤﻜﻥ ﺍﻻﺴﺘﻔﺎﺩﺓ ﻤﻥ ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺴﺎﺒﻕ ﺕ: 3 (1ﻜل ﺩﺍﻟﺔ ﻤﻥ ﺍﻟﺩﻭﺍل h, g, fﻫﻲ ﺩﺍﻟﺔ ﺨﻁﻴﺔ ﻷﻥ ﻜﻼ ﻤﻥ ) (d3 ), (d 2 ), (d1ﻴﻤﺭ ﻤﻥ ﺍﻟﻤﺒﺩﺃ .0ﻭﻜﻼ ﻤﻨﻬﻤﺎ ﻻ ﻴﻭﺍﺯﻱ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻤﻥ d1ﻨﺠﺩ ﺃﻥ( )f (2) = 2 : (2ﻭﻤﻥ) (d2ﻨﺠﺩ ﺃﻥg(1) = −2 : ﻭﻤﻥ ) (d3ﻨﺠﺩ ﺃﻥh(−2) = −1: (3ﻋﺒﺎﺭﺓ ﻜل ﻤﻥ ) g(x), f (xﻭ ) h(xﺒﺩﻻﻟﺔ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ xﺒﺎﻻﺴﺘﻔﺎﺩﺓ ﻤﻥ ﺍﻟﻔﺭﻉ \" \"2ﻭﻜﺫﺍ)h(x = 1 x ﻭ ﻭ g(x) = −2x f (x) = x ﺍﻟﺘﻤﺭﻴﻥ ﺍﻷﻭل ﻨﺠﺩ: 2 ﺕ : 4ﺍﻟﺩﺍﻟﺔ fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻷﻥ5 > 0 : ﺍﻟﺩﺍﻟﺔ gﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻷﻥ− 3 < 0 : 1 > ﻷﻥ0 : ﺘﻤﺎﻤﺎ ﻤﺘﺯﺍﻴﺩﺓ ﺍﻟﺩﺍﻟﺔ h 2 ﺍﻟﺩﺍﻟﺔ kﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻷﻥ1 > 0 : f )(9 ≠ )f (2 ﺕ : 5ﺍﻟﺩﺍﻟﺔ f 9 ﻟﻴﺴﺕ ﺨﻁﻴﺔ ﻷﻥ2 :
ﺍﻟﺩﻭﺍل ﺍﻟﺘﺂﻟﻔﻴﺔ ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ : -ﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﺘﺂﻟﻔﻴﺔ -ﺩﺭﺍﺴﺔ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ -ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ -ﺘﻌﻴﻴﻥ ﺩﺴﺘﻭﺭ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﻤﻌﻴﻨﺔ ﺒﺼﻭﺭﺘﻲ ﻋﺩﺩﻴﻥ ﺃﻭ ﺒﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ .1ﺘﻌﺭﻴﻑ .2ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ .3ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ .4ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ .5ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ .6ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ
• ﺘﻌﺭﻴﻑ :ﻨﺴﻤﻲ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﻜل ﺩﺍﻟﺔ fﻤﻌﺭﻓﺔ ﺒﺩﺴﺘﻭﺭ ﻤﻥ ﺍﻟﺸﻜل f (x) = ax + bﺤﻴﺙ aﻭ bﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ﺜﺎﺒﺘﺎﻥ ﻭﻋﻨﺩﺌﺫ aﻴﺴﻤﻰ ﻤﻌﺎﻤل ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ f ﻭ bﻴﺴﻤﻰ ﺍﻟﻘﻴﻤﺔ ﻓﻲ ﺍﻟﻤﺒﺩﺃ ﻟﻠﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ f ﺃﻤﺜﻠﺔ : -ﺍﻟﺩﻭﺍل h, g, fﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﺩﺴﺎﺘﻴﺭ := ) h(xﻫﻲ ﺩﻭﺍل ﺘﺂﻟﻔﻴﺔ 2, )g(x = − 1 x − 7, f )(x = −3x + 5 2 ﻤﻌﺎﻤل ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ fﻫﻭ -3ﻭﻗﻴﻤﺘﻬﺎ ﻓﻲ ﺍﻟﻤﺒﺩﺃ ﻫﻲ 5-7 ﻭﻗﻴﻤﺘﻬﺎ ﻓﻲ ﺍﻟﻤﺒﺩﺃ ﻫﻲ − 1 ﻫﻭ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ g ﻤﻌﺎﻤل 2 h(x) = 0.x + 2ﻤﻨﻪ ﻤﻌﺎﻤل ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ hﻫﻭ 0ﻭﻗﻴﻤﺘﻬﺎ ﻓﻲ ﺍﻟﻤﺒﺩﺃ ﻫﻲ 2 -ﻜل ﺩﺍﻟﺔ ﺨﻁﻴﺔ x → axﻫﻲ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﻗﻴﻤﺘﻬﺎ ﻓﻲ ﺍﻟﻤﺒﺩﺃ 0 -ﺍﻟﺩﺍﻟﺔ kﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﺩﺴﺘﻭﺭ k(x) = x 2 + 3ﻟﻴﺴﺕ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ. ﺘﻤﺭﻴﻥ f :ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﺒﺤﻴﺙ f (1) = 3 :ﻭ f (−1) = 5 ﺍﻟﻤﻁﻠﻭﺏ : (1ﺇﻋﻁﺎﺀ ﻋﺒﺎﺭﺓ ) f (xﺒﺩﻻﻟﺔ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ x (2ﺤﺴﺎﺏ ) f (0ﻭ )f (−5 ﺍﻟﺤل: (1ﺒﻤﺎ ﺃﻥ fﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﻋﺒﺎﺭﺓ ) f (xﺒﺩﻻﻟﺔ xﻤﻥ ﺍﻟﺸﻜل f (x) = ax + bﺤﻴﺙ aﻭ b ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ﺜﺎﺒﺘﺎﻥ ﻭﻟﺘﻌﻴﻴﻥ ﻋﺒﺎﺭﺓ ) f (xﺒﺩﻻﻟﺔ xﻴﻜﻔﻲ ﻭﻴﻠﺯﻡ ﺃﻥ ﻨﺠﺩ aﻭ b ﻭﻟﺩﻴﻨﺎ f (1) = 3ﻭ f (−1) = 5ﻤﻥ ﺍﻟﻤﻌﻁﻴﺎﺕ ﻭﻟﺩﻴﻨﺎ ﻜﺫﻟﻙ f (1) = a.1 + b :ﻭ f (−1) = a(−1) + b ﻤﻨﻪ ﻭﺒﺎﻟﺘﻌﺩﻴﺔ a.1 + b = 3 :ﻭ a(−1) + b = 5 ﺇﺫﻥ (1)......a + b = 3 :
ﻭ (2)..... − a + b = 5 ﻤﻥ )a = 3 − b (1ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ )− (3 − b) + b = 5 (2 ﺇﺫﻥ − 3 + b + b = 5 : 2b − 3 = 52b = 5 + 32b = 8=b 8 2b=4 ﻭﻟﺩﻴﻨﺎ a = 3 − b : ﻤﻨﻪ a = 3 − 4 : ﺇﺫﻥ a = −1:ﻤﻨﻪ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ fﻤﻌﺭﻓﺔ ﺒﺎﻟﺩﺴﺘﻭﺭ f (x) = −x + 4 (2ﻟﺩﻴﻨﺎ f (x) = −x + 4ﻤﻨﻪ f (0) = −0 + 4 ﺇﺫﻥ f (0) = 4ﻭ f (−5) = −(−5) + 4 =5+4 f (−5) = 9
• ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ: ﻜﻥ aﻭ bﻋﺩﺩﻴﻥ ﺤﻘﻴﻘﻴﻴﻥ ﻭﻟﺘﻜﻥ fﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﺩﺴﺘﻭﺭ f (x) = ax + b ﻟﻨﺩﺭﺱ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ fﻋﻠﻰ ) Rﺍﻟﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﺒﺩﺴﺘﻭﺭ(ﻟﻴﻜﻥ uﻭ ϑﻋﺩﺩﻴﻥ ﺤﻘﻴﻘﻴﻴﻥ ﻜﻴﻔﻴﻴﻥ ﻟﻨﻔﺭﺽ ϑ > uﻭﻟﻨﻘﺎﺭﻥ ) f (uﻭ ) f (ϑﻭﻟﺫﻟﻙ ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ﺍﻟﻔﺭﻕ )f (ϑ) − f (u ﻟﺩﻴﻨﺎ f (ϑ) − f (u) = (aϑ + b) − (au + b) : = aϑ + b − au − b = aϑ − au ﻤﻨﻪf (ϑ) − f (u) = a(ϑ − u) : ﻭﺒﻤﺎ ﺃﻥ ϑ > uﻓﺈﻥ ϑ − u > 0ﻭﺇﺸﺎﺭﺓ ) f (ϑ) − f (uﻫﻲ ﻨﻔﺴﻬﺎ ﺇﺸﺎﺭﺓ a ﺇﺫﻥ :•ﻓﻲ ﺍﻟﺤﺎﻟﺔ : a > 0ﻴﻜﻭﻥ f (ϑ) − f (u) > 0ﺃﻱ ) f (ϑ) > f (uﻭﺍﻟﺩﺍﻟﺔ fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ R•ﻓﻲ ﺍﻟﺤﺎﻟﺔ : a < 0ﻴﻜﻭﻥ f (ϑ) − f (u) < 0ﺃﻱ ) f (u) > f (ϑﻭﺍﻟﺩﺍﻟﺔ fﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ R•ﻓﻲ ﺍﻟﺤﺎﻟﺔ : a = 0ﻴﻜﻭﻥ f (ϑ) − f (u) = 0ﺃﻱ ) f (ϑ) = f (uﻭﺘﻜﻭﻥ fﺜﺎﺒﺘﺔ ﻋﻠﻰ R ﺍﻟﻨﻅﺭﻴﺔ : ﻟﻴﻜﻥ aﻭ bﻋﺩﺩﻴﻥ ﺤﻘﻴﻘﻴﻴﻥ ﻭﻟﺘﻜﻥ fﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭ f (x) = ax + b ﺇﺫﺍ ﻜﺎﻥ : a > 0 ﺍﻟﺩﺍﻟﺔ fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ R ﺇﺫﺍ ﻜﺎﻥ : a < 0 ﺍﻟﺩﺍﻟﺔ fﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ R ﺇﺫﺍ ﻜﺎﻥ : a = 0 ﺍﻟﺩﺍﻟﺔ fﺜﺎﺒﺘﺔ ﻋﻠﻰ R ﻤﺜﺎﻻ :ﺤﺴﺏ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﺴﺎﺒﻘﺔ: ﺍﻟﺩﺍﻟﺔ fﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭ f (x) = 3x + 11ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ R
ﻷﻥ ﻋﺒﺎﺭﺓ ) f (xﻤﻥ ﺍﻟﺸﻜل ax + bﺤﻴﺙ a = 3ﻭ 3 > 0ﺍﻟﺩﺍﻟﺔ gﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭ g(x) = −7x + 9ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ Rﻷﻥ ﻋﺒﺎﺭﺓ )g(x ﻤﻥ ﺍﻟﺸﻜل ax + bﺤﻴﺙ a = −7ﻭ − 7 < 0 ﺍﻟﺩﺍﻟﺔ hﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭ h(x) = 2ﺜﺎﺒﺘﺔ ﻋﻠﻰ ) Rﻭﻫﺫﺍ ﻭﺍﻀﺢ ﻷﻥ ) h(xﻤﺴﺘﻘل ﻋﻥ ( x • ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ:ﻟﻴﻜﻥ aﻭ bﻋﺩﺩﻴﻥ ﺤﻘﻴﻘﻴﻴﻥ ﻭﻟﺘﻜﻥ fﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭ f (x) = ax + b ﺍﻋﺘﻤﺎﺩﺍ ﻋﻠﻰ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﺴﺎﺒﻘﺔ ﻴﻜﻭﻥ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ fﻜﻤﺎ ﻴﻠﻲ: ﻓﻲ ﺍﻟﺤﺎﻟﺔ : a > 0 ∞x − ∞+)f (x ﻓﻲ ﺍﻟﺤﺎﻟﺔ : a < 0 ∞x − ∞+)f (x ﻓﻲ ﺍﻟﺤﺎﻟﺔ : a = 0 ∞x − ∞+f (x) b b
• ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﺩﺍﻟﺔ ﺨﻁﻴﺔ: ﻨﻅﺭﻴﺔ : ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ) , (0,ir, rj * ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﻫﻭ ﻤﺴﺘﻘﻴﻡ ﻻ ﻴﻭﺍﺯﻱ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ**ﻜل ﻤﺴﺘﻘﻴﻡ ﻻ ﻴﻭﺍﺯﻱ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻫﻭ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ 0, ir, rj ﺍﻟﺒﺭﻫﺎﻥ : ( ), ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡﻟﻴﻜﻥ aﻭ bﻋﺩﺩﻴﻥ ﺤﻘﻴﻘﻴﻴﻥ ﻭﻟﺘﻜﻥ fﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭ f (x) = ax + b ﻭﻟﻴﻜﻥ) (Dﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ f ﻟﺩﻴﻨﺎ f (0) = bﻤﻨﻪ) (Dﻴﺸﻤل ﺍﻟﻨﻘﻁﺔ )B(0,b ﻜﺫﻟﻙ ,ﻤﺜﻼ f (1) = a + b ,ﻤﻨﻪ) (Dﻴﺸﻤل ﺍﻟﻨﻘﻁﺔ )C(1, a + b ﺍﻟﻨﻘﻁﺘﺎﻥ Bﻭ Cﻤﺘﻤﺎﻴﺯﺘﺎﻥ)ﻷﻥ ﻓﺎﺼﻠﺘﻬﻤﺎ ﻤﺘﻤﺎﻴﺯﺘﺎﻥ(ﺤﺘﻰ ﻨﺜﺒﺕ ﺃﻥ) (Dﻤﺴﺘﻘﻴﻡ ﻴﻜﻔﻲ ﺃﻥ ﻨﺜﺒﺕ ﺃﻥ) (Dﻫﻭ ﺍﻟﻤﺴﺘﻘﻴﻡ) (BCﻤﻬﻤﺎ ﺘﻜﻭﻥ ﺍﻟﻨﻘﻁﺔ )M (x, y Mﺘﻨﺘﻤﻲ ﺇﻟﻰ) (BCﻤﻌﻨﺎﻩ M , C, Bﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ ﻭﻫﺫﺍ ﻤﻌﻨﺎﻩ BC // BMﺃﻱ BC1a ﻭ BC1(a−+0b) − b BM x − b ﺃﻱ BM x − 0 ﻭﻟﻨﺎ y y − b BC // BMﻴﻌﻨﻲ )1.( y − b) − ax = 0ﺤﺴﺏ ﺸﺭﻭﻁ ﺘﻭﺍﺯﻱ ﺸﻌﺎﻋﻴﻥ( ﻤﻨﻪ Mﺘﻨﺘﻤﻲ ﺇﻟﻰ) (BCﻴﻌﻨﻲ y − b − ax = 0ﺃﻱ y = ax + b y = ax + b ﺍﻟﺘﻲ ﺘ)ﺤyﻘ,ﻕM (x ﻤﻨﻪ ) (BCﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ•B ﺃﻱ )y = f (x ﻤﻨﻪ ) (BCﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﺍﻟﺘﻲ ﺇﺤﺩﺍﺜﻴﺎﻫﺎ ﻤﻥ ﺍﻟﺸﻜل))(x, f (x•C
ﻭﻋﻠﻴﻪ ) = (D)(BCﻭ) (BCﻻ ﻴﻭﺍﺯﻱ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻷﻥ Bﻭ Cﻟﻴﺱ ﻟﻬﻤﺎ ﻨﻔﺱ ﺍﻟﻔﺎﺼﻠﺔ ﻤﻨﻪ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﻫﻭﻤﺴﺘﻘﻴﻡ ﻻ ﻴﻭﺍﺯﻱ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ** ﻟﻴﻜﻥ dﻤﺴﺘﻘﻴﻤﺎ ﻻ ﻴﻭﺍﺯﻱ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ) ( ) (dﻴﻘﻁﻊ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻓﻲ ﻨﻘﻁﺔ ' Bﻭﺍﻟﻤﺴﺘﻘﻴﻡ)(lﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﺍﻟﺘﻲ ﻓﺎﺼﻠﺘﻬﺎ 1ﻓﻲ ﻨﻘﻁﺔ '(d ) C'B 'C ﺇﺫﺍ ﺴﻤﻴﻨﺎ ' bﺘﺭﺘﻴﺏ ' Bﻭ ' cﺘﺭﺘﻴﺏ 'C rj •M ﻟﺩﻴﻨﺎ )' B'(0,bﻭ )' C'(1, cﻭ ) (B'C') = (d 0 ir ﻤﻨﻪ) (dﻫﻭ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ )M (x, y 'B M x − 0b' ﻭﻟﻨﺎ B'C' // B' M ﺒﺤﻴﺙ y −ﻭ 'B' M // B'C 'B M x − b' ﻭ ﺃﻱ B'C'1c'−b' ﻭ B'C'1c'−−b0' yﻴﻌﻨﻲ x(c'−b') − ( y − b') = 0ﺃﻱ (c'−b')x − y + b' = 0ﺃﻱ 'y = (c'−b')x + bﻤﻨﻪ) (dﻫﻭ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ)' M (x, (c'−b')x + bﺤﻴﺙ xﻋﺩﺩ ﺤﻘﻴﻘﻲ ﺃﻱ) (dﻫﻭ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ 'x → (c'−b')x + b ﻭﻋﻠﻴﻪ ﻜل ﻤﺴﺘﻘﻴﻡ ﻻ ﻴﻭﺍﺯﻱ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻫﻭ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ. ﺕ .ﺍﺴﺘﻨﺘﺎﺝ ,ﺘﻌﺭﻴﻑ : ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ) (0,ir, rjﻟﻴﻜﻥ aﻭ bﻋﺩﺩﻴﻥ ﺤﻘﻴﻘﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ) M (x, yﺍﻟﺘﻲ ﺘﺤﻘﻕ y = ax + bﻫﻲ ﻤﺴﺘﻘﻴﻡ ) (dﻻ ﻴﻭﺍﺯﻱ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻭ y = ax + bﺘﺴﻤﻰﻤﻌﺎﺩﻟﺔ ﻟﻠﻤﺴﺘﻘﻴﻡ ) (dﻭ aﻴﺴﻤﻰ ﻤﻌﺎﻤل ﺘﻭﺠﻴﻪ ﺍﻟﻤﺴﺘﻘﻴﻡ ) (dﻭ bﻴﺴﻤﻰﺍﻟﺘﺭﺘﻴﺏ ﻓﻲ ﺍﻟﻤﺒﺩﺃ ﻟﻠﻤﺴﺘﻘﻴﻡ )) (dﻫﺫﺍ ﻷﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ ) (dﻴﺸﻤل ﺍﻟﻨﻘﻁﺔ )B(0,b ﻭﻓﻲ ﻫﺫﻩ ﺍﻟﻨﻘﻁﺔ ) B (dﻴﻘﻁﻊ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ(
( )0,ir, rj ﺠـ .ﻁﺭﻴﻘﺔ ﻟﻺﻨﺸﺎﺀ : ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻟﻴﻜﻥ aﻭ bﻋﺩﺩﻴﻥ ﺤﻘﻴﻘﻴﻴﻥﻹﻨﺸﺎﺀ ﺍﻟﻤﺴﺘﻘﻴﻡ ) (dﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ ) y = ax + bﺃﻱ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ ,( x → ax + bﻴﻜﻔﻲ ﺃﻥ ﻨﻌﻴﻥ ﻨﻘﻁﺘﻴﻥ ﻤﺘﻤﺎﻴﺯﺘﻴﻥ ﻤﻥ ﻫﺫﺍ ﺍﻟﻤﺴﺘﻘﻴﻡ,ﻟﺫﺍ ﻨﺨﺘﺎﺭ ﻗﻴﻤﺘﻴﻥ x1ﻭ x2ﻤﺘﻤﺎﻴﺯﺘﻴﻥ ﻟـ xﻨﺤﺴﺏ y1ﻭ y2ﺤﻴﺙ y1 = ax1 + bﻭ y2ﺤﻴﺙ y2 = ax2 + bﺜﻡ ﻨﻌﻠﻡ ﺍﻟﻨﻘﻁﺘﻴﻥ Aﻭ Bﺤﻴﺙ ) A(x1, y1ﻭ ) B(x2 , y2 ( )0,ir,(rjﻫﻭ ﺍﻟﻤﺴﺘﻘﻴﻡ )AB ﻓﺎﻟﻤﺴﺘﻘﻴﻡ ) (d ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺜﻼ :ﺍﻟﻤﺴﺘﻭﻱf )(x = 1 x + ﺒﺎﻟﺩﺴﺘﻭﺭ 3 ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ R ﻟﻨﻨﺸﺊ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ f 2) f (xﻤﻥ ﺍﻟﺸﻜل ax + bﻤﻨﻪ fﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﻭﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ﻫﻭ ﺍﻟﻤﺴﺘﻘﻴﻡ ) (dﺫﻭ ﻟﻨﻌﻴﻥ ﻨﻘﻁﺘﻴﻥ ﻤﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ ) (d =y 1 x + 3 ﺍﻟﻤﻌﺎﺩﻟﺔ 2 ﻟﻤﺎ y = 3 : x = 0ﻤﻨﻪ ﺍﻟﻨﻘﻁﺔ ) A(0,3ﺘﻨﺘﻤﻲ ﺇﻟﻰ ) (dﺘﻨﺘﻤﻲ ﺇﻟﻰ ) (d ﻤﻨﻪ ﺍﻟﻨﻘﻁﺔ )B(2,4 ﺃﻱ y = 4 y = 1 .2 + 3: x = ﻟﻤﺎ 2 2 ﻭﻫﻜﺫﺍ ) (dﻫﻭ ﺍﻟﻤﺴﺘﻘﻴﻡ ) (ABﻤﻨﻪ ﺍﻟﺸﻜل: B A •
• ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ: ﺕ : 1ﻟﺘﻜﻥ fﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﺩﺴﺘﻭﺭ f (x) = −2x + 7ﺒﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ، gﺼﻭﺭﺓ ﺍﻟﻌﺩﺩ 1ﻫﻲ ﻨﻔﺴﻬﺎ ﺼﻭﺭﺘﻪ ﺒﺎﻟﺩﺍﻟﺔ fﻭﺼﻭﺭﺓ ﺍﻟﻌﺩﺩ -1ﻫﻲ ﻨﺼﻑ ﺼﻭﺭﺘﻪ ﺒﺎﻟﺩﺍﻟﺔ f (1ﺃﺤﺴﺏ ﻤﻌﺎﻤل ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ , gﻭﻗﻴﻤﺘﻬﺎ ﻓﻲ ﺍﻟﻤﺒﺩﺃ ﻭ -10ﺒﺎﻟﺩﺍﻟﺔ g 3 (2ﺃﺤﺴﺏ ﺼﻭﺭ ﺍﻷﻋﺩﺍﺩ2 ,2:( )0,ir, rj (3ﻤﺜل ﺒﻴﺎﻨﻴﺎ ﺍﻟﺩﺍﻟﺘﻴﻥ fﻭ gﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻨﻔﺱ ﺍﻟﻤﻌﻠﻡ ﺕ f : 2ﻭ gﺩﺍﻟﺘﺎﻥ ﺘﺂﻟﻔﻴﺘﺎﻥ ﺒﺤﻴﺙ: g )(−4 = − 3 ﻭ g(− )1 = ﻭ6 ﻭf (6) = −1 f (2) = 1 2 4 (1ﺃﻋﻁ ﻋﺒﺎﺭﺓ ) f (xﺒﺩﻻﻟﺔ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ x (2ﺃﻋﻁ ﻋﺒﺎﺭﺓ ) g(xﺒﺩﻻﻟﺔ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ x (3ﻋﻴﻥ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﻜل ﻭﺍﺤﺩﺓ ﻤﻥ ﺍﻟﺩﺍﻟﺘﻴﻥ fﻭ gﺍﻟ4ﻤ(ﻌﻠﻨﻡﺴﻤrjﻲ 0d,1ir,ﻭ , d 2ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ,ﺍﻟﺘﻤﺜﻴﻠﻴﻥ ﺍﻟﺒﻴﺎﻨﻴﻴﻥ ﻟﻠﺩﺍﻟﺘﻴﻥ fﻭ gﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻨﻔﺱ) ( ) () ( ﺃ -ﺃﻨﺸﺊ ) (d1ﻭ) (d2 ﺏ -ﻨﺴﻤﻲ Pﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ ) (d1ﻭ) (d 2 ﻋﻴﻥ ,ﺒﻴﺎﻨﻴﺎ ,ﺇﺤﺩﺍﺜﻴﻲ ﺍﻟﻨﻘﻁﺔ Pﺘﺄﻜﺩ ﺃﻥ ﻓﺎﺼﻠﺔ ﺍﻟﻨﻘﻁﺔ Pﻫﻲ ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ) f (x) = g(xﺤﻴﺙ xﻫﻭ ﺍﻟﻤﺠﻬﻭل. rj 1ﺕ(3ﺒﺎ:ﻟﻨﻨﺴﻌﺒﺘﺔﻤﺩﺇﻟﻰﻋﻠﺍﻟﻰﻤﺍﻌﻟﻠﻡﺸﻜل rjﺍ,ﻟﻤirﺠ,ﺎ0ﻭﺭ( ), 0 ir ) (Dﻫﻭ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ f) (D ﺃ -ﻫل fﺩﺍﻟﺔ ﺨﻁﻴﺔ ؟ ﻟﻤﺎﺫﺍ ؟ ﺏ -ﻫل fﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ؟ ﻟﻤﺎﺫﺍ ؟ ﺕ -ﺃﻋﻁ ﻋﺒﺎﺭﺓ ) f (xﺒﺩﻻﻟﺔ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ x
ﺒﺤﻴﺙﻴﻜﻭﻥ)( )(Dϑrﺍﻟﺸﻤﻌﺎﻌﻠﻋﻡﻴﻥur,uϑrrﻭﻋﻠﻰ ﺨﻴﺎﺭﻙ- ﻭﺭﻗﺔ ﺍﻹﺠﺎﺒﺔ ﺜﻡ ﻋﻴﻥ -ﻤﻌﻠﻼ (2ﺃﻨﻘل ﺍﻟﺸﻜل ﻋﻠﻰ D, ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ gﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ : x → g(x) = 3x − 4ﺕ : 4ﺃﺭﺍﺩ ﻁﺎﻟﺏ Aﺃﻥ ﻴﺠﻤﻊ ﻗﻠﻴﻼ ﻤﻥ ﺍﻟﻨﻘﻭﺩ ,ﻓﻁﻠﺏ ﻤﻥ ﺼﺎﺤﺏ ﺩ ﹼﻜﺎﻥ ﺃﻥ ﻴﺸﻐﻠﻪ ﻜﺒﺎﺌﻊ ﺃﺜﻨﺎﺀ ﺍﻟﻌﻁﻠﺔﺍﻟﺼﻴﻔﻴﺔ .ﻗﺒل ﺼﺎﺤﺏ ﺍﻟﺩ ﹼﻜﺎﻥ ﺃﻥ ﻴﺸﻐل ﺍﻟﻁﺎﻟﺏ Aﻁﻭﺍل ﺸﻬﺭ ﺃﻭﺕ ﻭﺍﻗﺘﺭﺡ ﻋﻠﻴﻪ ﺃﻥ ﻴﺨﺘﺎﺭ ﻭﺍﺤﺩﺓ ﻤﻥﺍﻟﻁﺭﻴﻘﺘﻴﻥ ﺍﻟﺘﺎﻟﻴﺘﻴﻥ ﻟﻺﻴﺠﺎﺭ ﺤﻴﺙ ﺍﻟﻤﺒﻠﻎ ﻤﻘﺩﺭ ﺒﺎﻟﺩﻨﺎﻨﻴﺭ ﻭﺤﻴﺙ xﻤﺩﺨﻭل ﺍﻟﻤﺒﻴﻌﺎﺕ ﻟﺸﻬﺭ ﺃﻭﺕ. ﺍﻟﻁﺭﻴﻘﺔ ﺍﻷﻭﻟﻰ: ﺃﻥ ﻴﻜﻭﻥ , S1ﺃﺠﺭ ﺍﻟﻁﺎﻟﺏ , Aﻫﻭ 10%ﻤﻥ x ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺜﺎﻨﻴﺔ: (1ﺃﻥ ﻴﻜﻭﻥ , S2ﺃﺠﺭ ﺍﻟﻁﺎﻟﺏ Aﻫﻭ 1500ﺯﺍﺌﺩ 6%ﻤﻥ x (2ﺃﺜﺒﺕ ﺃﻥ S1ﻫﻭ ﺼﻭﺭﺓ xﺒﺩﺍﻟﺔ ﺨﻁﻴﺔ fﻴﻁﻠﺏ ﺘﺤﺩﻴﺩﻫﺎ (3ﺃﺜﺒﺕ ﺃﻥ S2ﻫﻭ ﺼﻭﺭﺓ xﺒﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ gﻴﻁﻠﺏ ﺘﺤﺩﻴﺩﻫﺎﻴﺭﻴﺩ ﺍﻟﻁﺎﻟﺏ Aﺃﻥ ﻴﻜﺴﺏ 10000ﺩﻴﻨﺎﺭﺍ ﺃﻴﺔ ﺍﻟﻁﺭﻴﻘﺘﻴﻥ ﺘﻤﻨﺢ ﻟﻪ ﺃﻜﺒﺭ ﺤﻅ ﻟﺒﻠﻭﻍ ﺫﻟﻙ ؟
• ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ: ﺕ:1 gﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﺃﻱ g(x) = ax + b :ﺤﻴﺙ b, a :ﺜﺎﺒﺘﻴﻥ ﻤﻥ R ﻭﻟﻜﻥ ) f (1) = g(1ﻭ f (1) = 5 ﻭﻋﻠﻴﻪg(1) = 5 : ﻭ f (−1) = 9 = )g(−1 1 ﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯf (−1) : 2 )g (−1 = 9 ﻭﻋﻠﻴﻪ: 2 ﻋﻨﺩﺌﺫ: )g (−1 = 9 )g (1 ﻤﻥ = 5 ﻭ2 a + b = 5 9 ﻨﺠﺩ: − a + b = 2 b = 19 a = 1 ﻤﻨﻪ: ﻭ4 4 ﻭﻋﻠﻴﻪ: = )g(x 1 x + 19 ﻟﺩﻴﻨﺎ ﺍﻟﺤﺴﺎﺏ: (2 4 4 )g (2 = 21 4) (D g )g(3 = 41 2 ﻭ8 )g (−10 = 9 ﻭ 4 rj (3ﺍﻟﺘﻤﺜﻴل: 0 ir ) (D f
ﺕ:2 (1ﻋﺒﺎﺭﺓ ) f (xﺒﺩﻻﻟﺔ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ x fﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﺃﻱ f (x) = ax + b :ﺤﻴﺙ aﻭ bﺜﺎﺒﺘﺎﻥ ﻤﻥ R ﻟﺩﻴﻨﺎ f (2) = 1:ﻭ f (6) = −1 2a + b = 1 ﻭﻋﻠﻴﻪ6a + b = −1 : a = − 1 ﻭﻤﻨﻪ: 2 b = 2 f ( )x = − 1 x + 2 ﻭﺒﺎﻟﺘﺎﻟﻲ: 2 (2ﻋﺒﺎﺭﺓ ) g(xﺒﺩﻻﻟﺔ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ x gﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﺃﻱ g(x) = a' x + b' :ﺤﻴﺙ ' aﻭ ' bﺜﺎﺒﺘﺎﻥ ﻤﻥ R = )g(−4 − 3 ﻭ g(− 14) = 6 ﻟﺩﻴﻨﺎ: 2 − 1 'a'+b = 6 4 − ﻭﻋﻠﻴﻪ: 3 '4a'+b = − 2 ) (d1 = 'a 2 ﻭﺒﺎﻟﺘﺎﻟﻲ: P rj = 'b 13 2(d2 ) 0 ir g )(x = 2 x + 13 ﻭﻤﻨﻪ: 2 − 1 < 0 ﻷﻥ: ﺘﻤﺎﻤﺎ ﻤﺘﻨﺎﻗﺼﺔ f ﺍﻟﺩﺍﻟﺔ 2 ﻭﺍﻟﺩﺍﻟﺔ gﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻷﻥ2 > 0 : (4ﺃ .ﺇﻨﺸﺎﺀ ) (d1ﻭ) : (d 2 ﺏ .ﺍﻟﺘﺄﻜﺩ ﺃﻥ ﻓﺎﺼﻠﺔ ﺍﻟﻨﻘﻁﺔ Pﻫﻲ ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ )f (x) = g(x
x = − 9 ﻨﺠﺩ: )f (x) = g(x ﻤﻥ: 5 9 ﻭﻤﻨﻪ ﻓﺎﺼﻠﺔ ﺍﻟﻨﻘﻁﺔ P ﻫﻲ− 5 : ﺕ: 3 (1ﺃ .ﺍﻟﺩﺍﻟﺔ fﻟﻴﺴﺕ ﺨﻁﻴﺔ ﻷﻥ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ Dﻻ ﻴﻤﺭ ﻤﻥ ﺍﻟﻤﺒﺩﺃ ( )0ﺏ .ﺍﻟﺩﺍﻟﺔ fﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﻷﻥ ﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ Dﻋﺒﺎﺭﺓ ﻋﻥ ﻤﺴﺘﻘﻴﻡ ﻻ ﻴﻭﺍﺯﻱ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ) ( ﺠـ .ﺇﻋﻁﺎﺀ ﻋﺒﺎﺭﺓ ) f (xﺒﺩﻻﻟﺔ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ x ﺒﻤﺎ ﺃﻥ fﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﻓﺈﻥ: f (x) = ax + bﺤﻴﺙ b, aﺜﺎﺒﺘﺎﻥ ﻤﻥ R ﻤﻥ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ) (Dﻨﺠﺩ: f − 1 = − 7 ﻭ f (1) = 1 2 2 a + b = 1 = − 7 ﻭﻋﻠﻴﻪ: − 1 a + b 2 2 a = 3 ﻤﻨﻪb = −2 : ﻭﺒﺎﻟﺘﺎﻟﻲf (x) = 3x − 2 : (2ﻟﻴﻜﻥ xﻋﺩﺩﺍ ﺤﻘﻴﻘﻴﺎ ﻜﻴﻔﻴﺎ g(x) = 3x − 4 ﻟﺩﻴﻨﺎ= 3x − 2 − 2 : ﻤﻨﻪg(x) = f (x) − 2 : ﻤﻥ ﺍﻟﻌﻼﻗﺔ ﺍﻷﺨﻴﺭﺓ ﻨﻼﺤﻅ ﺃﻥ ﺍﻟﻔﻭﺍﺼل ﻴﻤﻜﻥ ﺘﺜﺒﻴﺘﻬﺎ ﺃﻤﺎ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻓﺘﺯﺍﺡ ﺒـ -2:( )B 0, ir, rj ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻌﻠﻡ ﻟﺫﺍCrﻴﻤ=ﻜﻨﻨﺎurﺍﺨﻭﺘﻴﺎBﺭAﻤﺎ ﻴ=ﻠﻲϑ:r ﺤﻴﺙ A(0,2) :ﻭ )B(0,3 Arj0 ir)(D
ﺕ:4 ﻤﻥ ﺍﻟﻤﻌﻁﻴﺎﺕ ﻨﺠﺩ: S1 = x × 10 100 = 1 x 10 f )(x = 1 x ﺤﻴﺙ : ﺒﺩﺍﻟﺔ ﺨﻁﻴﺔ f ﻫﻭ ﺼﻭﺭﺓ x ﻭﺒﺎﻟﺘﺎﻟﻲ S1 10 (2ﻤﻥ ﺍﻟﻤﻌﻁﻴﺎﺕ ﻨﺠﺩ : S2 = 1500 + x × 6 100 = 3 x + 1500 10g )(x = 3 x + 1500 ﺒﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ ﺤﻴﺙ : ﻫﻭ ﺼﻭﺭﺓ x ﻭﺒﺎﻟﺘﺎﻟﻲ S2 10 ﻨﺠﺩx = 100.000 : ﺒﺠﻌل S1 = 10000 ﻨﺠﺩx = 85333,33 : ﻭﺒﺠﻌل S2 = 10000 ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺜﺎﻨﻴﺔ ﺘﻤﻨﺢ ﻟﻪ ﺃﻜﺒﺭ ﺤﻅ ﻟﺒﻠﻭﻍ ﺫﻟﻙ.
ﺍﻟﺩﺍﻟﺔ ) x → x 2ﺃﻭ ﺍﻟﺩﺍﻟﺔ \"ﻤﺭﺒﻊ\"( ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ : -ﺩﺭﺍﺴﺔ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ x → x2 -ﺇﻨﺸﺎﺀ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ x → x2 -ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ x → x2 -ﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﺃﻋﺩﺍﺩ ﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﺼﻔﺭ -ﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﺩﺍﻟﺔ ﺯﻭﺠﻴﺔ ﻭﺘﻨﺎﻅﺭ ﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻟﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ -ﺍﺴﺘﻌﻤﺎل ﺍﻟﻤﻌﻠﻭﻤﺎﺕ ﺤﻭل ﺍﻟﺩﺍﻟﺔ x → x2 ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ .1ﺍﻟﺩﺭﺱ ﺍﻟﻨﻅﺭﻱ ﺤﻭل ﺍﻟﺩﺍﻟﺔ x → x 2 .2ﻤﺠﻤﻭﻋﺎﺕ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﺼﻔﺭ ,ﺍﻟﺩﻭﺍل ﺍﻟﺯﻭﺠﻴﺔ .3ﺘﻤﺎﺭﻴﻥ ﻟﻤﻌﺎﻟﺠﺔ ﺒﻌﺽ ﺍﻟﻤﺸﻜﻼﺕ .4ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ .5ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ
ﺍﻟﺩﺭﺱ ﺍﻟﻨﻅﺭﻱ ﺤﻭل ﺍﻟﺩﺍﻟﺔ : x → x2 ﺃ .ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ : x → x2ﻡﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ , fﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﺩﺴﺘﻭﺭ , f (x) = x2ﻫﻲ Rﻷﻨﻪ ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ xﻓﺈﻥ ) f (xﻤﻭﺠﻭﺩ ﻓﻲ R ﺩﺭﺍﺴﺔ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ : x → x2 ﻨﻅﺭﻴﺔ: ﻟﺘﻜﻥ fﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭ f (x) = x2ﻟﺩﻴﻨﺎ: * fﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل]]− ∞,0 ** fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل [∞[0,+ ﺍﻟﺒﺭﻫﺎﻥ : * ﻟﻴﻜﻥ uﻭ ϑﻋﺩﺩﻴﻥ ﺤﻘﻴﻘﻴﻴﻥ ﻜﻴﻔﻴﻴﻥ ﻟﻨﻔﺭﺽ ϑ > uﻭﻟﻨﻘﺎﺭﻥ ) f (uﻭ ) f (ϑﻟﺫﻟﻙ ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ﺍﻟﻔﺭﻕ )f (ϑ) − f (u ﻟﺩﻴﻨﺎ f (ϑ) − f (u) = ϑ 2 − u 2ﻤﻨﻪ )f (ϑ) − f (u) = (ϑ − u)(ϑ + u ﻭﺒﻤﺎ ﺃﻥ ﻴﻨﺘﻤﻲ ﺇﻟﻰ] ]− ∞,0ﻓﺈﻥ ϑ ≤ 0ﻭﺒﻤﺎﺃﻥ ϑ > uﻓﺈﻥ u < 0 ﻭﻟﻨﺎ ϑ ≤ 0ﻤﻨﻪ ϑ + u ≤ 0 + uﺃﻱ ϑ + u ≤ uﻭ u < 0ﻤﻨﻪ (1)...ϑ + u < 0 ﻭ ϑ > uﻤﻨﻪ (2)...ϑ − u > 0 ﻤﻥ ) (1ﻭ) (ϑ − u)(ϑ + u) < 0 :(2ﻤﻨﻪ f (ϑ) − f (u) < 0ﻤﻨﻪ )f (ϑ) < f (uﺇﺫﻥ :ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭﺍﻥ uﻭ ϑﻤﻥ ﺍﻟﻤﺠﺎل , − ∞,0ﺇﺫﺍ ﻜﺎﻥ ϑ > uﻓﺈﻥ )] ]f (u) > f (ϑ ﻭﻋﻠﻴﻪ ﺍﻟﺩﺍﻟﺔ fﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ] ]− ∞,0 ** ﻟﻴﻜﻥ uﻭ ϑﻋﻨﺼﺭﻴﻥ ﻜﻴﻔﻴﻴﻥ ﻤﻥ ﺍﻟﻤﺠﺎل ∞[ [0,+ ﻟﻨﻔﺭﺽ ϑ > uﻭﻟﻨﻘﺎﺭﻥ ) f (uﻭ )f (ϑ ﻤﺜل ﻤﺎ ﺴﺒﻕ ,ﻟﺩﻴﻨﺎ )f (ϑ) − f (u) = (ϑ − u)(ϑ + u
ﻭﻫﻨﺎ u ≥ 0ﻭ ϑ > uﻤﻨﻪ ϑ > 0ﻭﻟﻨﺎ u ≥ 0ﻤﻨﻪ u + ϑ ≥ 0 + ϑ ﺇﺫﻥ u + ϑ ≥ ϑﻭ ϑ > 0ﻤﻨﻪ (3)...u + ϑ > 0ﻭ (4)...ϑ − u > 0 ﻭﻤﻥ ) (3ﻭ) (ϑ − u)(ϑ + u) > 0 :(4ﺃﻱ f (ϑ) − f (u) > 0ﻤﻨﻪ )f (ϑ) > f (uﺇﺫﻥ :ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭﺍﻥ uﻭ ϑﻤﻥ ﺍﻟﻤﺠﺎل ∞ , 0,+ﺇﺫﺍ ﻜﺎﻥ ϑ > uﻓﺈﻥ )[ [f (ϑ) > f (u ﻭﻋﻠﻴﻪ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺩﺍﻟﺔ fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ∞[ [0,+ ﺍﺴﺘﻨﺘﺎﺝ :ﻤﻘﺎﺭﻨﺔ ﻤﺭﺒﻌﻲ ﻋﺩﺩﻴﻥ ﻤﻥ ﻨﻔﺱ ﺍﻹﺸﺎﺭﺓ : * ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﺩﺩﺍﻥ ﺍﻟﺴﺎﻟﺒﺎﻥ aﻭ (b > a) : bﻴﻜﺎﻓﺊ) (b2 < a2** ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﺩﺩﺍﻥ ﺍﻟﺤﻘﻴﻘﻴﺎﻥ ﺍﻟﻤﻭﺠﺒﺎﻥ aﻭ (b > a) : bﻴﻜﺎﻓﺊ) (b2 > a2 ﺠـ .ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ : x → x2 ﺍﺴﺘﻨﺘﺎﺠﺎ ﻟﻠﻨﺘﺎﺌﺞ ﺍﻟﻭﺍﺭﺩﺓ ﻓﻲ ﻨﺹ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﺴﺎﺒﻘﺔ ﻴﻜﻭﻥ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ fﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ R ﺒﺎﻟﺩﺴﺘﻭﺭ f (x) = x 2ﻜﻤﺎ ﻴﻠﻲ:x −∞ 0 ∞+)f (x 0 ﻟﺩﻴﻨﺎf (0) = 0 : ir,:rjx →x ﻟﻠﺩﺍﻟﺔ 2 ﺍﻟﺒﻴﺎﻨﻲ ﺍﻟﺘﻤﺜﻴل ( )0, ﻤﺘﻌﺎﻤﺩ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﻨﺴﻭﺏ ﺍﻟﻤﺴﺘﻭﻱ ﻟﻴﻜﻥ ) (Cﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ fﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭ f (x) = x 2 ﺇﻋﺘﻤﺎﺩﺍ ﻋﻠﻰ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ : f ﻓﻲ\" ﺍﻟﻤﻨﻁﻘﺔ \" (C) :\" x ≤ 0ﻴﻨﺯل ﻓﻲ\" ﺍﻟﻤﻨﻁﻘﺔ \" (C) :\" x ≥ 0ﻴﺼﻌﺩﻭ ) (Cﻫﻭ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﺍﻟﺘﻲ ﺇﺤﺩﺍﺜﻴﺎﻫﺎ x, x 2ﺤﻴﺙ xﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻟﻨﻌﻠﻡ ﺒﻌﺽ ﺍﻟﻨﻘﻁ \"ﺍﻟﻤﺴﺎﻋﺩﺓ\") ( ﻤﻥ ﺍﻟﻤﻨﺤﻨﻲ ) (Cﺇﻋﺘﻤﺎﺩﺍ ﻋﻠﻰ ﺍﻟﺠﺩﻭل ﺍﻟﺘﺎﻟﻲ:
x -2 − 3 -1 − 1 01 132 2 2 2 2x2 4 9 1 1 01 194 44 4 4 ﻤﻨﻪ ﺍﻟﻨﻘﻁ D',C', B', A',O,C, B, Aﺤﻴﺙ: B'(− 3 , 9 ), 'A (−2,4), O(o, o), (D 1 , 1 ), C (1,1), (B 3 , 9 ), )A(2,4 2 4 2 4 2 4 ﺇﻟﻰ ﺍﻟﻤﻨﺤﻨﻲ ) (ρ ﺘﻨﺘﻤﻲ 'D (− 1 , 1 ), C ' (−1,1), 2 4 ﻤﻨﻪ ﺍﻟﺸﻜل: ) M '(−x, x 2 ) // // M (x, x 2 )(C rj x 0 ir −x ▲ ﺍﻟﻤﻨﺤﻨﻲ ) (Cﻫﻭ ﻗﻁﻊ ﻤﻜﺎﻓﺊ ﺍﻟﻨﻘﻁﺔ 0ﻤﺒﺩﺃ ﺍﻟﻤﻌﻠﻡ ﻫﻲ ﺫﺭﻭﺓ ﺍﻟﻘﻁﻊ ﺍﻟﻤﻜﺎﻓﺊ )(C ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻫﻭ ﻤﺤﻭﺭ ﺘﻨﺎﻅﺭ ﻟﻠﻘﻁﻊ ﺍﻟﻤﻜﺎﻓﺊ )(C
• ﻤﺠﻤﻭﻋﺎﺕ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﺼﻔﺭ ,ﺍﻟﺩﻭﺍل ﺍﻟﺯﻭﺠﻴﺔ: ﺘﻤﻬﻴﺩ : ﺍﻟﻤﺠﺎل − 1,1ﻴﻤﺜل ﻋﻠﻰ ﻤﺴﺘﻘﻴﻡ ﻤ ّﺩﺭﺝ ﻜﻤﺎ ﻴﻠﻲ] [:-1 − x 0 x 1 ﺍﻟﻤﺠﻤﻭﻋﺔ] [− 4,−1[ ∪ ]1,4ﺘﻤﺜل ﻋﻠﻰ ﻤﺴﺘﻘﻴﻡ ﻤ ّﺩﺭﺝ ﻜﻤﺎ ﻴﻠﻲ: -1 0 ﻤ4ﺴﺘﻘﻴﻡxﻤ ّﺩﺭﺝ[ ]-4 − x 1 ﻴﻠﻲ: ﻋﻠﻰ ﻴﻤﺜل ﺍﻟﻤﺠﺎل − 5,1 ﻜﻤﺎ -5 ﻴﻜﻥxx 01 −x− 1,1 0ﻷﻥ ﻤﻬﻤﺎ ﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔﻓﺈﻥ − xﻴﻨﺘﻤﻲ[ ] [ ] ﻓﻲ ﺇﻟﻰ * ﺍﻟﻤﺠﻤﻭﻋﺔ − 1,1 ﺇﻟﻰ []− 1,1 * ﻜﺫﻟﻙ ﺍﻟﻤﺠﻤﻭﻋﺔ] [− 4,−1[ ∪ ]1,4ﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ 0 * ﺍﻟﻤﺠﻤﻭﻋﺔ − 5,1ﻟﻴﺴﺕ ﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ [ ]0 .2ﺘﻌﺭﻴﻑ: ﻟﺘﻜﻥ Eﻤﺠﻤﻭﻋﺔ ﺠﺯﺌﻴﺔ ﻤﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ Rﺘﻜﻭﻥ Eﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﺼﻔﺭ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ: ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭ xﻤﻥ Eﻓﺈﻥ − xﻴﻨﺘﻤﻲ ﺇﻟﻰ E* ﺍﻟﻤﺠﻤﻭﻋﺔ E1ﺤﻴﺙ ∞ E1 = − ∞,0 ∪ 0,+ﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ 0ﻷﻥ E1ﻫﻲ ﻤﺠﻤﻭﻋﺔ[ ] [ ]ﺍﻷﻋﺩﺍﺩ ﺍﻟﺤﻘﻴﻘﻴﺔ ﻏﻴﺭ ﺍﻟﻤﻌﺩﻭﻤﺔ ﻭﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭ xﻤﻥ E1ﻟﺩﻴﻨﺎ x ≠ 0 ﻭﺒﺎﻟﺘﺎﻟﻲ − x ≠ 0ﺇﺫﻥ − xﻴﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ . E1
* ﺍﻟﻤﺠﻤﻭﻋﺔ E2ﺤﻴﺙ E2 = − 2,2ﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ 0ﻷﻥ ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭ xﻤﻥ [ ]E2 ﻟﺩﻴﻨﺎ 2 ≥ x ≥ −2ﻭﻤﻨﻪ )) − 2 ≤ −x ≤ −(−2ﺒﻀﺭﺏ ﻁﺭﻓﻲ ﻜل ﻤﺘﺒﺎﻴﻨﺔ ﻓﻲ -1ﺍﻹﺘﺠﺎﻩ ﻴﺘﻐﻴﺭ‼( ﻤﻨﻪ − 2 ≤ −x ≤ 2ﺇﺫﻥ − xﻴﻨﺘﻤﻲ ﺇﻟﻰ E2* ﺍﻟﻤﺠﻤﻭﻋﺔ − 3,1ﻟﻴﺴﺕ ﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ 0,ﻤﺜﺎل ﻤﻀﺎﺩ -2 :ﻴﻨﺘﻤﻲ ﺇﻟﻰ − 3,1ﻭﻟﻜﻥ )[ ] [ ](-2 ﻻ ﻴﻨﺘﻤﻲ ﺇﻟﻰ][− 3,1 .3ﺘﻌﺭﻴﻑ: لﺘﻜﻥ Fﺩﺍﻟﺔ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻔﻬﺎ D ﺘﻜﻭﻥ ﺍﻟﺩﺍﻟﺔ Fﺯﻭﺠﻴﺔ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﺘﺤﻘﻕ ﺍﻟﺸﺭﻁﺎﻥ : D (1ﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﺼﻔﺭ (2ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭ xﻤﻥ f (−x) = f (x) : D ﺃﻤﺜﻠﺔ :)h(x = )3x + 1, g(x = 3x 1 + 1 , f )(x = x2 : ﺤﻴﺙ h, g, f ﺍﻟﺩﻭﺍل ﻟﺘﻜﻥ 2 * ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﻜل ﻤﻥ h, g, fﻫﻲ Rﻭ Rﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﺼﻔﺭ * ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭ xﻤﻥ : R f (−x) = (−x)2ﺇﺫﻥ f (−x) = x 2ﺇﺫﻥ ) f (−x) = f (xﻭﺒﺎﻟﺘﺎﻟﻲ ﺍﻟﺩﺍﻟﺔ fﺯﻭﺠﻴﺔﻭﺒﺎﻟﺘﺎﻟﻲ = )g(−x ﺇﺫﻥ )g(x = )g(−x 1 = )g(−x 1 +1 ﺇﺫﻥ 3x 2 + 1 3(− x) 2 ﺍﻟﺩﺍﻟﺔ gﺯﻭﺠﻴﺔ h(−x) = 3(−x) + 1ﺇﺫﻥ h(−x) = −3x + 1ﻭ ) h(−xﻻ ﻴﺴﺎﻭﻱ ) h(xﻤﻥ ﺃﺠل ﻜل ﻗﻴﻡ , xﻤﺜﺎل ﻤﻀﺎﺩ h(−1) = −2 :ﺒﻴﻨﻤﺎ h(1) = 4ﻭﺒﺎﻟﺘﺎﻟﻲ ﺍﻟﺩﺍﻟﺔ hﻟﻴﺴﺕ ﺯﻭﺠﻴﺔ. ﻤﻼﺤﻅﺔ :ﺍﻟﺩﺍﻟﺔ f : x → f (x) = x 2ﻫﻲ ﺩﺍﻟﺔ ﺯﻭﺠﻴﺔﻭ)ﺤﺴﺏ ﻤﻼﺤﻅﺔ ﺒﻴﺎﻨﻴﺔ ﺴﺎﺒﻘﺔ( ﺍﻟﻤﻨﺤﻨﻲ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ , fﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ,ﻴﻘﺒل ﺤﺎﻤل ﻤﺤﻭﺭﺍﻟﺘﺭﺍﺘﻴﺏ ﻜﻤﺤﻭﺭ ﺘﻨﺎﻅﺭ ﻭﻫﺫﺍ ﻟﻴﺱ ﺨﺎﺼﺎ ﺒﺎﻟﺩﺍﻟﺔ ﻤﺭﺒﻊ ﻭﺍﻟﻨﺘﻴﺠﺔ ﺍﻟﻌﺎﻤﺔ ﻓﻲ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻤﻭﺍﻟﻴﺔ )ﺍﻟﻤﻘﺒﻭﻟﺔ ﺒﺩﻭﻥ ﺒﺭﻫﺎﻥ(.
.3ﻨﻅﺭﻴﺔ :ﻟﺘﻜﻥ Fﺩﺍﻟﺔ ﻭﻟﻴﻜﻥ ) (Cﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ fﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ) (0,ir, rjﺤﺘﻰ ﻴﻜﻭﻥ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻤﺤﻭﺭ ﺘﻨﺎﻅﺭ ﻟﻠﻤﻨﺤﻨﻲ ) ,(Cﻴﻠﺯﻡ ﻭﻴﻜﻔﻲ ﺃﻥ ﺘﻜﻭﻥ ﺍﻟﺩﺍﻟﺔ Fﺯﻭﺠﻴﺔ. • ﺘﻤﺎﺭﻴﻥ ﻟﻤﻌﺎﻟﺠﺔ ﺒﻌﺽ ﺍﻟﻤﺸﻜﻼﺕ: . 0, ir, rjﻓﻲ) ( ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﺍﻟﺘﻤﺭﻴﻥ ﺍﻷﻭل :ﺘﻤﺜﻴل ﺩﺍﻟﺔ ﺯﻭﺠﻴﺔ ﺒﻴﺎﻨﻴﺎ fﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﻭ ) (Cﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ f ﺍﻟﺸﻜل ﺍﻟﻤﻭﺍﻟﻲ ﺭﺴﻡ ) (C1ﻤﺠﻤﻭﻋﺔ ﻨﻘﻁ ) (Cﺍﻟﺘﻲ ﻓﻭﺍﺼﻠﻬﺎ ﻤﻭﺠﺒﺔ ﺍﻟﻤﻁﻠﻭﺏ :ﺇﺘﻤﺎﻡ ﺇﻨﺸﺎﺀ ) (Cﻋﻠﻤﺎ ﺃﻥ ﺍﻟﺩﺍﻟﺔ fﺯﻭﺠﻴﺔ )(C rj 0 ir ﺍﻟﺤل:ﺍﻟﺩﺍﻟﺔ fﺯﻭﺠﻴﺔ ﻭﺍﻟﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ,ﻤﻨﻪ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻭﻋﻠﻴﻪ ) (C2ﻤﺠﻤﻭﻋﺔ ﻨﻘﻁ ) (Cﺍﻟﺘﻲ ﻓﻭﺍﺼﻠﻬﺎ ﺴﺎﻟﺒﺔ ﻫﻲ ﻨﻅﻴﺭﺓ ) (C1ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ
)(C ﻤﻨﻪ ﺍﻟﺸﻜل:rj0 irﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻨﻲ :ﻁﺭﻴﻘﺔ ﻟﺩﺭﺍﺴﺔ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺩﺍﻟﺔ ﺍﻋﺘﻤﺎﺩﺍ ﻋﻠﻰ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺩﻭﺍل ﻤﺭﺠﻌﻴﺔ ﻟﺘﻜﻥ fﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭ f (x) = −x 2 − 6x + 2ﺘﺄﻜﺩ ﺃﻨﻪ ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ xﻓﺈﻥf (x) = −(x + 3)2 + 11ﺃﺜﺒﺕ ﺃﻥ ﺍﻟﺩﺍﻟﺔ fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل − ∞,−3ﻭﺃﻨﻬﺎ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ] ] ﺍﻟﻤﺠﺎل[∞[− 3,+( )0,ir, rj ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﺃﻨﺸﺊ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ f ﻤﺜل ﺒﻴﺎﻨﻴﺎ ﺍﻟﺩﺍﻟﺔ fﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﻨﺴﻭﺏ ﺍﻟﺤل: ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ − (x + 3)2 + 11 = −(x 2 + 6x + 9) + 11, x = −x 2 − 6x − 9 + 11 = −x2 − 6x + 2 )= f (x ﻤﻨﻪf (x) = −(x + 3)2 + 11 ﻟﻨﺒﺭﻫﻥ ﺃﻥ fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ] ]− ∞,−3 ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭﺍﻥ uﻭ ϑﻤﻥ ﺍﻟﻤﺠﺎل] ]− ∞,−3ﻟﻨﺎ u ≤ −3ﻭ ϑ ≤ −3 ﻤﻨﻪ u + 3 ≤ 0ﻭ ϑ + 3 ≤ 0ﻭﺇﺫﺍ ﻜﺎﻥ ϑ > uﻴﻜﻭﻥ ϑ + 3 > u + 3 ﺴﺎﻟﺒﺎﻥ ﻭﻴﻜﻭﻥ ) (ϑ + 3)2 < (u + 3)2ﻤﻘﺎﺭﻨﺔ ﻤﺭﺒﻌﻲ ﻋﺩﺩﻴﻥ ﺴﺎﻟﺒﻴﻥ(ﻭﻴﻜﻭﻥ ) − ϑ + 3 2 > −(u + 3)2ﺒﻀﺭﺏ ﻁﺭﻓﻲ ﻤﺘﺒﺎﻴﻨﺔ ﻓﻲ -1ﺍﻹﺘﺠﺎﻩ ﻴﺘﻐﻴﺭ‼() ( ﻭﻴﻜﻭﻥ) − (ϑ + 3)2 + 11 > (− u + 3)2 + 11ﻟﻘﺩ ﺍﻓﺘﺭﻀﻨﺎ (ϑ > u ﻭﻴﻜﻭﻥ ) f (ϑ) > f (uﻭﻋﻠﻴﻪ ﺍﻟﺩﺍﻟﺔ fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ] ]− ∞,−3 * ﻟﻨﺒﺭﻫﻥ ﺃﻥ ﺍﻟﺩﺍﻟﺔ fﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ∞[ [− 3,+ ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭﺍﻥ uﻭ ϑﻤﻥ ﺍﻟﻤﺠﺎل[∞ [− 3,+ﻟﻨﺎ u ≥ −3ﻭ ϑ ≥ −3
ﻤﻨﻪ u + 3 ≥ 0ﻭ ϑ + 3 ≥ 0ﻭﺇﺫﺍ ﻜﺎﻥ ϑ > uﻴﻜﻭﻥ ϑ + 3 > u + 3 ﻤﻭﺠﺒﺎﻥ ﻭﻴﻜﻭﻥ ) (ϑ + 3)2 > (u + 3)2ﻤﻘﺎﺭﻨﺔ ﻤﺭﺒﻌﻲ ﻋﺩﺩﻴﻥ ﻤﻭﺠﺒﻴﻥ( ﻭﻴﻜﻭﻥ − (ϑ + 3)2 < −(u + 3)2 ﻭﻴﻜﻭﻥ − (ϑ + 3)2 + 11 < (− u + 3)2 + 11 ﻭﻴﻜﻭﻥ )) f (ϑ) < f (uﻟﻘﺩ ﺍﻓﺘﺭﻀﻨﺎ (ϑ > u ﻭﻋﻠﻴﻪ ﺍﻟﺩﺍﻟﺔ fﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ∞[ [− 3,+ﻤﻼﺤﻅﺔ :ﻟﻤﻘﺎﺭﻨﺔ ) f (uﻭ ) f (ϑﺇﻨﻁﻼﻗﺎ ﻤﻥ ϑ > uﻟﻡ ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ﺍﻟﻔﺭﻕ )f (ϑ) − f (uﺒل ﺍﻨﺘﻘﻠﻨﺎ ﻤﻥ ϑ > uﻟﻠﻭﺼﻭل ﺇﻟﻰ ﻤﻘﺎﺭﻨﺔ ) f (uﻭ ) f (ϑﻷﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ) f (xﺇﻨﻁﻼﻗﺎ ﻤﻥ xﻴﺘﻡ ﺤﺴﺏ ﺍﻟﻤﺨﻁﻁ ﺍﻟﺘﺎﻟﻲ :)x → (x + 3) → (x + 3)2 → −(x + 3)2 → −(x + 3)2 + 11 = f (x)(1) (2 )(3 )(4 ﻭﻨﺤﺴﻥ \"ﺘﺼﺭﻑ\" ﺍﻟﻤﺘﺒﺎﻴﻨﺎﺕ ﻤﻊ ﻜل ﻤﻥ ﺍﻟﻤﺭﺍﺤل )(4) ,(3) ,(2) ,(1 ﻓﻲ ﺍﻟﻤﺭﺤﻠﺔ ) :(1ﺇﻀﺎﻓﺔ ﻨﻔﺱ ﺍﻟﻌﺩﺩ ﺇﻟﻰ ﻁﺭﻓﻲ ﻤﺘﺒﺎﻴﻨﺔ )ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ(ﻓﻲ ﺍﻟﻤﺭﺤﻠﺔ ) :(2ﺘﺭﺒﻴﻊ ﻁﺭﻓﻲ ﻤﺘﺒﺎﻴﻨﺔ ﻁﺭﻓﺎﻫﺎ ﻤﻥ ﻨﻔﺱ ﺍﻹﺸﺎﺭﺓ )ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ ( x → x 2 ﻓﻲ ﺍﻟﻤﺭﺤﻠﺔ ) :(3ﻀﺭﺏ ﻁﺭﻓﻲ ﻤﺘﺒﺎﻴﻨﺔ ﻓﻲ ﻋﺩﺩ ﻏﻴﺭ ﻤﻌﺩﻭﻡ )ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺩﺍﻟﺔ ﺨﻁﻴﺔ( ﻓﻲ ﺍﻟﻤﺭﺤﻠﺔ ) :(4ﺇﻀﺎﻓﺔ ﻨﻔﺱ ﺍﻟﻌﺩﺩ ﺇﻟﻰ ﻁﺭﻓﻲ ﻤﺘﺒﺎﻴﻨﺔ )ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ( ∞x − ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ : f )f (x ∞ -3 + 11 f (−3) = 11
ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ 0, ir, rj ﺘﻤﺜﻴل ﺍﻟﺩﺍﻟﺔ fﺒﻴﺎﻨﻴﺎ :ﻟﻠﺤﺼﻭل ﻋﻠﻰ) ( ﻟﻴﻜﻥ ) (Cﺍﻟﻤﻨﺤﻨﻲ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ f \"ﻨﻘﻁ ﻤﺴﺎﻋﺩﺓ ﻤﻥ ﺍﻟﻤﻨﺤﻨﻲ) ,(Cﻨﻨﺸﺊ ﺠﺩﻭﻻ ﻟﻠﻘﻴﻡ :x -8 -7 -6 -5 -4 -3 -2 -1 0 1 2f (x) -14 -5 2 7 10 11 10 7 2 -5 -14 )ﺒﺎﺴﺘﻌﻤﺎل( f (x) = −(x + 3)2 + 11: ﻤﻨﻪ ﺍﻟﺸﻜل: rj 0 ir )(C ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻟﺙ :ﺍﻟﺤل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻤﻌﺎﺩﻟﺔ ﺃﻭ ﻤﺘﺭﺍﺠﺤﺔ ﺤل ,ﻓﻲ , Rﺍﻟﻤﻌﺎﺩﻟﺔ (1)....x 2 = xﺤﻴﺙ xﻫﻭ ﺍﻟﻤﺠﻬﻭل ﺤل ,ﻓﻲ , Rﺍﻟﻤﺘﺭﺍﺠﺤﺔ (2)....x 2 > xﺤﻴﺙ xﻫﻭ ﺍﻟﻤﺠﻬﻭل ﻟﺘﻜﻥ fﻭ gﺍﻟﺩﺍﻟﺘﻴﻥ ﺍﻟﻤﻌﺭﻓﺘﺎﻥ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭﻴﻥ f (x) = xﻭ g(x) = x2ﻭ gﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻨﻔﺱ) () ( ﻭ) (Cﻟﻠﺩﺍﻟﺘﻴﻥ f D rjﺍﻟﺒ,ﻴﺎirﻨﻴ,ﻴ0ﻥ ﺍﻟﺘﻤﺜﻴﻠﻴﻥ ﺃﻨﺸﺊ ﻤﺘﻌﺎﻤﺩ ﺍﻟﻤﻌﻠﻡ ﺘﺄﻜﺩ ﺒﻴﺎﻨﻴﺎ ﻤﻥ ﻨﺘﻴﺠﺘﻲ ﺇﺠﺎﺒﺘﻙ ﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﺴﺅﺍﻟﻴﻥ (1ﻭ(2 ﺍﻟﺤل :ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﺘﻜﺎﻓﺊ x 2 − x = 0 ﺃﻱ x(x −1) = 0 ﺃﻱ x = 0ﺃﻭx = 1 ﻤﻨﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﻫﻲ}{0,1 ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ) (2ﺘﻜﺎﻓﺊ x 2 − x > 0 ﺃﻱ x(x −1) > 0 :
ﻟﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ) (2ﻨﻌﻴﻥ ﻤﺠﻤﻭﻋﺔ ﻗﻴﻡ xﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻤﻥ ﺃﺠﻠﻬﺎ ﺍﻟﺠﺩﺍﺀ )x(x − 1ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ﻭﻟﺫﻟﻙ ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ﻫﺫﺍ ﺍﻟﺠﺩﺍﺀ ﻭﻋﻠﻴﻨﺎ ﺃﻥ ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ﻜل ﻤﻥ ﻋﺎﻤﻠﻲ ﺍﻟﺠﺩﺍﺀ -ﺤﺴﺏ ﻗﻴﻡ - x ﺇﺸﺎﺭﺓ xﺘﻠﺨﺹ ﻓﻲ ﺍﻟﺠﺩﻭل ﺍﻟﺘﺎﻟﻲ: ∞x −∞ 0 + ﺇﺸﺎﺭﺓ x - 0+ ﻴﻜﻭﻥ x − 1 > 0ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥx > 1 ﻴﻜﻭﻥ x −1 < 0ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥx < 1 ﻴﻜﻭﻥ x − 1 = 0ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥx = 1 ﻭﺇﺸﺎﺭﺓ - x − 1ﺤﺴﺏ ﻗﻴﻡ - xﺘﻠﺨﺹ ﻓﻲ ﺍﻟﺠﺩﻭل ﺍﻟﺘﺎﻟﻲ: ∞x −∞ 1 + ﺇﺸﺎﺭﺓx − 1 - 0+ x ∞− ﻨﺠﻤﻊ ﺍﻟﻨﺘﺎﺌﺞ ﻓﻲ ﺠﺩﻭل ﻭﺍﺤﺩ: ﺇﺸﺎﺭﺓ xﺇﺸﺎﺭﺓx − 1 -0 ∞0 1 + - ++ - 0+ﺇﺸﺎﺭﺓ )x(x − 1 + 0 - 0+ ﻤﻨﻪ ﺤﺘﻰ ﻴﻜﻭﻥ x(x − 1) > 0ﻴﻠﺯﻡ ﻭﻴﻜﻔﻲ ﺃﻥ ﻴﻜﻭﻥ x < 0ﺃﻭx > 1 ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ) (2ﻫﻲ]− ∞,0[ ∪ ]1,+∞[:ﺍﻟﺩﺍﻟﺔ fﺩﺍﻟﺔ ﺨﻁﻴﺔ ﻤﻨﻪ) (Dﻤﺴﺘﻘﻴﻡ ﻴﺸﻤل 0ﻤﺒﺩﺃ ﺍﻟﻤﻌﻠﻡ ﻭﻟﻨﺎ f (1) = 1ﻤﻨﻪ ﻴﺸﻤل ﺍﻟﻨﻘﻁﺔ )A(1,1) (Cﻫﻭ ﺍﻟﻤﻨﺤﻨﻲ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ x → x 2ﻭﻫﻭ ﺍﻟﻘﻁﻊ ﺍﻟﻤﻜﺎﻓﺊ ,ﺍﻟﻤﻌﺭﻭﻑ ,ﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ y = x 2ﻭﻫﻭ ﻜﺫﻟﻙ ﻴﺸﻤل ﺍﻟﻨﻘﻁﺘﻴﻥ 0ﻭ Aﻷﻥ g(0) = 0ﻭg(1) = 1 ﻤﻨﻪ ﺍﻟﺸﻜل(C) : )(D )rj A(1,1 0 ir
ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﻗﻴﻡ xﺒﺤﻴﺙ ﻴﻜﻭﻥ ) f (x) = g(xﻭﻫﻲ ﻤﺠﻤﻭﻋﺔ ﻗﻴﻡ xﺍﻟﺘﻲ ﻤﻥ ﺃﺠﻠﻬﺎ \"ﺘﺭﺘﻴﺏ ﻨﻘﻁﺔ Dﺍﻟﺘﻲ ﻓﺎﺼﻠﺘﻬﺎ xﻴﺴﺎﻭﻱ ﺘﺭﺘﻴﺏ ﻨﻘﻁﺔ ) (Cﺍﻟﺘﻲ ﻓﺎﺼﻠﺘﻬﺎ \" xﻭﻫﻲ) (ﻭ ) (Cﻭﺘﻭﺠﺩ ﺒﺎﻟﻀﺒﻁ ﻨﻘﻁﺘﺎﻥ ﻤﺠﻤﻭﻋﺔ ﻗﻴﻡ xﺍﻟﺘﻲ ﻫﻲ ﻓﻭﺍﺼل ﻨﻘﻁ ﻤﺸﺘﺭﻜﺔ ﻟﻠﻤﻨﺤﻨﻴﻴﻥ ( )Dﻤﺸﺘﺭﻜﺘﺎﻥ ﻟﻠﻤﻨﺤﻨﻴﻴﻥ) (Dﻭ)℘( ﻭﻫﻤﺎ ) A(1,1ﻭ ) O(0,0ﻤﻨﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﻫﻲ {0,1}:ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ) (2ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﻗﻴﻡ xﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻤﻥ ﺃﺠﻠﻬﺎ ) g(x) > f (xﻭﻫﻲﻤﺠﻤﻭﻋﺔ ﻗﻴﻡ xﺍﻟﺘﻲ ﺘﺤﻘﻕ \":ﺘﺭﺘﻴﺏ ﻨﻘﻁﺔ ) (Cﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺔ xﺃﻜﺒﺭ ﺘﻤﺎﻤﺎ ﻤﻥ ﺘﺭﺘﻴﺏ ﻨﻘﻁﺔ Dﺫﺍﺕ) ( ﺍﻟﻔﺎﺼﻠﺔ \" x ﻭﻫﻲ ﻤﺠﻤﻭﻋﺔ ﻗﻴﻡ xﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻤﻥ ﺃﺠﻠﻬﺎ\" ) (Cﻓﻭﻕ ﺘﻤﺎﻤﺎ ( )\" D ﻭﺤﺴﺏ ﺍﻟﺸﻜل ﻫﺫﻩ ﺍﻟﻤﺠﻤﻭﻋﺔ ﻫﻲ]1,+∞[ ∪ ]− ∞,0[: ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺭﺍﺒﻊ :ﺘﻌﻴﻴﻥ ﺤﺼﺭ ﻟﻠﻌﺩﺩ x 2ﺇﻋﺘﻤﺎﺩﺍ ﻋﻠﻰ ﺤﺼﺭ ﻟﻠﻌﺩﺩ x 1 ≤ x ≤ ﻋﻴﻥ ﺤﺼﺭﺍ ﻟﻠﻌﺩﺩ x 2ﻋﻠﻤﺎ ﺃﻥ 30 2 ﻋﻴﻥ ﺤﺼﺭ ﻟﻠﻌﺩﺩ x 2ﻋﻠﻤﺎ ﺃﻥ −1,9 ≤ x ≤ −0,8ﺍﺴﺘﻌﻤل ﺍﻟﻘﻁﻊ ﺍﻟﻤﻜﺎﻓﺊ ﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ y = x 2ﻟﺤﺼﺭ x 2ﻋﻠﻤﺎ ﺃﻥ − 7 ≤ x ≤ 2 ﺍﻟﺤل: ≥x 1 x ≤ ﻭ 30 ﻤﻭﺠﺏ ﻤﻨﻪ x 1 ≤ x ≤ ﻟﺩﻴﻨﺎ 30 ﻭ2 2 1 2)ﻤﻘﺎﺭﻨﺔ ﻤﺭﺒﻌﻲ ﻋﺩﺩﻴﻥ ﻤﻭﺠﺒﻴﻥ() ( 2 2 x2 ≥ ﻭ x2 ≤ 30 ﻭﺒﺎﻟﺘﺎﻟﻲ x2 ≥ 1 ﻭ x2 ﺇﺫﻥ ≤ 30 4 1 ≤ x2 ﻓﺈﻥ ≤ 30 1 ≤ ≤x ﻭﻋﻠﻴﻪ :ﺇﺫﺍ ﻜﺎﻥ 30 4 2ﻟﺩﻴﻨﺎ −1,9 ≤ x ≤ −0,8ﻤﻨﻪ xﺴﺎﻟﺏ ﻭ x ≤ −0,8ﻭ x ≥ −1,9ﻭﺒﺎﻟﺘﺎﻟﻲ x 2 ≥ (− 0,8)2ﻭ ) x 2 ≤ (− 1,9)2ﻤﻘﺎﺭﻨﺔ ﻤﺭﺒﻌﻲ ﻋﺩﺩﻴﻥ ﺴﺎﻟﺒﻴﻥ( ﺇﺫﻥ x 2 ≥ 0,64ﻭ x 2 ≤ 3,61 ﻭﻋﻠﻴﻪ :ﺇﺫﺍ ﻜﺎﻥ −1,9 ≤ x ≤ −0,8ﻓﺈﻥ0,64 ≤ x 2 ≤ 3,61
(3ﻟﻨﻨﺸﺊ ﺍﻟﻘﻁﻊ ﺍﻟﻤﻜﺎﻓﺊ ) (Cﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ y = x 2ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ( )0, ir, rj A7 )(C )(℘1 2 B 2−7 rj 0 irﻟﺩﻴﻨﺎ − 7 2 = 7ﻭ 2 2 = 2ﻤﻨﻪ ﺍﻟﻨﻘﻁﺘﺎﻥ ) A(− 7,7ﻭ ) B( 2,2ﺘﻨﺘﻤﻴﺎﻥ ﺇﻟﻰ )( ) ( )(C x 2ﻫﻭ ﺘﺭﺘﻴﺏ ﻨﻘﻁﺔ ) (Cﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺔ xﻭﻟﻤﺎ − 7 ≤ x ≤ 2ﻨﻘﻁﺔ ℘ ﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺔ ( )xﺘﻨﺘﻤﻲ ﺇﻟﻰ ) (C1ﺠﺯﺀ ℘ ﺍﻟﺫﻱ \"ﻁﺭﻓﺎﻩ\" Aﻭ Bﻭﺤﺴﺏ ﺍﻟﺸﻜل ,ﺘﺭﺘﻴﺏ ﻫﺫﻩ ﺍﻟﻨﻘﻁﺔ ﻤﻭﺠﺏ ﻭﺃﻗل ﻤﻥ) ( ﺃﻭ ﻴﺴﺎﻭﻱ 7ﻭﻋﻠﻴﻪ 7 ≥ x 2 ≥ 0 ﻤﻨﻪ :ﺇﺫﺍ ﻜﺎﻥ − 7 ≤ x ≤ 2ﻓﺈﻥ 0 ≤ x 2 ≤ 7ﻤﻼﺤﻅﺔ :ﻟﺘﻌﻴﻴﻥ ﺤﺼﺭ ﻟﻠﻌﺩﺩ x 2ﻋﻠﻤﺎ ﺃﻥ − 7 ≤ x ≤ 2ﻴﻤﻜﻥ ﺍﺴﺘﻌﻤﺎل ﻁﺭﻴﻘﺔ ﺠﺒﺭﻴﺔ ﺒﺩﻻ ﻤﻥ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺒﻴﺎﻨﻴﺔ ﺍﻟﻤﻌﺭﻭﻀﺔ ﺴﺎﺒﻘﺎ:ﻟﺩﻴﻨﺎ ) − 7 ≤ x ≤ 2ﺘﺭﺒﻴﻊ ﻁﺭﻓﻲ ﻜل ﻤﺘﺒﺎﻴﻨﺔ ﻏﻴﺭ ﻤﻤﻜﻥ ﻷﻥ ﺍﻷﻋﺩﺍﺩ ﻟﻴﺴﺕ ﻤﻥ ﻨﻔﺱ ﺍﻹﺸﺎﺭﺓ( ﻭﻫﻨﺎﻙ ﺤﺎﻟﺘﺎﻥ: ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ 0 ≤ x ≤ 2 ﺍﻟﺤﺎﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ − 7 ≤ x ≤ 0≤ ) 02 ≤ x 2ﻤﻘﺎﺭﻨﺔ ﻤﺭﺒﻌﻲ ﻋﺩﺩﻴﻥ ﻤﻭﺠﺒﻴﻥ() (2 2 ≤0 x ≤ * ﺇﺫﺍ ﻜﺎﻥ 2 ﻴﻜﻭﻥ ﻤﻨﻪ 0 ≤ x 2 ≤ 2** ﺇﺫﺍ ﻜﺎﻥ − 7 ≤ x ≤ 0ﻴﻜﻭﻥ ) − 7 2 ≥ x 2 ≥ 02ﻤﻘﺎﺭﻨﺔ ﻤﺭﺒﻌﻲ ﻋﺩﺩﻴﻥ ﺴﺎﻟﺒﻴﻥ() (
ﻭﻴﻜﻭﻥ 7 ≥ x 2 ≥ 0ﺍﻟﺨﻼﺼﺔ ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ , 0 ≤ x 2 ≤ 2ﻭﺒﻤﺎ ﺃﻥ 2 ≤ 7ﻓﺈﻥ 0 ≤ x 2 ≤ 7ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ 0 ≤ x 2 ≤ 7 ﻤﻨﻪ ﻓﻲ ﺠﻤﻴﻊ ﺍﻟﺤﺎﻻﺕ :ﺇﺫﺍ ﻜﺎﻥ − 7 ≤ x ≤ 2ﻓﺈﻥ 7 ≥ x 2 ≥ 0
• ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ:ﺕ : 1ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﻘﻁﻊ ﺍﻟﻤﻜﺎﻓﺊ ﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ y = x 2ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ , 0, ir, rjﺃﻋﻁ ﺤﺼﺭﺍ) ( ﻟﻠﻌﺩﺩ x 2ﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻥ ﺍﻟﺤﺎﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ: − 4 ≤ x ≤ −1 (1 1 < ≤x 3 (2 2 2 − 3 < x ≤ 2 (3 ﺕ : 2ﺒﺎﺴﺘﻌﻤﺎل ﻁﺭﻴﻘﺔ ﺠﺒﺭﻴﺔ ,ﻋﻴﻥ ﺤﺼﺭﺍ ﻟﻠﻌﺩﺩ x 2ﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻥ ﺍﻟﺤﺎﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ : x (1ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺠﺎل][− 5,−2 x (2ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺠﺎل[[3,6 x (3ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺠﺎل]]− 4,8 x (4ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺠﺎل ] [− 6,2ﺕ : 3ﺃﺩﺭﺱ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ fﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ Iﻭ Jﻓﻲ ﻜل ﺤﺎﻟﺔ ﺤﺎﻟﺔ ﻤﻥ ﺍﻟﺤﺎﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ: f (x) = x2 − 5 (1ﻭ[∞J = ]− ∞,0], I = [0,+ f (x) = −x2 + 10 (2ﻭ[∞J = ]− ∞,0], I = [0,+ f (x) = (x −1)2 − 3 (3ﻭ [∞J = ]− ∞,1], I = [1,+J = − ∞,− 1 , I = − 1 ,+∞ ﻭ f (x) = (2x + 1)2 − 2 (4 2 2
)ir(,Crjﻫ,ﻭ0ﺍﻟﺘﻤﺜﻴل)( )(C ﺕ : 4ﻓﻲ ﺍﻟﺸﻜل ﺍﻟﻤﺠﺎﻭﺭ, ﺍﻟﺒﻴﺎﻨﻲ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﻤﻌﻠﻡﻟﺩﺍﻟﺔ fﻤﻌﺭﻓﺔ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭ rj f (x) = (x − a)2 + bﺤﻴﺙ aﻭ 0 ir b ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ﺜﺎﺒﺘﺎﻥ ﺇﻋﺘﻤﺎﺩﺍ ﻋﻠﻰ ﺍﻟﺸﻜل ,ﻋﻴﻥ ) f (3), f (2), f (1), f (0؟ ﻫل ﺍﻟﺩﺍﻟﺔ fﺯﻭﺠﻴﺔ ؟ ﻟﻤﺎﺫﺍ ؟ﺃﺜﺒﺕ ﺃﻥ bﻫﻭ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺼﻐﺭﻯ ﻟﻠﺩﺍﻟﺔ fﻋﻠﻰ Rﺜﻡ ﻋﻴﻥ ﻗﻴﻤﺔ ﺍﻟﻤﺘﻐﻴﺭ ﺍﻟﺘﻲ ﺘﺅﺨﺫ ﻤﻥ ﺃﺠﻠﻬﺎ ﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺼﻐﺭﻯ ﺃﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﻟﻠﻤﻨﺤﻨﻲ)(Cﺕ : 5ﻗل ﻤﻌﻠﻼ ﻋﻠﻰ ﺤﻜﻤﻙ ,ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺩﺍﻟﺔ fﺯﻭﺠﻴﺔ ﺃﻭ ﻏﻴﺭ ﺯﻭﺠﻴﺔ ﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻥ ﺍﻟﺤﺎﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ: f (x) = 3x2 + 7 (1 f (x) = −5x2 −1 (2 f (x) = (x − 3)2 (3 )f (x = 3x 2 (4 x4 + 2 f (x) = 2 (5 f (x) = 3x3 + 1 (6f )(x = −2x ﻟfﺘﻜ ﻓﻥﻲfﺍﻟﻤﺍﻟﺴﺩﺘﺍﻟﻭﺔﻱﺍﻟﺍﻤﻟﻌﻤﻨﺭﻓﺴﺔﻭﺏﻋﻠﺇﻟﻰﻰRﻤﻌﺒﻠﺎﻡﻟﺩﻤﺴﺘﺘﻌﺎﻭﻤﺭﺩ2ir,+rj2ﻭﻟﻴﻜﻥ) (Cﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ) (0, ﺕ: 6 ﻟﻠﺩﺍﻟﺔ ﺃﺜﺒﺕ ﺃﻥ ﺍﻟﺩﺍﻟﺔ fﺯﻭﺠﻴﺔ .ﻤﺎﺫﺍ ﺘﺴﺘﻨﺘﺞ ﻓﻴﻤﺎ ﻴﺨﺹ ﺍﻟﻤﻨﺤﻨﻰ)(C ﺃﺩﺭﺱ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ fﻋﻠﻰ ﺍﻟﻤﺠﺎل ] ]− ∞,0 ﺃﺘﻤﻡ ﺍﻟﺠﺩﻭل ﺍﻟﻤﻭﺍﻟﻲ :
x 0 − 1 -1 − 3 -2 2 2 )f (x ﺃﻨﺸﺊ)’(Cﻤﺠﻤﻭﻋﺔ ﻨﻘﻁ) (Cﺍﻟﺘﻲ ﻓﻭﺍﺼﻠﻬﺎ ﺴﺎﻟﺒﺔ ﺃﻨﺸﺊ ﺍﻟﻤﻨﺤﻨﻰ)(C ﺍﺴﺘﻨﺘﺞ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ f ﺕ :7ﻟﺘﻜﻥ fﻭ gﺍﻟﺩﺍﻟﺘﻴﻥ ﺍﻟﻤﻌﺭﻓﺘﻴﻥ ﻋﻠﻰ Rﺒﺎﻟﺩﺴﺘﻭﺭﻴﻥf (x) = 2x 2 − 5x + 1ﻭg f ∆( ﻭ) ﺍﻟ−ﻤﻌﻠ=ﻡ)ﺍﻟxﻤﺘ(ﻌﺎgﻤﺩﻭﻟﻴrjﻜ,ﻥ)0(C, ir x+ ﻭ1ﺒﺎﻟﻨﺴﺒﺔ) (ﻟﻠﺩﺍﻟﺘﻴﻥﺍﻟﺒﻴﺎﻨﻴﻴﻥ ﺍﻟﺘﻤﺜﻴﻠﻴﻥ ﺍﻟﺘﺭﺘﻴﺏ- -ﻋﻠﻰ ﻨﻔﺱ ﺇﻟﻰ ﻋﻴﻥ ﺇﺤﺩﺍﺜﻴﺎﺕ ﺍﻟﻨﻘﻁ ﺍﻟﻤﺸﺘﺭﻜﺔ ﻟﻠﻤﻨﺤﻨﻴﻴﻥ) (Cﻭ ∆) ( f )(x = (2 x − 5 ) 2 − 17 ﺘﺄﻜﺩ ﺃﻨﻪ :ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ x 4 8 ﻓﺈﻥ 5 ,+∞ ﻭ − ∞, 5 ﺍﻟﻤﺠﺎﻟﻴﻥ ﻤﻥ ﻜل ﻋﻠﻰ ﺃﺩﺭﺱ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ f 4 4 ﺃﻨﺸﺊ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ f ﺃﻨﺸﺊ ﺍﻟﻤﻨﺤﻨﻴﻴﻥ)∆( ﻭ ) ( C ﻟﺘﻜﻥ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ (1)...x 2 − 2x < 0ﺤﻴﺙ xﻫﻭ ﺍﻟﻤﺠﻬﻭل ﺤل ,ﻓﻲ , Rﺍﻟﻤﺘﺭﺍﺠﺤﺔ )(1 ﺘﺄﻜﺩ ﺃﻥ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ) (1ﺘﻜﺎﻓﺊ ) g(x) > f (xﺜﻡ ﺤل ﺒﻴﺎﻨﻴﺎ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ).(1
• ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ: ﺕ:1 (1ﺒﻤﺎ ﺃﻥ − 4 ≤ x ≤ −1ﻓﺈﻥ1 ≤ x2 ≤ 16 :1 ≤ x2 ≤ 9 1 < ≤x 34 ﻓﺈﻥ4 : 2 (2ﺒﻤﺎ ﺃﻥ 2 (3ﺒﻤﺎ ﺃﻥ − 3 < x ≤ 2ﻓﺈﻥ0 ≤ x 2 ≤ 4 : ﺕ2(1ﻤﻥ ﺃﺠل x ∈ [− 5,−2] :ﻨﺠﺩ4 ≤ x 2 ≤ 25 : (2ﻤﻥ ﺃﺠل x ∈ [3,6[ :ﻨﺠﺩ9 ≤ x 2 < 36 : (3ﻤﻥ ﺃﺠل x ∈ ]− 4,8] :ﻨﺠﺩ0 ≤ x2 ≤ 64 : (4ﻤﻥ ﺃﺠل x ∈ − 6,2 :ﻨﺠﺩ] [0 ≤ x2 ≤ 6 : ﺕ:3 f (x) = x 2 − 5 (1ﺍﻟﺩﺍﻟﺔ fﻋﻠﻰ ﺍﻟﻤﺠﺎل Iﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻭﻋﻠﻰ ﺍﻟﻤﺠﺎل Jﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ f (x) = −x2 + 10 (2ﺍﻟﺩﺍﻟﺔ fﻋﻠﻰ ﺍﻟﻤﺠﺎل Iﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻭﻋﻠﻰ ﺍﻟﻤﺠﺎل Jﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ f (x) = (x −1)2 − 3 (3ﺍﻟﺩﺍﻟﺔ fﻋﻠﻰ ﺍﻟﻤﺠﺎل Iﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻭﻋﻠﻰ ﺍﻟﻤﺠﺎل Jﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ f (x) = (2x + 1)2 − 2 (4ﺍﻟﺩﺍﻟﺔ fﻋﻠﻰ ﺍﻟﻤﺠﺎل Iﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻭﻋﻠﻰ ﺍﻟﻤﺠﺎل Jﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ
ﺕ: 4 ﺒﺎﻻﻋﺘﻤﺎﺩ ﻋﻠﻰ ﺍﻟﺸﻜل ﻨﺠﺩ: f (0) = 3ﻭ f (1) = 0ﻭ f (2) = −1ﻭ f (3) = 0 ﻟﺩﺍﻟﺔ fﻟﻴﺴﺕ ﺩﺍﻟﺔ ﺯﻭﺠﻴﺔ ﻷﻥ ﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ﻻ ﻴﻘﺒل ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻜﻤﺤﻭﺭ ﺘﻨﺎﻅﺭ. ﻟﻴﻜﻥ xﻋﺩﺩﺍ ﺤﻘﻴﻘﻴﺎ ﻜﻴﻔﻴﺎ ﻟﺩﻴﻨﺎ(x − a)2 ≥ 0 : ﻭﻋﻠﻴﻪ(x − a)2 + b ≥ b : ﻭﺒﺎﻟﺘﺎﻟﻲf (x) ≥ b : ﻭﻤﻨﻪ bﻫﻭ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺼﻐﺭﻯ ﻟﻠﺩﺍﻟﺔ fﻋﻠﻰ R ﻭﻟﻜﻥf (a) = b : ﻭﻤﻨﻪ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺼﻐﺭﻯ bﺘﺅﺨﺫ ﻤﻥ ﺃﺠل ﺍﻟﻘﻴﻤﺔ aﻟﻠﻤﺘﻐﻴﺭ ﻤﻥ ﺍﻟﺸﻜل ﻨﺠﺩ:ﺍﻟﻘﻴﻤﺔ ﺍﻟﺼﻐﺭﻯ ﻟﻠﺩﺍﻟﺔ fﻋﻠﻰ Rﻫﻲ -1ﻭﻗﻴﻤﺔ ﺍﻟﻤﺘﻐﻴﺭ ﺍﻟﺘﻲ ﺘﺅﺨﺫ ﻤﻥ ﺃﺠﻠﻬﺎ ﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ ﻫﻲ 2 : ﻤﻥ ﺍﻟﻔﺭﻋﻴﻥ \" \"3ﻭ\" \"4ﻨﺠﺩ ﺃﻥ : a = 2ﻭb = −1 ﻭﻤﻨﻪf (x) = (x − 2)2 −1: = x2 − 4x + 3 ﺕ:5 f (x) = 3x2 + 7 (1 ﺍﻟﺩﺍﻟﺔ fﺯﻭﺠﻴﺔ ﻷﻨﻬﺎ ﺘﺤﻘﻕ ﺍﻟﺘﻌﺭﻴﻑ (2ﻨﻔﺱ ﺍﻟﺸﺊ ﻤﻥ ﺃﺠلf (x) = −5x 2 − 1: f (x) = (x − 3)2 (3 ﺍﻟﺩﺍﻟﺔ fﻟﻴﺴﺕ ﺯﻭﺠﻴﺔ ﻷﻥ: f (1) = 4ﻭ f (−1) = 16ﻭ 4 ≠ 16
)f (x = 3x 2 (4 x4 + 2 ﺍﻟﺩﺍﻟﺔ fﺯﻭﺠﻴﺔ ﻷﻨﻬﺎ ﺘﺤﻘﻕ ﺍﻟﺘﻌﺭﻴﻑ (5ﻨﻔﺱ ﺍﻟﺸﺊ ﻤﻥ ﺃﺠلf (x) = 2 : f (x) = 3x3 + 1 (6 ﺍﻟﺩﺍﻟﺔ fﻟﻴﺴﺕ ﺯﻭﺠﻴﺔ ﻷﻥ: f (2) = 25ﻭ f (−2) = −23ﻭ 25 ≠ −23 ﺕ (1 : 6ﺍﻟﺩﺍﻟﺔ fﺯﻭﺠﻴﺔ ﻷﻨﻬﺎ ﺘﺤﻘﻕ ﺍﻟﺘﻌﺭﻴﻑ ﻭﻨﺴﺘﻨﺘﺞ ﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ) ( C ﻴﻘﺒل ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻜﻤﺤﻭﺭ ﺘﻨﺎﻅﺭ (2ﺍﻟﺩﺍﻟﺔ fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ] ]− ∞,0 (3ﺍﻟﺠﺩﻭل ﺒﻌﺩ ﺇﺘﻤﺎﻤﻪ ﻴﺼﺒﺢ : x 0 − 1 -1 − 3 -2 2 2 )f (x 23 0 − 5 -6 2 2 rj rj(C) 0 ir 0 ir )(C
ﺍﺴﺘﻨﺘﺞ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ f ∞x −∞ 0 + f (x) 2 ﺕ:7) f (x) = g(xﺘﻌﻨﻲ2x2 − 5x + −x + 1: ﺍﻟﻨﻘﺎﻁ ﺍﻟﻤﺸﺘﺭﻜﺔ ﻟﻠﻤﻨﺤﻨﻴﻴﻥ ) ( C ﻭ)∆( ﺤﻠﻬﺎ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔf (x) = g(x) : ﻭﻫﺫﺍ ﻴﻌﻨﻲ2x2 −4x = 0: ﻭﻫﺫﺍ ﻴﻌﻨﻲ2x(x − 2) = 0 : ﻭﻫﺫﺍ ﻴﻌﻨﻲ x = 0 :ﺃﻭ x = 2 ﻭﻋﻠﻴﻪ ) ( C ﻭ)∆( ﻴﺸﺘﺭﻜﺎﻥ ﻓﻲ ﻨﻘﻁﺘﻴﻥ Aﻭ Bﺤﻴﺙ: )) A(0, g(0ﻭ ))B(2, g(2 ﺃﻱ A(0,1) :ﻭ )B(2,−1 (2ﻟﻴﻜﻥ xﻋﺩﺩﺍ ﺤﻘﻴﻘﻴﺎ ﻜﻴﻔﻴﺎ 2 x − 5 2 − 17 = 2 x2 − 5 x + 25 − 17 : ﻟﺩﻴﻨﺎ 4 8 2 16 8 = 2x2 − 5x + 25 − 17 8 8 = 2x2 − 5x +1 )= f (x 5 ,+∞ ﺍﻟﻤﺠﺎل ﻋﻠﻰ ﺘﻤﺎﻤﺎ ﻤﺘﺯﺍﻴﺩﺓ ﻭﻫﻲ − ∞, 5 ﺍﻟﻤﺠﺎل ﻋﻠﻰ ﺘﻤﺎﻤﺎ ﻤﺘﻨﺎﻗﺼﺔ f (3 4 4
(4ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﻟﻠﺩﺍﻟﺔ f ∞x − ∞5 + )f (x 4 − 17 8 )(C ﺇﻨﺸﺎﺀ)∆( ﻭ ) : ( C)∆( rj 0 ir ﺃ .ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ) (1ﻓﻲ Rﻫﻲ] [0,2 : ﺏ.ﺍﻟﻤﺘﺭﺍﺠﺤﺔ )g(x) > f (x ﻭﺘﻜﺎﻓﺊ− x + 1 > 2x2 − 5x + 1 : ﺘﻜﺎﻓﺊ2x 2 − 4x < 0 : ﺘﻜﺎﻓﺊx 2 − 2x < 0 :ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ) (1ﺒﻴﺎﻨﻴﺎ ﻴﻌﻨﻲ ﺍﻟﺒﺤﺙ ﻋﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻔﻭﺍﺼل ﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻤﻥ ﺃﺠﻠﻬﺎ ∆ ﻓﻭﻕ ) ( )( C ﺘﻤﺎﻤﺎ ﻭﻤﻥ ﺍﻟﺸﻜل ﺍﻟﺴﺎﺒﻕ ﻨﺠﺩ ﺃﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﻁﻠﻭﺒﺔ ﻫﻲ] [0,2 :
\"ﻤﻘﻠﻭﺏ\"( ﺍﻟﺩﺍﻟﺔ )ﺃﻭ x → 1 ﺍﻟﺩﺍﻟﺔ x ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ: x → 1 ﺍﻟﺩﺍﻟﺔ ﺘﻌﺭﻴﻑ ﻤﺠﻤﻭﻋﺔ - I x x → 1 ﺍﻟﺩﺍﻟﺔ ﺘﻐﻴﺭ ﺇﺘﺠﺎﻩ ﺩﺭﺍﺴﺔ - II x x → 1 ﺍﻟﺩﺍﻟﺔ ﺘﻐﻴﺭﺍﺕ ﺠﺩﻭل ﺇﻨﺸﺎﺀ - III x - IVﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﺩﺍﻟﺔ ﻓﺭﺩﻴﺔ ﻭﺘﻨﺎﻅﺭ ﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﺒﺩﺃ ﺍﻟﻤﻌﻠﻡﻤﺸﻜﻼﺕ ﻟﻤﻌﺎﻟﺠﺔ →x 1 ﺍﻟﺩﺍﻟﺔ ﺤﻭل ﺍﻟﻤﻌﻠﻭﻤﺎﺕ ﺍﺴﺘﻌﻤﺎل -V x ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ x → 1 ﺍﻟﺩﺍﻟﺔ ﺤﻭل ﺍﻟﻨﻅﺭﻱ ﺍﻟﺩﺭﺱ .1 x .2ﺍﻟﺩﻭﺍل ﺍﻟﻔﺭﺩﻴﺔ .3ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ .4ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭﺍﻟﻤﺸﻜﻼﺕ
: x → 1 ﺍﻟﺩﺍﻟﺔ ﺤﻭل ﺍﻟﻨﻅﺭﻱ ﺍﻟﺩﺭﺱ • x : x → 1 ﺍﻟﺩﺍﻟﺔ ﺘﻌﺭﻴﻑ ﻤﺠﻤﻭﻋﺔ xﻭﺍﻟﻌﺩﺩ 1 ﻫﻭ ﻤﻘﻠﻭﺏ ﻭﻫﺫﺍ ﺍﻟﻤﻘﻠﻭﺏ ﺍﻟﻤﺠﻤﻭﻋﺔ - R ﻟﻪ – ﻓﻲ ﻤﻌﺩﻭﻡ x ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻏﻴﺭ ﻨﻌﻠﻡ ﺃﻨﻪ : x ﺼﻔﺭ ﻟﻴﺱ ﻟﻪ ﻤﻘﻠﻭﺏ. ﻤﻨﻪ:ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺠﻤﻴﻊ )f (x = 1 ﺒﺎﻟﺩﺴﺘﻭﺭ ﺍﻟﻤﻌﺭﻓﺔ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ f xﺍﻷﻋﺩﺍﺩ ﺍﻟﺤﻘﻴﻘﻴﺔ ﻏﻴﺭ ﺍﻟﻤﻌﺩﻭﻤﺔ ﻭﻫﺫﻩ ﺍﻟﻤﺠﻤﻭﻋﺔ ﻫﻲ[∞]− ∞,0[∪ ]0,+ ﻭﻴﺭﻤﺯ )ﺃﻴﻀﺎ( ﺇﻟﻰ ﻫﺫﻩ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺒﺎﻟﺭﻤﺯ * Rﺃﻭ ﺒﺎﻟﺭﻤﺯ}R − {0 : x → 1 ﺍﻟﺩﺍﻟﺔ ﺘﻐﻴﺭ ﺇﺘﺠﺎﻩ ﺩﺭﺍﺴﺔ x 1 < 1 ﻴﻜﺎﻓﺊ (b > ﻓﺈﻥ )a ab > ﺒﺤﻴﺙ 0 ﺤﻘﻴﻘﻴﻴﻥ ﻋﺩﺩﻴﻥ ﻭb ﻜﺎﻥ a ﺇﺫﺍ ﺃﻨﻪ ﻨﻌﻠﻡ ﺘﺫﻜﻴﺭ: b a ﻟﻨﺩﺭﺱ ﺇﺘﺠﺎﻩ[ ] [ ]f)(x 1 ﻟﺘﻜﻥ f = x ﺒﺎﻟﺩﺴﺘﻭﺭ , − ∞,0 ∪ ∞0,+ ﺍﻟﻤﺠﻤﻭﻋﺔ ﻋﻠﻰ ﺍﻟﻤﻌﺭﻓﺔ, ﺍﻟﺩﺍﻟﺔ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ fﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ [ ]− ∞,0ﻭ [∞]0,+ ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭﺍﻥ uﻭ ϑﻤﻥ ﺍﻟﻤﺠﺎل [ ]− ∞,0ﻟﻨﺎ u < 0ﻭ ϑ < 0ﻤﻨﻪ u.ϑ > 0ﻭﺇﺫﺍ ﻤﻨﻪ: 1 < 1 ﻴﻜﻭﻥ ϑ > u ﻜﺎﻥ ϑ u ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭﺍﻥ uﻭ ϑﻤﻥ ﺍﻟﻤﺠﺎل [ ]− ∞,0ﺇﺫﺍ ﻜﺎﻥ ϑ > uﻓﺈﻥ )f (u) > f (ϑ ﻭﺒﺎﻟﺘﺎﻟﻲ ﺍﻟﺩﺍﻟﺔ fﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ] [− ∞,0
ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭﺍﻥ uﻭ ϑﻤﻥ ﺍﻟﻤﺠﺎل ∞ 0,+ﻟﻨﺎ u > 0ﻭ ϑ > 0ﻤﻨﻪ u.ϑ > 0ﻭﺇﺫﺍ[ ] ﻤﻨﻪ: 1 < 1 ﻴﻜﻭﻥ ϑ > u ﻜﺎﻥ ϑ uﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭﺍﻥ uﻭ ϑﻤﻥ ﺍﻟﻤﺠﺎل [∞ ]0,+ﺇﺫﺍ ﻜﺎﻥ ϑ > uﻓﺈﻥ )f (u) > f (ϑﻭﺒﺎﻟﺘﺎﻟﻲ ﺍﻟﺩﺍﻟﺔ fﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ∞ 0,+ﻭﻫﻜﺫﺍ ﻟﻘﺩ ﺒﺭﻫﻨﺎ ﻋﻠﻰ[ ] ﺍﻟﻨﻅﺭﻴﺔ:f )(x = 1 ﺒﺎﻟﺩﺴﺘﻭﺭ ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ * R ﺍﻟﺩﺍﻟﺔ f x .Iﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل[َ ]− ∞,0ﻭ .IIﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل [∞]0,+ : x → 1 ﺍﻟﺩﺍﻟﺔ ﺘﻐﻴﺭﺍﺕ ﺠﺩﻭل x )f (x = 1 ﺒﺎﻟﺩﺴﺘﻭﺭ *R ﺍﻟﻤﺠﻤﻭﻋﺔ ﻋﻠﻰ ﺍﻟﻤﻌﺭﻓﺔ ﺍﻟﺩﺍﻟﺔ ﻟﺘﻜﻥ f xﺇﺼﻁﻼﺡ 0 :ﻫﻭ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ﺍﻟﻭﺤﻴﺩ ﺍﻟﺫﻱ ﻟﻴﺴﺕ ﻟﻪ ﺼﻭﺭﺓ ﺒﺎﻟﺩﺍﻟﺔ fﻭﻟﻠﺘﻌﺒﻴﺭ ﻋﻠﻰ ﻫﺫﺍ ,ﻓﻲ ﺠﺩﻭلﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ , fﻨﻀﻊ ﺘﺤﺕ \" 0ﺨﻁ ﻤﺴﺘﻘﻴﻡ ﺸﺎﻗﻭﻟﻲ ﻤﻀﺎﻋﻑ\" ﻭﺍﻋﺘﻤﺎﺩﺍ ﻋﻠﻰ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﻭﺍﺭﺩﺓ ﻓﻲ ﻨﺹﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﺴﺎﺒﻘﺔ ﻭﻋﻠﻰ ﺍﻻﺼﻁﻼﺡ ﺍﻟﺴﺎﺒﻕ ﻴﻜﻭﻥ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ fﻜﻤﺎ ﻴﻠﻲ: x −∞ 0 ∞+ )f (x : x → 1 ﻟﻠﺩﺍﻟﺔ ﺍﻟﺒﻴﺎﻨﻲ ﺍﻟﺘﻤﺜﻴل xﻭﻟﻴﻜﻥ )(C f )(x = 1 ﺒﺎﻟﺩﺴﺘﻭﺭ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ * R ﻟﺘﻜﻥ f x ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﻌﻠﻡ 0, ir, rj ﻟﻠﺩﺍﻟﺔ fﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ) (ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻏﻴﺭ ﻤﻌﺩﻭﻡ. ﺤﻴﺙ x x, 1 ﺇﺤﺩﺍﺜﻴﺎﻫﺎ ﺍﻟﺘﻲ ﺍﻟﻨﻘﻁ ﻤﺠﻤﻭﻋﺔ ﻫﻭ )(C x
\"ﻓﻲ ﺃﻗﺼﻰ M x, 1 ﺍﻟﻨﻘﻁﺔ ﻭﺘﻜﻭﻥ 0 ﻤﻥ ﻗﺭﻴﺒﺎ 1 \"ﻜﺒﻴﺭ ﺠﺩﺍ\" ﻭ x > 0 ﻟﻤﺎ ﻴﻜﻭﻥ x x ﻴﻜﻭﻥ xﺍﻟﻴﻤﻴﻥ\" )ﻷﻥ ﻓﺎﺼﻠﺘﻬﺎ ﻤﻭﺠﺒﺔ ﻭﻜﺒﻴﺭﺓ ﺠﺩﺍ( ﻭ\"ﻗﺭﻴﺒﺔ ﻤﻥ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل\" )ﻷﻥ ﺘﺭﺘﻴﺒﻬﺎ ﻗﺭﻴﺏ ﻤﻥ (0\"ﻗﺭﻴﺒﺔ ﻤﻥ M x, 1 ﺍﻟﻨﻘﻁﺔ ﻭﺘﻜﻭﻥ \" ﺠﺩﺍ ﻜﺒﻴﺭ \" 1 \"ﻗﺭﻴﺏ ﻤﻥ \"0ﻭ x > 0 ﻟﻤﺎ ﻴﻜﻭﻥ x x ﻴﻜﻭﻥ x ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ\")ﻷﻥ ﻓﺎﺼﻠﺘﻬﺎ ﻗﺭﻴﺒﺔ ﻤﻥ (0ﻭ\"ﻋﺎﻟﻴﺔ ﺠﺩﺍ\" ﻷﻥ ﺘﺭﺘﻴﺒﻬﺎ ﻜﺒﻴﺭ ﺠﺩﺍ ﻭﻤﻭﺠﺏ.x ﻨﺴﺘﻌﻴﻥ ﺒﺠﺩﻭل ﻗﻴﻡ ﻟﺘﻌﻴﻴﻥ ﺒﻌﺽ ﺍﻟﻨﻘﻁ ﻤﻥ ﺍﻟﻤﻨﺤﻨﻲ )(C -4 -2 -1 − 1 1 1 11 2 4 2 −4 4 21 − 1 − 1 -1 -2 -4 4 2 1 1 1x 4 2 2 4 ( )0,ir, rj ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﻤﻌﻠﻡ →x 1 ﻭﻓﻲ ﺍﻟﺸﻜل ﺍﻟﻤﻭﺍﻟﻲ) (Cﺍﻟﻤﻨﺤﻨﻲ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ x )(C rj 0 ir ▲ ﺍﻟﻤﻨﺤﻨﻲ) (Cﻫﻭ ﻗﻁﻊ ﺯﺍﺌﺩ ﺍﻟﻨﻘﻁﺔ 0ﻤﺒﺩﺃ ﺍﻟﻤﻌﻠﻡ ﻫﻲ ﻤﺭﻜﺯ ﺘﻨﺎﻅﺭ ﻟﻠﻘﻁﻊ ﺍﻟﺯﺍﺌﺩ)(C
• ﺍﻟﺩﻭﺍل ﺍﻟﻔﺭﺩﻴﺔ : ﺘﻌﺭﻴﻑ : ﻟﺕﻜﻥ Fﺩﺍﻟﺔ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻔﻬﺎ D ﺘﻜﻭﻥ ﺍﻟﺩﺍﻟﺔ Fﻓﺭﺩﻴﺔ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﺘﺤﻘﻕ ﺍﻟﺸﺭﻁﺎﻥ : D (1ﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﺼﻔﺭ (2ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭ xﻤﻥ F(−x) = −F(x) : D ﺃﻤﺜﻠﺔ :ﻟﺘﻜﻥ g, fﻭ hﺍﻟﺩﻭﺍل ﺤﻴﺙ: h(x) = −5x + 2, g(x) = x , f (x) = x 3 x2 + 1 ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﻜل ﻤﻥ ﺍﻟﺩﻭﺍل h, g, fﻫﻲ Rﻭ Rﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﺼﻔﺭ. ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ : x f (−x) = (−x)3ﻤﻨﻪ ) f (−x) = −(x3ﺃﻱ ) f (−x) = − f (xﻭﺒﺎﻟﺘﺎﻟﻲ ﺍﻟﺩﺍﻟﺔ fﻓﺭﺩﻴﺔﻭﺒﺎﻟﺘﺎﻟﻲ ﺍﻟﺩﺍﻟﺔ g ﺃﻱ )g(−x) = −g(x = )g(−x −x = )g(−x −x ﻤﻨﻪ x 2 + 1 (−x)2 +1 ﻓﺭﺩﻴﺔ h(−x) = −5(−x) + 2ﻤﻨﻪ h(−x) = 5x + 2ﻭ ) h(−xﻻ ﻴﺴﺎﻭﻱ ) − h(xﻤﻥ ﺃﺠل ﻜل ﻗﻴﻡ xﻤﺜﺎل ﻤﻀﺎﺩ h(−3) = 17 :ﻭ h(3) = −13ﻭ (−13) ≠ 17 ﻤﻨﻪ ﺍﻟﺩﺍﻟﺔ hﻟﻴﺴﺕ ﻓﺭﺩﻴﺔ. ﻤﻼﺤﻅﺔ:ﺘﻨﺎﻅﺭﻴﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻫﻲ ﺩﺍﻟﺔ ﻓﺭﺩﻴﺔ ﻷﻥ * R = )f (x 1 ﺒﺎﻟﺩﺴﺘﻭﺭ *,R ﻋﻠﻰ ﺍﻟﻤﻌﺭﻓﺔ ﻟﺩﺍﻟﺔ f xf )(−x = − 1 ﻤﻨﻪ f )(−x = 1 , ﻤﻥ * R ﺍﻟﺼﻔﺭ ﻭﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﻨﺼﺭ x x −x ﻤﻨﻪ )f (−x) = − f (x
ﻭ)ﺤﺴﺏ ﻤﻼﺤﻅﺔ ﺒﻴﺎﻨﻴﺔ ﺴﺎﺒﻘﺔ( ﺍﻟﻤﻨﺤﻨﻲ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ fﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﻌﻠﻡ 0, ir, rjﻴﻘﺒل ﺍﻟﻨﻘﻁﺔ ( )0ﻤﺒﺩﺃ ﺍﻟﻤﻌﻠﻡ ﻜﻤﺭﻜﺯ ﺘﻨﺎﻅﺭ ﻭﻫﺫﺍ ﻟﻴﺱ ﺨﺎﺼﺎ ﺒﺎﻟﺩﺍﻟﺔ \"ﻤﻘﻠﻭﺏ\" ﻭﺍﻟﻨﺘﻴﺠﺔ ﺍﻟﻌﺎﻤﺔ ﻓﻲ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻤﻭﺍﻟﻴﺔ )ﺍﻟﻤﻘﺒﻭﻟﺔ ﺒﺩﻭﻥ ﺒﺭﻫﺎﻥ( ﻨﻅﺭﻴﺔ : ﻟﺘﻜﻥ Fﺩﺍﻟﺔ ﻭﻟﻴﻜﻥ) (Cﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ Fﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﻌﻠﻡ) (0,ir, rjﺤﺘﻰ ﺘﻜﻭﻥ ﺍﻟﻨﻘﻁﺔ ) 0ﻤﺒﺩﺃ ﺍﻟﻤﻌﻠﻡ( ﻤﺭﻜﺯ ﺘﻨﺎﻅﺭ ﻟﻠﻤﻨﺤﻨﻲ) (Cﻴﻠﺯﻡ ﻭﻴﻜﻔﻲ ﺃﻥ ﺘﻜﻭﻥ ﺍﻟﺩﺍﻟﺔ Fﻓﺭﺩﻴﺔ ﻤﻼﺤﻅﺔ ﻫﺎﻤﺔ : ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻋﺩﺩ ﺼﺤﻴﺢ nﻟﻨﺎ ﺇﻤﺎ nﻓﺭﺩﻱ ﻭﺇﻤﺎ nﺯﻭﺠﻲ ﻭﻟﻜﻥ ﻫﺫﺍ ﻏﻴﺭ ﺼﺤﻴﺢ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺩﺍﻟﺔﻓﻤﺜﻼ ﻴﻤﻜﻥ ﺃﻥ ﻨﺘﺄﻜﺩ ﺒﺴﻬﻭﻟﺔ ﺃﻥ ﺍﻟﺩﺍﻟﺔ fﺤﻴﺙ f (x) = 2x + 5 :ﻟﻴﺴﺕ ﻓﺭﺩﻴﺔ ﻭﻟﻴﺴﺕ ﺯﻭﺠﻴﺔ ﻭﺃﻥ ﺍﻟﺩﺍﻟﺔ g(x) = 0 : gﻫﻲ ﻓﺭﺩﻴﺔ ﻭﺯﻭﺠﻴﺔ ﻓﻲ ﺁﻥ ﻭﺍﺤﺩ‼
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