دروس مادة الرياضيات للفصل الاول للشعب العلمية سنة ثالثة ثانوي

‫ﻭ ﻋﻠﻴﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ‪ a‬ﻭ ‪b‬‬ ‫‪ x‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ‬ ‫ﻨﺘﺎﺌﺞ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ﺍﻟﺩﺍﻟﺔ ‪e x‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬‫‪ a > b‬ﺘﻜﺎﻓﺊ ‪ea > eb :‬‬ ‫‪ a = b‬ﺘﻜﺎﻓﺊ ‪ ea = eb :‬؛‬‫‪ x > 0‬ﺘﻜﺎﻓﺊ ‪e x > 1 :‬‬ ‫‪ a < b‬ﺘﻜﺎﻓﺊ ‪ ea < eb :‬؛‬ ‫‪:x‬‬ ‫‪ x < 0‬ﺘﻜﺎﻓﺊ ‪e x < 1 :‬‬ ‫‪ -10‬ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺩﺍﻟﺔ )‪e g( x‬‬ ‫ﻤﺒﺭﻫﻨﺔ ‪:‬‬‫ﺇﺫﺍ ﻜﺎﻨﺕ ‪ g‬ﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ‪ I‬ﻭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪ I‬ﻓﺈﻥ ﺍﻟﺩﺍﻟﺔ )‪x e g( x‬‬ ‫ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪ I‬ﻭ ﺩﺍﻟﺘﻬﺎ ﺍﻟﻤﺸﺘﻘﺔ ﻫﻲ ﺍﻟﺩﺍﻟﺔ )‪( ). x g′ x e g( x‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬‫ﺍﻟﺩﺍﻟﺔ )‪ x e g( x‬ﻫﻲ ﻤﺭﻜﺏ ﺍﻟﺩﺍﻟﺘﻴﻥ ‪ g‬ﺍﻟﺘﻲ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪ I‬ﻭ ﺍﻟﺩﺍﻟﺔ ‪x e x‬‬ ‫ﺍﻟﺘﻲ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ﻭ ﻋﻠﻴﻪ ﺍﻟﺩﺍﻟﺔ )‪ x e g( x‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪ I‬ﻭ ﺩﺍﻟﺘﻬﺎ‬ ‫ﺍﻟﻤﺸﺘﻘﺔ ﻫﻲ ﺍﻟﺩﺍﻟﺔ )‪( )x g′ x e g( x‬‬ ‫ﻤﺜﺎل ‪:‬‬‫ﺍﻟﺩﺍﻟﺔ ‪ x e x2 −4 x : f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ﻷﻨﻬﺎ ﻤﺭﻜﺏ ﻟﻠﺩﺍﻟﺔ ‪x x2 − 4 x‬‬ ‫ﺍﻟﺘﻲ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ﻭ ﺍﻟﺩﺍﻟﺔ ‪ x e x‬ﺍﻟﺘﻲ‬ ‫ﻭ ﺩﺍﻟﺘﻬﺎ ﺍﻟﻤﺸﺘﻘﺔ ﻤﻌﺭﻓﺔ ﺒﺎﻟﻌﺒﺎﺭﺓ ‪:‬‬ ‫ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ‬‫‪f ′( x) = (2x − 4)ex2−4x‬‬ ‫‪ -11‬ﺍﻟﺩﺍﻟﺔ ﺍﻷﺼﻠﻴﺔ ﻟﻠﺩﺍﻟﺔ ‪: f‬‬‫ﻟﺩﻴﻨﺎ ‪ x g′ x e g( x) :‬ﺤﻴﺙ ‪ g‬ﺩﺍﻟﺔ ﻤﺴﺘﻤﺭﺓ ﻭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ‪( )I‬‬ ‫ﻫﻲ ﺍﻟﺩﺍﻟﺔ ‪ x e g( x) : h‬ﻷﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪ x‬ﻤﻥ ‪: I‬‬

‫)‪f ′( x) = g′( x)eg(x) = f ( x‬‬ ‫ﻤﺜﺎل ‪:‬‬ ‫ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺩﻭﺍل ﺍﻷﺼﻠﻴﺔ ﻟﻠﺩﺍﻟﺔ ‪: f‬‬ ‫‪x xe x2 −4‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﺴﺘﻤﺭﺓ ﻋﻠﻰ ﻷﻨﻬﺎ ﻤﺭﻜﺏ ﻭ ﺠﺩﺍﺀ ﺩﻭﺍل ﻤﺴﺘﻤﺭﺓ ﻋﻠﻰ‬‫ﻭ ﻋﻠﻴﻪ ﺘﻘﺒل ﺩﻭﺍل‬ ‫ﺃﺼﻠﻴﺔ ‪. h‬‬ ‫‪1‬‬ ‫‪.2 x.e x2−4‬‬ ‫ﻟﺩﻴﻨﺎ ‪f x = xe x2 −4 :‬‬‫‪( ) ( )h‬‬ ‫‪x‬‬ ‫=‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫ﺤﻴﺙ ‪ c‬ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﺜﺎﺒﺕ ‪( ).‬‬‫‪f‬‬ ‫‪x‬‬ ‫=‬ ‫‪1‬‬ ‫‪.e x2 −4‬‬ ‫‪+‬‬ ‫‪c‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪:‬‬ ‫‪2‬‬

‫ﺘﻜﻨﻭﻟﻭﺠﻴﺎ ﺍﻹﻋﻼﻡ ﻭ ﺍﻻﺘﺼﺎل‬‫ﺒﺎﺴﺘﻌﻤﺎل ﻁﺭﻴﻘﺔ ‪Euler‬‬ ‫ﺍﻟﺘﻁﺒﻴﻕ ‪:‬‬ ‫ﺃﻨﺸﺊ ﺘﻤﺜﻴﻼ ﺘﻘﺭﻴﺒﻴﺎ ﻟﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ‪ yc y‬ﻭ ‪y(0) = 1‬‬ ‫ﺒﻤﺠﺩﻭل ‪ Excel‬ﻓﻲ ﺍﻟﻤﺠﺎل ‪ −2;2‬ﻭﺍﻟﺨﻁﻭﺓ‪[ ]h = 0, 005 .‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫ﻟﺩﻴﻨﺎ‪ 'y | f c( x).'x :‬ﻭﻤﻨﻪ ‪ f ( x  h)  f ( x) | f c( x).h‬ﺃﻭ‬ ‫‪ f ( x  h)  f ( x) |  f c( x).h‬ﻤﻊ ‪h > 0‬‬ ‫ﻭﺒﻤﺎ ﺃﻥ ‪ yc y‬ﻓﺈﻥ )‪ f ( x) f c( x‬ﻓﻨﺤﺼل ﻋﻠﻰ‬ ‫)‪ f ( x + h) ≈ f ( x)(1 + h‬ﺃﻭ )‪. f ( x − h) ≈ f ( x)(1 − h‬‬‫ﻨﺘﺤﺼل ﺒﺎﻟﻌﺒﺎﺭﺓ ﺍﻷﻭﻟﻰ )‪ f ( x + h) ≈ f ( x)(1 + h‬ﻗﻴﻡ ﺍﻟﺩﺍﻟﺔ )ﺍﻟﺤل( ﻤﻥ ﺃﺠل ‪x > 0‬‬ ‫ﻭﺘﻌﻁﻲ ﺍﻟﻌﺒﺎﺭﺓ ﺍﻟﺜﺎﻨﻴﺔ )‪ f ( x − h) ≈ f ( x)(1 − h‬ﻗﻴﻡ ﺍﻟﺩﺍﻟﺔ )ﺍﻟﺤل( ﻤﻥ ﺃﺠل‬‫ﺍﻟﻘﻴﻡ ‪ x < 0‬ﻭﺫﻟﻙ ﺒﺎﻋﺘﺒﺎﺭ ‪ f (0) = 1‬ﻓﻲ ﺍﻻﻨﻁﻼﻗﺔ ﻭﺠﻌل ‪ h‬ﺼﻐﻴﺭﺍ ﺒﺎﻟﻘﺩﺭ ﺍﻟﺫﻱ ﻴﻀﻤﻥ ﺘﻘﺭﻴﺒﺎ‬ ‫ﺠﻴﺩﺍ‪.‬‬ ‫ﻨﺴﺘﺨﺩﻡ ﻤﺠﺩﻭل ‪ Excel‬ﻟﻤﻘﺎﺭﺒﺔ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ ﺍﻟﺤل ‪.‬‬ ‫ﺤﺠﺯ ﺍﻷﻋﺩﺍﺩ‪:‬‬ ‫ﻨﺤﺠﺯ ﺍﻟﺨﻁﻭﺓ ‪ h‬ﻓﻲ ﺍﻟﺨﺎﻨﺔ ‪ A3‬ﻤﺜﻼ‪.‬‬ ‫ﻋﻠﻰ ﺍﻟﺠﺯﺀ @‪>2;0‬‬ ‫ﻨﺤﺠﺯ ﻓﻲ ﺍﻟﺨﺎﻨﺔ ‪ A4‬ﻗﻴﻤﺔ ﺇﺒﺘﺩﺍﺌﻴﺔ ﻟﻠﻤﺘﻐﻴﺭ ﻭﻫﻲ ‪0‬‬‫ﻨﺤﺠﺯ ﻓﻲ ﺍﻟﺨﺎﻨﺔ ‪ A5‬ﺍﻟﻘﺎﻋﺩﺓ ‪ = x − h‬ﺍﻟﺘﻲ ﺘﻌﻁﻲ ﻗﻴﻡ ﺍﻟﻤﺘﻐﻴﺭ ‪ x‬ﺍﻟﺘﻲ ﻫﻲ ﻗﺒل ‪ 0‬ﺒﻁﺭﺡ‬‫ﺍﻟﺨﻁﻭﺓ ﻓﻲ ﻜل ﻤﺭﺓ ﺤﺘﻰ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﺍﻟﻌﺩﺩ ‪ 2‬ﻓﻨﺤﺠﺯ‪ A4  A$3 :‬ﺜﻡ ﻨﻌﻤﻡ ﻋﻠﻰ ﺒﺎﻗﻲ‬ ‫ﺍﻟﺨﺎﻨﺎﺕ ﻤﻥ ﻋﻤﻭﺩ ‪ A‬ﺇﻟﻰ ﻏﺎﻴﺔ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﺍﻟﻘﻴﻤﺔ ‪ 2‬ﺃﻭ ﺃﻗﺭﺏ ﻗﻴﻤﺔ ﻟﻬﺎ‪.‬‬ ‫ﻨﺤﺠﺯ ﻓﻲ ﺍﻟﺨﺎﻨﺔ ‪ B4‬ﺍﻟﻌﺩﺩ ‪ 1‬ﻭﻫﻭ ﻗﻴﻤﺔ ﺍﻟﺩﺍﻟﺔ ﻤﻥ ﺃﺠل ‪ 0‬ﻷﻥ ‪f (0) 1‬‬ ‫ﻨﺤﺠﺯ ﻓﻲ ﺍﻟﺨﺎﻨﺔ ‪ B5‬ﺍﻟﻘﻴﻤﺔ ﺍﻟﺘﻘﺭﻴﺒﻴﺔ ﻟﻠﻌﺩﺩ )‪ y = f ( x − h‬ﻭﻟﺩﻴﻨﺎ‬ ‫)‪ f ( x − h) ≈ f ( x).(1 − h‬ﻓﻨﺤﺠﺯ ‪ B 4 * (1 A$3) :‬ﺜﻡ ﻨﻌﻤﻡ ﻋﻠﻰ ﺒﺎﻗﻲ‬ ‫ﺍﻟﺨﺎﻨﺎﺕ ﻤﻥ ﻋﻤﻭﺩ ‪ B‬ﺤﺘﻰ ﺍﻟﻭﺼﻭل ﺇﻟﻰ ﺁﺨﺭ ﻗﻴﻤﺔ ﻟﻠﻤﺘﻐﻴﺭ ﻤﻥ ﺍﻟﻌﻤﻭﺩ ‪. A‬‬

‫ﻋﻠﻰ ﺍﻟﺠﺯﺀ @‪>0;2‬‬ ‫ﻨﺤﺠﺯ ﻓﻲ ﺍﻟﺨﺎﻨﺔ ‪ C4‬ﻗﻴﻤﺔ ﺍﺒﺘﺩﺍﺌﻴﺔ ﻟﻠﻤﺘﻐﻴﺭ ﻭﻫﻲ ‪0‬‬ ‫ﻨﺤﺠﺯ ﻓﻲ ﺍﻟﺨﺎﻨﺔ ‪ C5‬ﺍﻟﻘﺎﻋﺩﺓ ‪ = x + h‬ﺍﻟﺘﻲ ﺘﻌﻁﻲ ﻗﻴﻡ ﺍﻟﻤﺘﻐﻴﺭ ‪ x‬ﺍﻟﺘﻲ ﻫﻲ ﺒﻌﺩ ‪ 0‬ﺒﺈﻀﺎﻓﺔ‬ ‫ﺍﻟﺨﻁﻭﺓ ﻓﻲ ﻜل ﻤﺭﺓ ﺤﺘﻰ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﺍﻟﻌﺩﺩ ‪ 2‬ﻓﻨﺤﺠﺯ‪ C4  A$3 :‬ﺜﻡ ﻨﻌﻤﻡ ﻋﻠﻰ ﺒﺎﻗﻲ‬ ‫ﺍﻟﺨﺎﻨﺎﺕ ﻤﻥ ﻋﻤﻭﺩ ‪ C‬ﺇﻟﻰ ﻏﺎﻴﺔ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﺍﻟﻘﻴﻤﺔ ‪ 2‬ﺃﻭ ﺃﻗﺭﺏ ﻗﻴﻤﺔ ﻟﻬﺎ‪.‬‬ ‫ﻨﺤﺠﺯ ﻓﻲ ﺍﻟﺨﺎﻨﺔ ‪ D4‬ﺍﻟﻌﺩﺩ ‪ 1‬ﻭﻫﻭ ﻗﻴﻤﺔ ﺍﻟﺩﺍﻟﺔ ﻤﻥ ﺃﺠل ‪ 0‬ﻷﻥ ‪f (0) 1‬‬ ‫ﻨﺤﺠﺯ ﻓﻲ ﺍﻟﺨﺎﻨﺔ ‪ D5‬ﺍﻟﻘﻴﻤﺔ ﺍﻟﺘﻘﺭﻴﺒﻴﺔ ﻟﻠﻌﺩﺩ )‪ y = f ( x + h‬ﻭﻟﺩﻴﻨﺎ‬‫)‪ f ( x + h) ≈ f ( x).(1 + h‬ﻓﻨﺤﺠﺯ ‪ D 4 * (1 A$3) :‬ﺜﻡ ﻨﻌﻤﻡ ﻋﻠﻰ ﺒﺎﻗﻲ ﺍﻟﺨﺎﻨﺎﺕ‬ ‫ﻤﻥ ﻋﻤﻭﺩ ‪ D‬ﺤﺘﻰ ﺍﻟﻭﺼﻭل ﺇﻟﻰ ﺁﺨﺭ ﻗﻴﻤﺔ ﻟﻠﻤﺘﻐﻴﺭ ﻤﻥ ﺍﻟﻌﻤﻭﺩ ‪. B‬‬ ‫ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ‪:‬‬ ‫ﻨﺨﺘﺎﺭ ﺍﻟﻌﻤﻭﺩﻴﻥ ‪ A‬ﻭ‪ B‬ﻨﻀﻐﻁ ﻋﻠﻰ ﺍﻟﻤﺴﺎﻋﺩ ﺍﻟﺒﻴﺎﻨﻲ‬‫ﺜﻡ ﺍﻟﻤﻨﺤﻨﻰ ﻤﻥ ﺍﻟﻨﻭﻉ ‪،‬‬ ‫ﻭﻨﺨﺘﺎﺭ‬‫ﺜﻡ ﺍﺨﺘﻴﺎﺭ ﺍﻟﺴﻠﺴﻠﺔ ﺒﺎﻟﻀﻐﻁ ﻋﻠﻰ ﻨﺠﺩ‬ ‫ﻨﻭﺍﺼل ﺍﻟﻌﻤﻠﻴﺔ ﺒﺎﻟﻀﻐﻁ ﻋﻠﻰ‬‫ﺍﻟﺴﻠﺴﻠﺔ ﺍﻷﻭﻟﻰ ﺍﻟﺘﻲ ﺘﺨﺹ ﺘﻤﺜﻴل ﺍﻟﺩﺍﻟﺔ )ﺍﻟﺤل( ﻋﻠﻰ ﺍﻟﻤﺠﺎل ﺍﻷﻭل ‪ 2;0‬ﻤﺤﺠﻭﺯﺓ@ >‬‫ﻹﻀﺎﻓﺔ ﺍﻟﺴﻠﺴﻠﺔ ﺍﻟﺜﺎﻨﻴﺔ ﺍﻟﺘﻲ ﺘﻌﻁﻲ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ‬ ‫‪ .‬ﺜﻡ ﻨﻀﻐﻁ ﻋﻠﻰ‬ ‫ﺒﺎﺴﻡ‬‫ﺜﻡ ﻨﺤﺠﺯ ﻗﻴﻡ ﺍﻟﻌﻤﻭﺩ ‪ C‬ﺒﺎﻟﻀﻐﻁ ﺒﺎﻟﻔﺄﺭﺓ ﻤﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻷﻭﻟﻰ ﻓﻲ‬‫ﺜﻡ ﻨﺤﺠﺯ ﻗﻴﻡ ﺍﻟﻌﻤﻭﺩ ‪ D‬ﺒﺎﻟﻀﻐﻁ ﺒﺎﻟﻔﺄﺭﺓ ﻤﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻷﻭﻟﻰ ﻓﻲ‬ ‫ﻋﻠﻰ ﺍﻟﻤﺠﺎل ﺍﻟﺜﺎﻨﻲ ‪ 0;2‬ﻜﻤﺎ ﻴﻠﻲ‪> @:‬‬ ‫ﻨﻀﻊ ﻤﺅﺸﺭ ﺍﻟﻜﺘﺎﺒﺔ ﻋﻠﻰ ﺨﺎﻨﺔ ﻗﻴﻡ ‪x‬‬ ‫‪C4‬ﺇﻟﻰ ﺁﺨﺭ ﻗﻴﻤﺔ ﻤﻥ ﻨﻔﺱ ﺍﻟﻌﻤﻭﺩ‪.‬‬ ‫ﻨﻀﻊ ﻤﺅﺸﺭ ﺍﻟﻜﺘﺎﺒﺔ ﻋﻠﻰ ﺨﺎﻨﺔ ﻗﻴﻡ ‪y‬‬ ‫‪ D4‬ﺇﻟﻰ ﺁﺨﺭ ﻗﻴﻤﺔ ﻤﻥ ﻨﻔﺱ ﺍﻟﻌﻤﻭﺩ‪.‬‬‫ﻓﻴﻅﻬﺭ ﺍﻟﻤﻨﺤﻨﻴﺎﻥ ﻤﻜﻤﻼﻥ ﻟﺒﻌﻀﻬﻤﺎ ﺒﻠﻭﻨﻴﻥ ﻤﺨﺘﻠﻔﻴﻥ‬ ‫ﻨﻀﻐﻁ ﺒﻌﺩﻫﺎ ﻋﻠﻰ ﺍﻟﺘﺎﻟﻲ‬‫‪،‬ﺤﻴﺙ ﻴﺸﻜﻼﻥ ﻤﻨﺤﻨﻲ ﺍﻟﺩﺍﻟﺔ )ﺍﻟﺤل( ﻋﻠﻰ ﺍﻟﻤﺠﺎل ‪ ، 2;2‬ﺜﻡ ﺍﻹﻨﻬﺎﺀ@ >‬



‫ﺘﻤـﺎﺭﻴﻥ ﻭ ﻤﺸﻜﻼﺕ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 1‬‬ ‫ﻀﻊ ﺍﻟﻌﻼﻤﺔ √ ﺃﻤﺎﻡ ﻜل ﺠﻤﻠﺔ ﺼﺤﻴﺤﺔ ﻭ ﺍﻟﻌﻼﻤﺔ × ﺃﻤﺎﻡ ﻜل ﺠﻤﻠﺔ ﺨﺎﻁﺌﺔ‬ ‫‪e−3 < 0‬‬ ‫‪(1‬‬ ‫‪e−5 = −e5 (2‬‬ ‫‪ (3‬ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ ‪ x e x‬ﻴﻘﻁﻊ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ‪.‬‬ ‫‪e2x = ex‬‬ ‫‪(4‬‬ ‫‪lim‬‬ ‫‪x‬‬ ‫‪=0‬‬ ‫‪(5‬‬ ‫‪ex‬‬ ‫∞‪x→+‬‬ ‫‪ (6‬ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺩﺍﻟﺔ ‪ x e3 x‬ﻫﻲ ‪x e3 x‬‬ ‫‪lim e x‬‬ ‫=‬ ‫∞‪+‬‬ ‫‪(7‬‬ ‫‪xx> →0‬‬ ‫‪ (8‬ﺇﺫﺍ ﻜﺎﻥ ‪ x ≥ 2‬ﻓﺈﻥ ‪e x ≥ e2‬‬ ‫‪ (9‬ﺇﺫﺍ ﻜﺎﻥ ‪ x ≤ 0‬ﻓﺈﻥ ‪e x ≤ 0‬‬‫‪ (11‬ﺇﺸﺎﺭﺓ ‪:‬‬ ‫‪ (10‬ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪e x .e− x = 1 : x‬‬ ‫‪(13‬‬ ‫‪ x − 1 e x‬ﻫﻲ ﻨﻔﺱ ﺇﺸﺎﺭﺓ ‪( )x − 1‬‬ ‫‪ (12‬ﻤﻥ ﺃﺠل ‪e x−1 < 0 : x < 1‬‬ ‫‪lim xe x = 0‬‬ ‫∞‪x→+‬‬ ‫‪lim‬‬ ‫‪ex‬‬ ‫=‬ ‫∞‪+‬‬ ‫‪(14‬‬ ‫‪x3‬‬ ‫∞‪x→+‬‬ ‫‪e x − e = e x−1 (15‬‬ ‫‪( )e2x − 2e x + 1 = e x − 1 2 (16‬‬ ‫‪ (17‬ﺍﻟﺩﺍﻟﺔ ‪ x e−1 x + 3 :‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ‬

‫‪ (18‬ﻟﻴﺱ ﻟﻠﻤﺘﺭﺍﺠﺤﺔ ‪ e x2 −4 x < 0‬ﺤل ﻓﻲ‬ ‫‪ (19‬ﺍﻟﺩﺍﻟﺔ ‪ x e x2 −4 :‬ﻫﻲ ﺤل ﻟﻠﻤﻌﺎﺩﻟﺔ ‪y′ = y :‬‬ ‫‪e‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪<1‬‬ ‫‪(20‬‬ ‫‪2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 2‬‬ ‫ﺒﺴﻁ ﻤﺎ ﻴﻠﻲ ‪:‬‬‫‪1) e3 x .e−2 x‬‬ ‫‪2) e x .e2‬‬ ‫)‪( )3‬‬ ‫‪1‬‬ ‫‪e-x 2‬‬‫)‪4‬‬ ‫‪e4 x+1‬‬ ‫)‪( )5‬‬ ‫‪e−4x+5‬‬ ‫‪( )6) ex e−x + e2x‬‬ ‫‪e2 x .e−1‬‬ ‫‪ex-2 2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 3‬‬ ‫ﺒﻴﻥ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪ x‬ﻓﺈﻥ ‪:‬‬ ‫‪ex − e−x‬‬ ‫=‬ ‫‪e2x − 1‬‬ ‫‪ex + e−x‬‬ ‫‪e2x + 1‬‬ ‫‪:‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬‫= )‪1) f ( x‬‬ ‫ﺒﻴﻥ ﺃﻥ ﺍﻟﺩﻭﺍل ‪ f‬ﺍﻟﻤﻌﺭﻓﺔ ﻜﻤﺎ ﻴﻠﻲ ﻫﻲ ﺩﻭﺍل ﻤﻌﺭﻓﺔ ﻋﻠﻰ‬ ‫‪ex + e−x‬‬ ‫)‪2‬‬ ‫)‪f (x‬‬ ‫‪1‬‬ ‫‪= e2x + 1‬‬‫‪3) f ( x) = 1‬‬ ‫)‪4‬‬ ‫)‪f (x‬‬ ‫‪ex‬‬ ‫‪= ex + 3‬‬ ‫‪ex + 4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬ ‫‪( )f x‬‬ ‫‪x‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪f‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ‬ ‫‪= ex +1‬‬ ‫ﺇﺫﺍ ﻋﻠﻤﻨﺎ ﺃﻥ‪ 0 ≤ x ≤ 1 :‬ﻋﻴﻥ ﺤﺼﺭﺍ ﻟﻠﻌﺒﺎﺭﺓ ‪( ). f x‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ‪. 6‬‬ ‫‪1) e2 x−1 = 1‬‬ ‫ﺤل ﻓﻲ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫‪( )3) x2 − 4x + 3 e x = 0‬‬ ‫‪2) e x2−1 = e‬‬ ‫‪4) e2x − 2e x + 1 = 0‬‬ ‫‪( )5) x2 − 4 e x = (9x + 10) e x 6) ex − e−x = 0‬‬ ‫)‪7‬‬ ‫‪e3x‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫=‬ ‫‪0‬‬ ‫‪ex‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 7‬‬ ‫ﺍﻟﻤﺘﺭﺍﺠﺤﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫ﺤل ﻓﻲ‬ ‫‪1) xe2x − x2e2x ≤ 0‬‬ ‫‪2) e2x−1 ≤ 1‬‬ ‫‪3) e x2 −4 ≤ e7 x−16‬‬ ‫‪5) e2x − 2e x + 1 ≤ 0‬‬ ‫)‪4‬‬ ‫‪e x2−2‬‬ ‫≤‬ ‫‪1‬‬ ‫‪ex‬‬ ‫‪6) x2e4x + 4e x > 0‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 8‬‬‫ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻭ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻲ ﺘﻘﺒل ﻓﻴﻬﺎ ﺍﻻﺸﺘﻘﺎﻕ ﺜﻡ ﻋﻴﻥ ﺩﺍﻟﺘﻬﺎ ﺍﻟﻤﺸﺘﻘﺔ ‪f ′‬‬ ‫‪1) f ( x ) =e2x -e− x‬‬ ‫ﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫‪2) f ( x ) =xe x−1‬‬ ‫)‪3‬‬ ‫‪f‬‬ ‫(‬ ‫=)‪x‬‬ ‫‪ex‬‬ ‫‪+‬‬ ‫‪x‬‬ ‫)‪4‬‬ ‫‪f‬‬ ‫(‬ ‫)‪x‬‬ ‫=‬ ‫‪ex‬‬ ‫‪1‬‬ ‫‪ex‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪x2 −‬‬‫‪( )5) f ( x) = x2 − 4x + 5 e x‬‬ ‫‪6) f ( x) = ex − 1‬‬ ‫‪1‬‬ ‫‪8) f ( x ) = sin xecos x‬‬ ‫‪7) f ( x) = e x‬‬‫)‪9‬‬ ‫‪f‬‬ ‫(‬ ‫=)‪x‬‬ ‫‪ex‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪10) f ( x ) = e2x − 2e x + 1‬‬ ‫‪ex‬‬ ‫‪+‬‬ ‫‪1‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ‪. 9‬‬ ‫ﺍﺤﺴﺏ ﻨﻬﺎﻴﺎﺕ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻤﺎ ﻴﻠﻲ ﻋﻨﺩ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ ‪:‬‬ ‫‪1) f ( x) = x + e x‬‬ ‫‪2) f ( x) = e− x+4‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪3) f ( x) = e x2‬‬ ‫‪4) f ( x) = xe x‬‬ ‫)‪5‬‬ ‫= )‪f (x‬‬ ‫‪ex‬‬ ‫)‪6‬‬ ‫‪f‬‬ ‫(‬ ‫)‪x‬‬ ‫=‬ ‫‪e−x‬‬ ‫‪1‬‬ ‫‪x2‬‬ ‫‪ex +‬‬ ‫)‪7‬‬ ‫‪f‬‬ ‫(‬ ‫)‪x‬‬ ‫=‬ ‫‪e2x −‬‬ ‫‪1‬‬ ‫‪8) f ( x) = ( x - 1)ex‬‬ ‫‪x‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 10‬‬‫‪‬‬ ‫‪( x) = e2x‬‬ ‫‪−‬‬ ‫‪e‬‬ ‫‪x‬‬‫‪‬‬ ‫‪(0) = 0‬‬ ‫‪x‬‬‫‪‬‬ ‫‪f‬‬ ‫‪, x≠0‬‬‫‪‬‬ ‫‪f‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻭ ﺍﻟﻤﻌﺭﻓﺔ ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬‫‪‬‬ ‫‪ – 1‬ﺍﺩﺭﺱ ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻋﻨﺩ ‪. 0‬‬ ‫‪ – 2‬ﻋﻴﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻤﻥ ﺃﺠل ‪. x ≠ 0‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 11‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺤﻴﺙ ‪( ) ( )f x = 4 x + 3 e x :‬‬ ‫ﺍﺤﺴﺏ ﻜل ﻤﻥ ﺍﻟﻤﺸﺘﻘﺎﺕ ‪. f (4) , f (3) , f ′′ , f ′ :‬‬ ‫ﺜﻡ ﺍﺴﺘﻨﺘﺞ ﻋﺒﺎﺭﺓ ‪( ). f (n) x‬‬ ‫ﻭ ﺒﺭﻫﻥ ﻋﻠﻰ ﺍﻻﺴﺘﻨﺘﺎﺝ ﺒﺎﻟﺘﺭﺍﺠﻊ ﻤﻥ ﺃﺠل ‪. n ≥ 1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 12‬‬ ‫ﺃﺩﺭﺱ ﺍﻟﺘﻐﻴﺭﺍﺕ ﻟﻜل ﻤﻥ ﺍﻟﺩﻭﺍل ﺍﻟﻤﻌﺭﻓﺔ ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫‪1) f ( x) = xe x‬‬ ‫‪2) f ( x) = x − ex‬‬

‫)‪3‬‬ ‫= )‪f (x‬‬ ‫‪ex‬‬ ‫)‪4‬‬ ‫‪f‬‬ ‫(‬ ‫)‪x‬‬ ‫=‬ ‫‪ex‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪x‬‬ ‫‪ex‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 13‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺤﻴﺙ ‪( )f x = e x + e− x :‬‬‫‪ (1‬ﺍﺩﺭﺱ ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪ f‬؛ ﺜﻡ ﺍﺴﺘﻨﺘﺞ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪x‬‬ ‫ﻓﺈﻥ ‪. f ( x ) ≥ 2 :‬‬‫‪( )g‬‬‫‪x‬‬ ‫=‬ ‫‪ex‬‬ ‫‪1‬‬ ‫‪−1‬‬ ‫ﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﻌﺒﺎﺭﺓ ‪:‬‬ ‫‪g‬‬ ‫‪ (2‬ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ‬ ‫‪+ e−x‬‬ ‫‪ -‬ﺍﺩﺭﺱ ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪. g‬‬ ‫‪ -‬ﺒﺎﺴﺘﻌﻤﺎل ﺁﻟﺔ ﺒﻴﺎﻨﻴﺔ ﺃﻨﺸﺊ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ‪ c‬ﻟﻠﺩﺍﻟﺔ ‪( ). g‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 14‬‬‫‪ h (I‬ﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ﻜﻤﺎ ﻴﻠﻲ ‪( )h x = 1 − x − e− x :‬‬ ‫‪ (1‬ﺍﺩﺭﺱ ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪. h‬‬ ‫‪ (2‬ﺍﺴﺘﻨﺘﺞ ﺇﺸﺎﺭﺓ ‪ h x‬ﻋﻠﻰ ‪( ).‬‬ ‫‪( )f‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪ (II‬ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ ‪f‬‬ ‫‪= ex −1‬‬ ‫‪ (1‬ﺍﺤﺴﺏ ‪ f ′ x‬ﺜﻡ ﺍﺴﺘﻨﺘﺞ ﺇﺸﺎﺭﺘﻬﺎ ﻭ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ ‪( ). f‬‬ ‫‪ (2‬ﺍﺤﺴﺏ ﻨﻬﺎﻴﺎﺕ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻋﻨﺩ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻔﻬﺎ ‪ .‬ﺜﻡ ﺍﺴﺘﻨﺘﺞ ﺠﺩﻭل‬ ‫ﺘﻐﻴﺭﺍﺘﻬﺎ ‪.‬‬‫‪ (3‬ﺃﻨﺸﺊ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍل ‪ f‬ﻓﻲ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﺘﺠﺎﻨﺱ ‪( )O; i ; j‬‬ ‫ﺒﺎﺴﺘﻌﻤﺎل ﺁﻟﺔ ﺒﻴﺎﻨﻴﺔ ‪.‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ‪. 15‬‬‫‪( )f‬‬ ‫‪x‬‬ ‫‪2e x‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺤﻴﺙ ‪:‬‬ ‫‪= ex −1‬‬ ‫‪ (1‬ﺍﺩﺭﺱ ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪. f‬‬‫‪ (2‬ﻟﻴﻜﻥ ‪ C‬ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻓﻲ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﺘﺠﺎﻨﺱ ‪ . O; i , j‬ﺒﻴﻥ ﺃﻥ) ( ) (‬ ‫‪ C‬ﻴﻘﺒل ﺜﻼﺜﺔ ﻤﺴﺘﻘﻴﻤﺎﺕ ﻤﻘﺎﺭﺒﺔ ‪( ).‬‬‫‪ (3‬ﺒﻴﻥ ﺃﻥ ﺍﻟﻨﻘﻁﺔ ‪ w 0;1‬ﻤﺭﻜﺯ ﺘﻨﺎﻅﺭ ﻟﻠﻤﻨﺤﻨﻰ ‪ C‬ﺜﻡ ﺃﻨﺸﺌﻪ ‪( ) ( ).‬‬‫‪( )g‬‬‫‪x‬‬ ‫‪2e x‬‬ ‫‪ (4‬ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ ‪ g‬ﺤﻴﺙ ‪:‬‬ ‫‪= ex −1‬‬ ‫‪ -‬ﺍﻜﺘﺏ ‪ g x‬ﺩﻭﻥ ﺭﻤﺯ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ ‪( ).‬‬‫‪ -‬ﺃﻨﺸﺊ ‪ δ‬ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ ‪ g‬ﺒﺎﺴﺘﺨﺩﺍﻡ ‪( ) ( ). C‬‬‫‪ -‬ﻨﺎﻗﺵ ﺒﻴﺎﻨﻴﺎ ﺘﺒﻌﺎ ﻟﻘﻴﻡ ﺍﻟﻭﺴﻴﻁ ﺍﻟﺤﻘﻴﻘﻲ ‪ m‬ﻋﺩﺩ ﻭ ﺇﺸﺎﺭﺓ ﺤﻠﻭل‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ ﺫﺍﺕ ﺍﻟﻤﺠﻬﻭل ﺍﻟﺤﻘﻴﻘﻲ ‪ m‬ﺒﺤﻴﺙ ‪:‬‬ ‫‪( m − 3) e x − 1 = 2e x‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 16‬‬‫‪ f (I‬ﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﺒﺎﻟﻌﺒﺎﺭﺓ ‪( )( ) ( )C ، f x = 2 x2 − 3 x e x :‬‬ ‫ﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ﻓﻲ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﺘﺠﺎﻨﺱ ‪( ). O; i , j‬‬ ‫‪ (1‬ﺍﺩﺭﺱ ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪. f‬‬ ‫‪ (2‬ﺍﺩﺭﺱ ﺍﻟﻔﺭﻭﻉ ﺍﻟﻼﻨﻬﺎﺌﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ ‪( ). C‬‬ ‫‪ (3‬ﺒﻴﻥ ﺃﻥ ‪ C‬ﻴﻘﺒل ﻨﻘﻁﺘﻲ ﺍﻨﻌﻁﺎﻑ ﻴﻁﻠﺏ ﺘﻌﻴﻴﻥ ﻓﺎﺼﻠﺘﻴﻬﻤﺎ ‪( ).‬‬ ‫‪ (4‬ﻋﻴﻥ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ∆ ﻟﻠﻤﻨﺤﻨﻲ ‪ C‬ﻋﻨﺩ ﺍﻟﻨﻘﻁﺔ ﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺔ ‪( ) ( ). 0‬‬ ‫‪ (5‬ﺍﺭﺴﻡ ) ∆ ( ﻭ ) ‪. (C‬‬

‫‪ ( II‬ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ ‪ g‬ﺤﻴﺙ ‪( ) ( )g x = 2 x2 + ax + b e x :‬‬‫‪ (1‬ﻋﻴﻥ ﺍﻟﻌﺩﺩﻴﻥ ﺍﻟﺤﻘﻴﻘﻴﻴﻥ ‪ a‬ﻭ ‪ b‬ﺤﺘﻰ ﺘﻜﻭﻥ ﺍﻟﺩﺍﻟﺔ ‪ g‬ﺃﺼﻠﻴﺔ ﻟﻠﺩﺍﻟﺔ ‪. f‬‬ ‫‪ (2‬ﺍﻭﺠﺩ ﺍﻟﺸﺭﻁ ﺍﻟﺫﻱ ﻴﺤﻘﻘﻪ ‪ a‬ﻭ ‪ b‬ﺤﺘﻰ ﺘﻘﺒل ﺍﻟﺩﺍﻟﺔ ‪ g‬ﻗﻴﻤﺔ ﺤﺩﻴﺔ‬ ‫ﻜﺒﺭﻯ ﻭ ﺃﺨﺭﻯ ﺼﻐﺭﻯ ‪.‬‬

‫ﺍﻟﺤـﻠــــــﻭل‬ 1 ‫ﺍﻟﺘﻤﺭﻴﻥ‬ . √ (4 . × (3 . × (2 . × (1 . √ (8 √ . × (12 √ (7 . × (6 √ (5 . √ (16 . × . √ (20 . × (11 √ (10 . × (91) e3 x .e−2 x = e3 x−2 x = e x (15 √ (14 . × (13 (19 . × (18 √ (17 2 ‫ﺍﻟﺘﻤﺭﻴﻥ‬ : ‫ﺍﻟﺘﺒﺴﻴﻁ‬ 2) e x .e2 = e x+2 ( )3) 1 2 = 1 = e2x e− x e−2 x4) e4 x+2 = e4x+1.e−2x .e1 = e4 x−2 x+1 = e2x+2 e2x .e−1( )5)e−4 x+5 e−4 x+5 = e−4 x+5 .e−2 x+4 = e−6 x+9 ex−2 2 = e2x−4( )6) e x e− x + e2 x = e x .e− x + e x .e2 x = e x− x + e x+2 x = 1 + e3n 3 ‫ﺍﻟﺘﻤﺭﻴﻥ‬ ex − e−x = e2x −1 : ‫ﻨﺒﻴﻥ ﺃﻥ‬ ex + e−x e2x +1

‫‪ex − e−x‬‬ ‫‪ex‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪e2x − 1‬‬ ‫‪e2x −‬‬ ‫‪ex‬‬ ‫‪e2x‬‬‫‪ex + e−x‬‬ ‫‪ex‬‬ ‫‪+‬‬ ‫‪ex‬‬ ‫‪ex‬‬ ‫‪e2x +‬‬ ‫‪e2x‬‬ ‫=‬ ‫‪1‬‬ ‫=‬ ‫‪ex‬‬ ‫=‬ ‫‪1‬‬ ‫×‬ ‫=‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪ex‬‬ ‫‪e2x +‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪ex‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪4‬‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪( )f x = e x + e− x :‬‬ ‫ﻭﻤﻨﻪ ‪{ }Df = x ∈ : e x + e− x ≥ 0 :‬‬ ‫ﻭ ﻫﻲ ﻤﺤﻘﻘﺔ ﺩﻭﻤﺎ ﻷﻥ ‪ e x > 0‬ﻭ ‪D f = ، e− x > 0‬‬ ‫‪f‬‬ ‫(‬ ‫)‪x‬‬ ‫=‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪e2x +‬‬ ‫‪{ }Df = x ∈ : e2x + 1 ≠ 0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫= ‪. Df‬‬ ‫‪ e2 x + 1 = 0‬ﺘﻜﺎﻓﺊ ‪ e2 x = −1‬ﻭ ﻫﺫﺍ ﻤﺴﺘﺤﻴل ﺇﺫﻥ ‪:‬‬ ‫‪f (x) = 1‬‬ ‫‪ (3‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪ex + 4‬‬ ‫ﻭﻤﻨﻪ ‪{ }Df = x ∈ : e x + 4 > 0 :‬‬ ‫= ‪. Df‬‬ ‫‪ e x + 4 > 0‬ﻤﺤﻘﻘﺔ ﺩﻭﻤﺎ ﻷﻥ ‪ e x > 0‬ﻭﻤﻨﻪ ‪:‬‬ ‫‪f‬‬ ‫(‬ ‫)‪x‬‬ ‫=‬ ‫‪ex‬‬ ‫‪3‬‬ ‫‪ (4‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪ex +‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪{ }Df = x ∈ : e x + 3 ≠ 0‬‬ ‫= ‪. Df‬‬ ‫‪ e x + 3 = 0‬ﺘﻜﺎﻓﺊ ‪ e x = −3 :‬ﻭﻫﺫﺍ ﻤﺴﺘﺤﻴل ﻭﻤﻨﻪ ‪:‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪5‬‬ ‫ﺘﻌﻴﻴﻥ ﺤﺼﺭﺍ ﻟﻠﻌﺒﺎﺭﺓ ‪( )f x :‬‬ ‫ﻟﺩﻴﻨﺎ ‪ 0 ≤ x ≤ 1 :‬ﻭﻤﻨﻪ ‪e0 ≤ e x ≤ e1 :‬‬

‫‪1 11‬‬ ‫‪2≤ ex +1≤1+ e‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪1 + e ≤ e x + 1 ≤ 2 :‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫×‪0‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪e‬‬ ‫≤‬ ‫‪x‬‬ ‫×‬ ‫‪ex +‬‬ ‫‪1‬‬ ‫≤‬ ‫‪1(e‬‬ ‫‪+‬‬ ‫)‪1‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫ﺇﺫﻥ ‪. 0 ≤ f ( x ) ≤ 1 + e :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪6‬‬ ‫ﺤل ﺍﻟﻤﻌﺎﺩﻻﺕ ‪:‬‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪ e2 x−1 = 1 :‬ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪e2 x−1 = e0 :‬‬ ‫‪s‬‬ ‫=‬ ‫‪‬‬ ‫‪1‬‬ ‫‪‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪x‬‬ ‫=‬ ‫‪1‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫ﻭﻋﻠﻴﻪ ‪2 x − 1 = 0 :‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪ e2 x−1 = e :‬ﻭ ﻫﻲ ﺘﻜﺎﻓﺊ ‪ x2 − 1 = 1 :‬ﻭﻤﻨﻪ ‪x2 = 2 :‬‬ ‫ﺇﺫﻥ ‪ x = 2 :‬ﺃﻭ ‪ x = − 2‬ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪{ }s = − 2; 2‬‬ ‫‪ (3‬ﻟﺩﻴﻨﺎ ‪( )x2 − 4 x + 3 x = 0 :‬‬ ‫ﻭ ﻫﻲ ﺘﻜﺎﻓﺊ ‪∆′ = 4 , x2 − 4 x + 3 = 0 :‬‬ ‫ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪ x2 = 3 , x1 = 1 :‬ﺇﺫﻥ ‪{ }s = 1;3 :‬‬ ‫‪ (4‬ﻟﺩﻴﻨﺎ ‪e2 x − 2e x + 1 = 0 :‬‬‫‪( y − 1)2 = 0‬‬ ‫‪ y2 − 2y + 1 = 0‬‬ ‫= ‪ e x‬ﻨﺠﺩ ‪:‬‬ ‫‪y‬‬‫‪‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪‬‬ ‫ﺒﻭﻀﻊ‬‫‪‬‬ ‫‪y‬‬ ‫=‬ ‫‪e‬‬ ‫‪x‬‬ ‫‪‬‬ ‫‪y‬‬ ‫=‬ ‫‪e‬‬ ‫‪x‬‬ ‫= ‪{ }s‬‬ ‫‪0‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪x=0‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪y=1‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪y=1‬‬ ‫‪e x = e0‬‬ ‫ﻭ ﻤﻨﻪ ‪e x = 1 :‬‬ ‫‪ (5‬ﻟﺩﻴﻨﺎ ‪( ) ( )x2 − 4 e x = 9 x − 18 e x :‬‬ ‫ﻭ ﻫﻲ ﺘﻜﺎﻓﺊ ‪ x2 − 4 = 9 x − 18 :‬ﻭ ﻤﻨﻪ ‪x2 − 9 x + 14 = 0 :‬‬‫‪ ∆ = 25‬ﻭ ﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪ x1 = 2‬ﻭ ‪ x2 = 7‬ﺇﺫﻥ ‪{ }s = 2;7 :‬‬

‫‪ex‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪=0‬‬ ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪ex − ex = 0‬‬ ‫‪ (6‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪ex‬‬‫‪e2x = 1‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪e2x − 1 = 0‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪e2x − 1‬‬ ‫‪=0‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪ex‬‬‫ﺃﻱ ‪ e2 x = e0 :‬ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ 2 x = 0 :‬ﻭﻤﻨﻪ ‪ x = 0 :‬ﺇﺫﻥ ‪{ }s 0 :‬‬‫‪e4x −‬‬ ‫‪1‬‬ ‫=‬ ‫‪0‬‬ ‫ﻭ ﻫﻲ ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪e3x‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪=0‬‬ ‫‪ (7‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪ex‬‬ ‫‪ex‬‬‫ﻭﻤﻨﻪ ‪ e4 x − 1 = 0 :‬ﺇﺫﻥ ‪ e4 x = 1 :‬ﻭﻋﻠﻴﻪ ‪4 x = 0 :‬‬ ‫ﻭ ﻤﻨﻪ ‪ x = 0 :‬ﺇﺫﻥ ‪{ }s 0 :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪7‬‬ ‫ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺎﺕ ‪:‬‬‫‪ (1‬ﻟﺩﻴﻨﺎ ‪ xe2 x − x2e2 x ≤ 0 :‬ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪( )e2 x x − x2 ≤ 0 :‬‬‫ﻭﻋﻠﻴﻪ ‪ x − x2 ≤ 0 :‬ﻷﻥ ‪ e2 x > 0 :‬ﺇﺫﻥ ‪( )x − x + 1 ≤ 0 :‬‬‫ﻭﻋﻠﻴﻪ ‪ x ∈ ]−∞;0] ∪ [1;+∞[ :‬ﺇﺫﻥ ‪s ∈ ]−∞;0] ∪ [1;+∞[ :‬‬ ‫‪x‬‬ ‫∞‪−‬‬ ‫‪01‬‬ ‫∞‪+‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪+‬‬ ‫‪+‬‬ ‫‪+‬‬ ‫‪+‬‬ ‫‪-‬‬ ‫‪−x + 1‬‬ ‫‪-‬‬ ‫‪+‬‬ ‫‪-‬‬‫)‪x(− x + 1‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪e2 x−1 ≤ 1 :‬‬ ‫ﻭ ﻫﻲ ﺘﻜﺎﻓﺊ ‪ e2 x−1 ≤ e0 :‬ﻭ ﻤﻨﻪ ‪2 x − 1 ≤ 0 :‬‬

‫‪s‬‬ ‫‪‬‬ ‫;∞‪−‬‬ ‫‪1‬‬ ‫‪‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪x‬‬ ‫≤‬ ‫‪1‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2‬‬‫‪ (3‬ﻟﺩﻴﻨﺎ ‪ e x2 −4 ≤ e7 x−16 :‬ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪x2 − 4 ≤ 7 x − 16 :‬‬ ‫ﻭﻋﻠﻴﻪ ‪x2 − 7 x + 12 ≤ 0 :‬‬ ‫ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ‪∆ = 1 ، x2 − 7 x + 12 :‬‬ ‫ﻴﻭﺠﺩ ﺠﺫﺭﺍﻥ ‪x2 = 4 ، x1 = 3 :‬‬‫∞‪x −‬‬ ‫‪3‬‬ ‫∞‪4 +‬‬ ‫‪-‬‬ ‫‪-‬‬‫‪x2 − 7 x + 12‬‬ ‫‪+‬‬ ‫ﺇﺫﻥ ‪S = [3 ; 4] :‬‬ ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪e x2 −2 .e x ≤ 1 :‬‬ ‫‪e x2−2‬‬ ‫≤‬ ‫‪1‬‬ ‫‪ (4‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪ex‬‬ ‫ﻭﻋﻠﻴﻪ ‪ e x2 + x−2 ≤ 1 :‬ﺇﺫﻥ ‪e x2 + x−2 ≤ e1 :‬‬‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ x2 + x − 2 ≤ 1 :‬ﺇﺫﻥ ‪x2 + x − 3 ≤ 0 :‬‬ ‫ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ‪∆ = 13 ، x2 + x − 3 :‬‬‫= ‪x2‬‬ ‫‪−1 +‬‬ ‫‪13‬‬ ‫‪،‬‬ ‫‪x1‬‬ ‫=‬ ‫‪−1‬‬ ‫‪−‬‬ ‫‪13‬‬ ‫ﻴﻭﺠﺩ ﺠﺫﺭﺍﻥ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬‫‪x −∞ x1‬‬ ‫∞‪x2 +‬‬‫‪x2 + x − 3‬‬ ‫‪+‬‬ ‫‪-+‬‬ ‫ﺇﺫﻥ ‪S =  x1; x2  :‬‬ ‫‪ (5‬ﻟﺩﻴﻨﺎ ‪e2 x − 2e x + 1 ≤ 0 :‬‬

‫‪ y2 − 2y + 1≤ 0‬‬ ‫ﻨﺠﺩ ‪:‬‬ ‫ﺒﻭﻀﻊ ‪e x = y :‬‬ ‫‪e x = y‬‬ ‫‪y −1= 0‬‬ ‫‪( y − 1)2 ≤ 0‬‬ ‫‪‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪:‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪e‬‬ ‫‪x‬‬ ‫=‬ ‫‪y‬‬ ‫‪‬‬ ‫‪e‬‬ ‫‪x‬‬ ‫=‬ ‫‪y‬‬ ‫ﺇﺫﻥ ‪ e x = 1 :‬ﻭﻋﻠﻴﻪ ‪ x = 0 :‬ﺇﺫﻥ ‪{ }S = 0 :‬‬‫‪ (6‬ﻟﺩﻴﻨﺎ ‪ x2e4 x + 4e x > 0 :‬ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪( )e x x2e3 x + 4 > 0‬‬ ‫ﻭﻫﻲ ﻤﺤﻘﻘﺔ ﺩﻭﻤﺎ ﻷﻥ ‪ e x > 0‬ﻭ ‪ e3 x > 0‬ﻭ ‪x2 > 0‬‬ ‫ﻭﻋﻠﻴﻪ ‪S = :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪8‬‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪( )f x = e2 x − e− x :‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﻌﺭﻓﺔ ﻋﻠﻰ ﻭ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪.‬‬ ‫ﺤﻴﺙ ‪( ). f ′ x = 2e2x + e− x :‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪( )f x = xe x−1 :‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﻌﺭﻓﺔ ﻋﻠﻰ ﻭ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪.‬‬‫ﺤﻴﺙ ‪ ( )f ′ x = 1.e x−1 + xe x−1 :‬ﻭﻤﻨﻪ ‪f ′( x) = (1 + )x e x−1 :‬‬ ‫‪( )f‬‬ ‫‪x‬‬ ‫=‬ ‫‪ex‬‬ ‫‪+‬‬ ‫‪x‬‬ ‫‪ (3‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪ex‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﻌﺭﻓﺔ ﻋﻠﻰ * ﻭ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ * ‪.‬‬ ‫‪( )( ( ) ) ( )f ′( x) = ex + 1‬‬ ‫‪ex −1 − ex ex + x‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪ex −1 2‬‬ ‫‪-1-xe x‬‬ ‫‪e2 x -1-e2 x −‬‬ ‫‪xe x‬‬ ‫‪= e x -1 2‬‬ ‫‪e x -1 2‬‬‫‪( ) ( )( ) ( )f ′‬‬‫‪x‬‬ ‫‪x‬‬ ‫‪ f ′‬ﻭﻋﻠﻴﻪ ‪:‬‬ ‫=‬ ‫ﻭﻤﻨﻪ ‪:‬‬

‫‪( )f‬‬ ‫‪x‬‬ ‫‪ex‬‬ ‫‪ (4‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪= x2 − 1‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﻌﺭﻓﺔ ﻭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪{ }. −1;1‬‬ ‫‪( )ex‬‬ ‫= )‪( )f ′( x‬‬ ‫‪x2 − 1 − 2 x.e x‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪x2 − 1 2‬‬ ‫= )‪( ( ) )f ′( x‬‬ ‫‪x2 − 2x − 1 ex‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪x2 − 1 2‬‬‫‪ (5‬ﻟﺩﻴﻨﺎ ‪ f x = x2 − 4 x + 5 e x :‬ﺤﻴﺙ ‪( )( )Df = :‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﻌﺭﻓﺔ ﻭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻠﻰ ﺤﻴﺙ ‪:‬‬ ‫‪( )f ′( x) = ( 2x − 4) e3x + x2 − 4x + 5 e x‬‬ ‫ﻭﻋﻠﻴﻪ ‪( )f ′( x ) = 2x − 4 + x2 − 4x + 5 e x :‬‬ ‫‪( )f ′( x) = x2 − 2x + 1 e x‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬‫‪ (6‬ﻟﺩﻴﻨﺎ ‪ f x = ex − 1 :‬ﺤﻴﺙ ‪{ } ( )Df = x ∈ : ex − 1 ≥ 0 :‬‬‫ﻭﻤﻨﻪ ‪ e x − 1 ≥ 0 :‬ﻭﻋﻠﻴﻪ ‪ e x ≥ 1 :‬ﻭﺒﺎﻟﺘﺎﻟﻲ ‪x ≥ 0 :‬‬ ‫ﺇﺫﻥ ‪Df = [0;+∞[ :‬‬ ‫‪ex‬‬ ‫ﺤﻴﺙ ‪e x − 1 :‬‬‫‪( ) ] [f ′‬‬‫‪x‬‬ ‫=‬ ‫∞‪0;+‬‬ ‫ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻠﻰ‬ ‫‪f‬‬ ‫ﺍﻟﺩﺍﻟﺔ‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪ (7‬ﻟﺩﻴﻨﺎ ‪( )f x = e x :‬‬‫= )‪f ′(x‬‬ ‫‪−1‬‬ ‫‪e‬‬ ‫‪1‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﻌﺭﻓﺔ ﻭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻠﻰ *‬ ‫‪x2‬‬ ‫‪x‬‬

( )f x = sin x.ecos x : ‫( ﻟﺩﻴﻨﺎ‬8 : ‫ ﻤﻌﺭﻓﺔ ﻭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻠﻰ ﺤﻴﺙ‬f ‫ﺍﻟﺩﺍﻟﺔ‬ f ′( x ) = cos x.ecos x + sin x ( − sin )x ecos x ( )f ′( x ) = ecos x cos x − sin2 x : ‫ﻭﻤﻨﻪ‬ ( )f ′( x ) = ecos x cos x − 1 − cos2 x  : ‫ﻭﻋﻠﻴﻪ‬ ( )( )f ′ x = cos2 x + cos x − 1 ecos x : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ ( )fx = ex − 1 : ‫( ﻟﺩﻴﻨﺎ‬9 ex + 1 : ‫ ﻤﻌﺭﻓﺔ ﻭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻠﻰ ﺤﻴﺙ‬f ‫ﺍﻟﺩﺍﻟﺔ‬( ( ) ( ) ) ( )( )f ′ex exx = ex +1 − ex −1 = e2x + ex − e2x + ex ex +1 2 ex −1 2 2e x ( )= e x + 1 2 ( )f x = e2x − 2e x + 1 : ‫( ﻟﺩﻴﻨﺎ‬10 { }Df = x ∈ : e2x − 2e x + 1 ≥ 0 : ‫ﺤﻴﺙ‬ y = e x : ‫ ﺒﻭﻀﻊ‬e2 x − 2e x + 1 ≥ 0 ‫ﻨﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ‬ ( )y − 1 2 ≥ 0 : ‫ ﻭ ﻋﻠﻴﻪ‬y2 − 2 y + 1 ≥ 0 : ‫ﻨﺠﺩ‬ ( ). D f = : ‫ ﺇﺫﻥ‬. ‫ ﻭ ﻫﻲ ﻤﺤﻘﻘﺔ ﺩﻭﻤﺎ‬e x − 1 2 ≥ 0 : ‫ﺃﻱ ﺃﻥ‬ : ‫ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ * ﺤﻴﺙ‬f ‫ﺍﻟﺩﺍﻟﺔ‬ f ′( x ) = 2e2x − 2e x = e2x − e x 2 e2 x − 2e x + 1 e2 x − 2e x + 1

9 ‫ﺍﻟﺘﻤﺭﻴﻥ‬] [ ( )D f = −∞;+∞ : ‫ ﺤﻴﺙ‬f x = x + e x : ‫( ﻟﺩﻴﻨﺎ‬1( )( )lim fx→−∞ x = lim x + e x = −∞ : ‫ﻭﻤﻨﻪ‬ x→−∞( )( )lim fx→+∞ x = lim x + e x = +∞ : ‫ﻭﻋﻠﻴﻪ‬ x→+∞] [ ( )D f = −∞;+∞ : ‫ ﺤﻴﺙ‬f x = e− x+4 : ‫( ﻟﺩﻴﻨﺎ‬2( )lim f x = lim e− x+4 = +∞ : ‫ﻭﻤﻨﻪ‬x→−∞ x→−∞( )lim f x = lim e− x+4 = 0 : ‫ﻭﻋﻠﻴﻪ‬x→+∞ x→+∞ 1Df = ]−∞;0[ ∪ ]0;+∞[ : ‫ ﺤﻴﺙ‬f ( x ) = e x2 : ‫( ﻟﺩﻴﻨﺎ‬3 1 lim f ( x ) = lim e x2 = 1 : ‫ﻭﻤﻨﻪ‬ x→−∞ x→−∞ 1 lim f ( x ) = lim e x2 = +∞ : ‫ﻭﻋﻠﻴﻪ‬ << x→0 x→0 1lim f ( x ) = lim e x2 = +∞ : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ >>x→0 x→0 1 lim f ( x ) = lim e x2 = 1 : ‫ﺇﺫﻥ‬ x→+∞ x→+∞ 1Df = ]−∞;0[ ∪ ]0;+∞[ : ‫ ﺤﻴﺙ‬f ( x ) = xe x : ‫( ﻟﺩﻴﻨﺎ‬4 1( )lim f x = lim xe x = −∞ : ‫ﻭﻤﻨﻪ‬x→−∞ x→−∞ 1lim f ( x) = lim xe x = 0 : ‫ﻭﻋﻠﻴﻪ‬ < <x→0 x→0

1 1( )lim fx = lim xe x = lim ex = lim et = +∞ : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ > > > 1 t x→0 x→+∞x→0 x→0 x ( )lim fx 1 : ‫ﺇﺫﻥ‬ .( 1 = t : ‫) ﻨﻀﻊ‬ x x→+∞ = lim xe x = +∞ x→+∞Df = ]−∞ ; 0[ ∪ ]0 ; + ∞[ : ‫ﺤﻴﺙ‬ f (x) = ex : ‫( ﻟﺩﻴﻨﺎ‬5 x2 ( )lim f x = lim ex = 0 x2 x→−∞ x→−∞ ( )lim fx = lim ex = +∞ < < x2 x→0 x→0 ( )lim fx = lim ex = +∞ > > x2 x→0 x→0 ( )lim fx = lim ex = +∞ x2 x→+∞ x→+∞ ] [ ( )Df = −∞;+∞ f x e−x : ‫( ﻟﺩﻴﻨﺎ‬6 : ‫ﺤﻴﺙ‬ = ex +1 ( )lim f x = lim e− x 1 = +∞ ex + x→−∞ x→−∞ ( )lim f x = lim e− x = 0 ex +1 x→+∞ x→+∞Df = ]−∞ ; 0[ ∪ ]0 ; [+ ∞ : ‫ﺤﻴﺙ‬ f ( x) = e2x − 1 : ‫( ﻟﺩﻴﻨﺎ‬7 x ( )lim f x = lim e2x − 1 = 0 x x→−∞ x→−∞

( )lim f x = lim e2 x− 1 = lim 2 × e2x − 1 = 2 < < x < 2 xx→0 x→0 x→0( )lim f x = lim e2 x− 1 = lim 2 × e2x − 1 = 2 > > x > 2 xx→0 x→0 x→0 e 2 x− 1 e2x 1 x x x( )lim f x = lim = lim −x→+∞ x→+∞ x→+∞ lim 2 e2x − 1 = +∞ x→+∞ 2 x xDf = ]−∞;+∞[ : ‫ ﺤﻴﺙ‬f ( x ) = ( x − 1) e x : ‫( ﻟﺩﻴﻨﺎ‬8lim f ( x) = lim ( x − 1)e x = lim xe x − e x = 0x→−∞ x→−∞ x→−∞ lim f ( x) = lim ( x − 1)ex = +∞x→+∞ x→+∞ 10 ‫ﺍﻟﺘﻤﺭﻴﻥ‬ Df = : 0 ‫ ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬-1( ) ( )lim fx −f 0 e2x − ex e2x − x < x−0 x2 = lim x = lim ex→0 < x < x→0 x→0 = lim ex × ex − 1 = −∞ < x x x→0 . ‫ ﻤﻥ ﺍﻟﻴﺴﺎﺭ‬0 ‫ ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬f ‫ﻭ ﻤﻨﻪ‬x −f( ) ( )lim f0 e 2 x− e x lim e x ex −1 x−0> x2 x> x = lim = × = +∞x→0 > x→0 x→0. 0 ‫ ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ‬f ‫ ﻤﻥ ﺍﻟﻴﻤﻴﻥ ﻭ ﺒﺎﻟﺘﺎﻟﻲ‬0 ‫ ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ‬f ‫ﻭ ﻤﻨﻪ‬ : x ≠ 0 ‫ ﺘﻌﻴﻴﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻤﻥ ﺃﺠل‬-2

( ) ( )f ′( x) = 2e2 x − e x x − 1 e2x − ex x2f ′( x) = 2 xe2 x − xe x − e2x + ex x2 ( 2 x − 1) e 2x − ( x − 1) e x x2f ′ ( x ) = 11 ‫ﺍﻟﺘﻤﺭﻴﻥ‬ f ( x) = (4x + 3)exf ′( x) = 4.e x + (4x + 3) e x = (4 + 4x + 3) e x f ′( x) = (4x + 7)exf ′′( x) = 4.e x + (4x + 7) e x = (4 + 4x + 7) e x f ′′( x) = (4x + 11)e xf (3) ( x ) = 4.e x + (4x + 11) e x = (4 + 4x + 11) e x f (3) ( x) = (4x + 15) e xf (4) ( x ) = 4.e x + (4x + 15) e x = (4 + 4x + 15) e x f (4) ( x) = (4x + 19) e x : f (n) ‫ﺍﺴﺘﻨﺘﺎﺝ ﻋﺒﺎﺭﺓ‬ f (n) ( x) = (4x + 3 + 4n)ex ( ) ( )f (n) x = 4 x + 3 + 4n e x : p(n) ‫ﺍﻟﺒﺭﻫﺎﻥ ﺒﺎﻟﺘﺭﺍﺠﻊ ﻋﻠﻰ ﺼﺤﺔ‬ ( ) ( ). ‫ ﻤﺤﻘﻘﺔ‬p(1) ‫ ﻭ ﻤﻨﻪ‬f (n) x = 4 x + 7 e x : n = 1 ‫ﻤﻥ ﺃﺠل‬

( ) ( )p k + 1 ‫ ﻭ ﻨﺒﺭﻫﻥ ﺼﺤﺔ‬p k ‫ﻨﻔﺭﺽ ﺼﺤﺔ‬ p(k ) : f (k) ( x) = (4x + 3 + 4k )ex p( k + 1) : f (k+1) ( x) = 4x + 3 + 4( k + 1) e x( )f (k+1) ( x) = f k ′ ( x) = 4.e x + (4x + 3 + 4k ) e x : ‫ﻟﺩﻴﻨﺎ‬ = (4x + 3 + 4k + 4) e x = 4x + 3 + 4( k + 1) e x ( ) ( )‫ ﺼﺤﻴﺤﺔ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ‬p n ‫ ﺼﺤﻴﺤﺔ ﻭ ﻋﻠﻴﻪ‬p k + 1 ‫ﻭﻤﻨﻪ‬ . n ≥ 1 ‫ ﺤﻴﺙ‬n ‫ﻁﺒﻴﻌﻲ‬ 12 ‫ﺍﻟﺘﻤﺭﻴﻥ‬ : f ‫ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ‬ ] [ ( )Df = −∞;+∞ : ‫ ﺤﻴﺙ‬f x = xe x : ‫( ﻟﺩﻴﻨﺎ‬1lim f ( x) = lim xe x = 0x→−∞ x→−∞lim f ( x) = lim xe x = +∞x→+∞ x→+∞ f ′( x) = e x + xe x = e x ( x + 1) ( ): ‫ ﻭ ﻟﺩﻴﻨﺎ‬x + 1 ‫ ﻟﻪ ﻨﻔﺱ ﺇﺸﺎﺭﺓ‬f ′ x ‫ﻭ ﻋﻠﻴﻪ‬ x −∞ −1 +∞ x+1 - +

‫ﺇﺫﻥ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ∞‪ −1;+‬ﻭ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ[ [‬ ‫ﻋﻠﻰ ‪ −∞;−1‬؛ ﻭ ﻋﻠﻴﻪ ﺠﺩﻭل ﺍﻟﺘﻐﻴﺭﺍﺕ ﻜﻤﺎ ﻴﻠﻲ ‪] ]:‬‬ ‫‪x‬‬ ‫∞‪−‬‬ ‫‪−1‬‬ ‫∞‪+‬‬ ‫‪+‬‬‫)‪f ′( x‬‬ ‫‪-‬‬ ‫‪−1‬‬ ‫‪0‬‬ ‫‪e‬‬ ‫∞‪+‬‬‫)‪f (x‬‬‫‪ (2‬ﻟﺩﻴﻨﺎ ‪ f x = x − e x :‬ﺤﻴﺙ ‪] ] ( )Df = −∞ ; + ∞ :‬‬‫∞‪lim f ( x) = lim x − e x = −‬‬‫∞‪x→−‬‬ ‫∞‪x→−‬‬‫‪( )lim f‬‬ ‫‪x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪x‬‬ ‫‪−‬‬ ‫‪ex‬‬ ‫=‬ ‫‪lim‬‬ ‫‪x‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪ex‬‬ ‫‪‬‬ ‫=‬ ‫∞‪−‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪‬‬‫∞‪x→+‬‬ ‫∞‪x→+‬‬ ‫∞‪x→+‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪f ′(x) = 1− ex‬‬‫‪ f ′ x = 0‬ﺘﻜﺎﻓﺊ ‪ 1 − e x = 0 :‬ﻭﻤﻨﻪ ‪ e x = 1 :‬ﻭﻋﻠﻴﻪ ‪( )x = 0 :‬‬‫‪ f ′ x > 0‬ﺘﻜﺎﻓﺊ ‪ 1 − e x > 0 :‬ﻭﻤﻨﻪ ‪ e x < 1 :‬ﺇﺫﻥ ‪( )x < 0 :‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ‪] ]. −∞;0‬‬‫‪ f ′ x < 0‬ﺘﻜﺎﻓﺊ ‪ x > 0‬ﻭﻤﻨﻪ ‪ f‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ‪] ] ( ). −∞;0‬‬

x −∞ 0 +∞ + -f ′( x)f ( x) −1 −∞ −∞Df = ]−∞;0[ ∪ ]0;+∞[ : ‫ﺤﻴﺙ‬ f ( x) = ex : ‫( ﻟﺩﻴﻨﺎ‬3 x ( )lim f x = lim ex =0 x x→−∞ x→−∞ ( )lim f x = lim ex = −∞ < < x x→0 x→0 ( )lim f x = lim ex = +∞ > > x x→0 x→0 ( )lim f x = lim ex = +∞ x x→+∞ x→+∞ f ′( x) = ex.x − ex = ex ( x − 1) x2 x2( ) ( )x = 1 : ‫ ﺘﻜﺎﻓﺊ‬f ′ x = 0 : ‫ ﺇﺫﻥ‬x − 1 ‫ ﻟﻪ ﻨﻔﺱ ﺇﺸﺎﺭﺓ‬f ′ x[ [ ( )1;+∞ ‫ ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ‬f : ‫ ﺇﺫﻥ‬x > 1 ‫ ﺘﻜﺎﻓﺊ‬f ′ x > 0 ] [ [ [. 0 ; 1 ‫ ﻭﻋﻠﻰ‬−∞;0 ‫ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ‬f ‫ﻭ ﻜﺫﻟﻙ‬

x −∞ 0 1 +∞ - - +f ′( x)f (x) 0 +∞ −∞ −∞ e{ } ( )Df = x ∈ ex + 1 :ex −1≠ 0 : ‫ﺤﻴﺙ‬ f x = ex − 1 : ‫( ﻟﺩﻴﻨﺎ‬4 . x = 0 : ‫ ﻭ ﻋﻠﻴﻪ‬e x = 1 ‫ ﺘﻜﺎﻓﺊ‬e x − 1 = 0 Df = ]−∞ ; 0[ ∪ ]0 ; + ∞[ :‫ﺇﺫﻥ‬( )• ex + 1lim f x = lim ex − 1 = −1x→−∞ x→−∞( )• ex + 1lim f x = lim ex − 1 = −∞ < <x→0 x→0 e x + 1 → 2  : ‫ﻷﻥ‬  e x − 1 <→( )• x lim ex + 1lim f = > ex − 1 = +∞ > x→0x→0 e x + 1 → 2  : ‫ﻷﻥ‬  ex − 1 >→

ex + 1 ex  1 + 1  ex − 1 ex  − ex ( )• 1 lim f x = lim = lim  ex  x →+∞ x →+∞ x→+∞ 1 = lim 1 + 1 =1 1 − ex x→+∞ 1 ex( ( ) )( )• ex ex −1 − ex ex +1 ex −1 2f ′(x) = ( )ex ( ) ( )= ex −1− ex −1 = −2e x ex −1 2 ex −1 2 f ′( x ) < 0 : ‫ﻭ ﻤﻨﻪ‬ ] [ ] [: 0;+∞ ‫ ﻭ‬−∞;0 ‫ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ‬f ‫ﺍﻟﺩﺍﻟﺔ‬ x −∞ 0 +∞ - -f ′( x) −1 +∞ 1f (x) −∞ • Df = ]−∞ ; + ∞[ 13 ‫ﺍﻟﺘﻤﺭﻴﻥ‬ : f ‫ ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ‬-1

‫∞‪• lim f ( x) = +‬‬ ‫∞‪• lim f ( x) = +‬‬ ‫∞‪x→−‬‬ ‫∞‪x→+‬‬‫‪• f ′( x) = ex − e−x‬‬‫‪ f ′ x = 0‬ﺘﻜﺎﻓﺊ ‪ e x − e− x = 0 :‬ﻭﻤﻨﻪ ‪( )e x = e− x :‬‬ ‫ﻭ ﻋﻠﻴﻪ‪ x = − x :‬ﻭ ﻤﻨﻪ‪ 2 x = 0 :‬ﺃﻱ‪. x = 0 :‬‬ ‫‪ f ′ x > 0‬ﺘﻜﺎﻓﺊ ‪ e x − e− x > 0 :‬ﻭﻤﻨﻪ ‪( )e x > e− x :‬‬ ‫ﺇﺫﻥ ‪ x > − x :‬ﻭﻤﻨﻪ ‪ 2 x > 0 :‬ﻭﻋﻠﻴﻪ ‪x > 0 :‬‬ ‫ﺇﺫﻥ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ∞ ‪ 0 ; +‬ﻭ ﺒﺎﻟﺘﺎﻟﻲ ﻓﻬﻲ ﻤﺘﻨﺎﻗﺼﺔ[ [‬ ‫ﺘﻤﺎﻤﺎ ﻋﻠﻰ ‪] ]. −∞;0‬‬‫∞‪x −∞ 0 +‬‬‫)‪f ′( x‬‬ ‫‪-‬‬ ‫‪+‬‬‫∞‪f ( x) +∞ 2 +‬‬‫ﻭ ﻋﻠﻴﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ‬ ‫ﺍﻻﺴﺘﻨﺘﺎﺝ ‪:‬‬ ‫ﻤﻥ ﺠﺩﻭل ﺍﻟﺘﻐﻴﺭﺍﺕ ﺍﻟﻘﻴﻤﺔ ‪ 2‬ﻫﻲ ﻗﻴﻤﺔ ﺤﺩﻴﺔ ﺼﻐﺭﻯ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻋﻠﻰ‬ ‫ﺤﻘﻴﻘﻲ ‪ x‬ﻓﺈﻥ ‪( )f x ≥ 2 :‬‬ ‫‪ -2‬ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ ‪: g‬‬‫‪{ }• Dg = x ∈ : e x + e x − 1 ≠ 0‬‬ ‫ﻨﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‪ e x + e− x − 1 = 0 :‬ﺃﻱ‪e x + e− x = 1 :‬‬ ‫ﺃﻱ‪ f x = 1 :‬ﻭ ﻫﺫﺍ ﻤﺴﺘﺤﻴل ﻷﻥ ‪( ) ( )f x ≥ 2 :‬‬ ‫ﺇﺫﻥ ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭل ﻭ ﻋﻠﻴﻪ ‪] [. D f = −∞;+∞ :‬‬‫‪• lim g ( x) = 0‬‬ ‫‪• lim g ( x) = 0‬‬ ‫∞‪x→−‬‬ ‫∞‪x→+‬‬

‫= )‪g′( x‬‬ ‫‪−ex + e−x‬‬ ‫‪ex + e−x − 1 2‬‬‫•) (‬‫‪ g′( x ) = 0‬ﺘﻜﺎﻓﺊ ‪ −e x + e− x = 0 :‬ﺃﻱ‪e− x = e x :‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪ − x = x :‬ﻭ ﻤﻨﻪ ‪ 2 x = 0 :‬ﺃﻱ ‪x = 0 :‬‬ ‫‪ g′ x = 0‬ﺘﻜﺎﻓﺊ ‪ −e x + e− x > 0 :‬ﺃﻱ‪( )e− x > e x :‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪ − x > x :‬ﻭﻤﻨﻪ ‪ 2 x > 0 :‬ﺃﻱ ‪x > 0 :‬‬ ‫ﺇﺫﻥ ‪ g‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ‪ −∞;0‬ﻭ ﻋﻠﻴﻪ ﻓﻬﻲ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻡ] ]‬ ‫ﻋﻠﻰ ]∞‪. [0;+‬‬ ‫∞‪x −‬‬ ‫‪0‬‬ ‫∞‪+‬‬‫)‪g′( x‬‬ ‫‪+‬‬ ‫‪-‬‬‫‪g(x) 0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪g(0) = 1‬‬ ‫ﺇﻨﺸﺎﺀ ‪( ): c‬‬ ‫‪y‬‬ ‫‪1,5‬‬ ‫‪1‬‬ ‫‪0,5‬‬‫‪-2,5 -2 -1,5 -1 -0,5 0‬‬ ‫‪0,5 1 1,5 2 2,5 x‬‬ ‫‪-0,5‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ ‪14‬‬ ‫‪ -1 (I‬ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ ‪] [Dh = −∞;+∞ : h‬‬‫(‪lim h‬‬ ‫)‪x‬‬ ‫=‬ ‫‪lim 1−‬‬ ‫‪x‬‬ ‫‪−‬‬ ‫‪e− x‬‬ ‫=‬ ‫)‪lim (−x‬‬ ‫‪ −1‬‬ ‫‪+‬‬ ‫‪1−‬‬ ‫‪e− x‬‬ ‫‪‬‬ ‫=‬ ‫∞‪−‬‬ ‫‪‬‬ ‫‪−x‬‬ ‫‪‬‬‫∞‪x→−‬‬ ‫∞‪x→−‬‬ ‫∞‪x→−‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪‬‬ ‫∞‪lim h( x) = lim 1 − x − e− x = −‬‬ ‫∞‪x→+‬‬ ‫∞‪x→+‬‬ ‫‪h′( x ) = −1 + e− x‬‬ ‫‪ h′ x = 0‬ﺘﻜﺎﻓﺊ ‪ −1 + e− x = 0 :‬ﻭﻤﻨﻪ ‪( )e− x = 1 :‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪ e− x = e0 :‬ﺃﻱ ‪x = 0 :‬‬ ‫‪ h′ x > 0‬ﺘﻜﺎﻓﺊ ‪ −1 + e− x > 0 :‬ﻭﻋﻠﻴﻪ ‪( )e− x > 1 :‬‬ ‫ﺃﻱ ﺃﻥ ‪ − x > 0 :‬ﻭ ﻤﻨﻪ ‪x < 0 :‬‬ ‫ﺇﺫﻥ ‪ h‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ‪ −∞;0‬ﻭ ﻋﻠﻴﻪ ﻓﻬﻲ ﻤﺘﻨﺎﻗﺼﺔ] ]‬ ‫ﺘﻤﺎﻤﺎ ﻋﻠﻰ ∞‪[ [. 0;+‬‬ ‫∞‪x −∞ 0 +‬‬ ‫)‪h′( x‬‬ ‫‪+‬‬ ‫‪-‬‬ ‫‪h( x) 0‬‬ ‫∞‪−∞ −‬‬ ‫‪h( x) < 0‬‬ ‫ﻭ ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪h(0) = 0‬‬ ‫‪ -2‬ﺍﺴﺘﻨﺘﺎﺝ ﺇﺸﺎﺭﺓ ‪( )h x‬‬ ‫ﻟﺩﻴﻨﺎ ‪ h x = 0 :‬ﺘﻜﺎﻓﺊ ‪( )x = 0 :‬‬ ‫ﻤﻥ ﺃﺠل ‪. x ∈ * :‬‬

( )D f = * : f ′ x ‫ ﺤﺴﺎﺏ‬-1 (II ( ( ) ) ( )f ′( x) = 1 ex −1 − exx = e x − 1 − xe x ex −1 2 ex −1 2 ( ) ( )ex ex( ) ( )f ′( x) = = 1− e−x − x 1− x − e−x ex −1 2 ex −1 2 e x .h( x) ex −1 2 ( )f ′( x) = x = 0 : ( )‫ ﻭﻤﻨﻪ‬h x = 0 ‫ ﺘﻜﺎﻓﺊ‬f ′( x ) = 0 ‫ﻭ ﻋﻠﻴﻪ‬ h( x ) < 0 ‫ ﻭ ﻫﻲ ﻤﺤﻘﻘﺔ ﻷﻥ‬h( x ) < 0 ‫ ﺘﻜﺎﻓﺊ‬f ( x ) < 0 . ‫ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ‬f ‫ ﺇﺫﻥ‬x ∈ * : ‫ﻤﻥ ﺃﺠل‬( )lim f x = lim x 1 = lim 1 = 0 ex − ex −1x→+∞ x→+∞ x→+∞ x : ‫ ﺤﺴﺎﺏ ﺍﻟﻨﻬﺎﻴﺎﺕ‬-2( )lim f x = lim e x 1 = +∞ x−x→−∞ x→−∞( )lim f x = lim x 1 = lim 1 1 = 1 < < ex − < ex −x→0 x→0 x→0 x( )lim fx = lim x = lim 1 = 1 > > ex −1 > ex −1x→0 x→0 x→0 x

‫ﺠﺩﻭل ﺍﻟﺘﻐﻴﺭﺍﺕ ‪:‬‬‫∞‪x −‬‬ ‫‪0‬‬ ‫∞‪+‬‬ ‫‪-‬‬‫)‪f ′( x‬‬ ‫‪-‬‬‫∞‪f (x) +‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪( ) ( )lim‬‬‫‪f‬‬‫‪x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪1‬‬ ‫=‬ ‫‪−1‬‬ ‫‪:‬‬ ‫‪cf‬‬ ‫‪ -3‬ﺇﻨﺸﺎﺀ‬ ‫‪x‬‬ ‫>‬ ‫‪x−‬‬ ‫∞‪x→−‬‬ ‫‪e‬‬ ‫‪1‬‬ ‫‪x→0‬‬‫‪( )lim f‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪x‬‬ ‫‪+ xe x −‬‬ ‫‪x‬‬ ‫>‬ ‫‪ex −‬‬ ‫‪ex −1‬‬‫∞‪x→−‬‬ ‫‪1‬‬ ‫∞‪x→−‬‬ ‫‪x→0‬‬ ‫=‬ ‫‪lim‬‬ ‫‪xe x‬‬ ‫‪ex −1‬‬‫ﺇﺫﻥ ﻴﻭﺠﺩ ﻤﺴﺘﻘﻴﻡ ﻤﻘﺎﺭﺏ ﻤﺎﺌل ﻤﻌﺎﺩﻟﺘﻪ ‪ y = − x‬ﻋﻨﺩ ∞‪ −‬ﻭ ﻟﺩﻴﻨﺎ ‪ y = 0‬ﻤﻌﺎﺩﻟﺔ ﻤﺴﺘﻘﻴﻡ‬ ‫ﻤﻘﺎﺭﺏ ﻋﻨﺩ ∞‪. +‬‬ ‫‪y‬‬ ‫‪3‬‬ ‫‪2,5‬‬ ‫‪2‬‬ ‫‪1,5‬‬ ‫‪1‬‬ ‫‪0,5‬‬‫‪-3 -2,5 -2 -1,5 -1 -0,5 0‬‬ ‫‪0,5 1 1,5 2 2,5 x‬‬ ‫‪-0,5‬‬ ‫‪-1‬‬

15 ‫ﺍﻟﺘﻤﺭﻴﻥ‬ { }D f = x ∈ : e x − 1 ≠ 0 : f ‫ ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ‬-1 Df = * ، Df = ]−∞;0[ ∪ ]0;+∞] ( )lim f x = lim 2e x = 0 ex −1 x→−∞ x→−∞ ( )lim fx = lim 2e x 1 = −∞ < < ex − x→0 x→0 x −∞ 0 +∞ ex −1 - +( )2e x → 2 x lim 2e xe x < : ‫ ﻷﻥ‬lim f = > ex − 1 = +∞ : ‫ﻭﻤﻨﻪ‬ > − 1→0 x→0 x→0 2e x → 2 lim f ( x) = 2  > : ‫ﻷﻥ‬ : ‫ﻭﻋﻠﻴﻪ‬ e x − x→+∞ 1→0 ( ) ( )2ex( ) ( )f ′( x) = e x − 1 − e x .2e x 2e x ex −1− ex = ex −1 2 ex −1 2 −2e x ex −1 2 ( )f ′( x) = :‫ﺇﺫﻥ‬ ( )‫ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻜل ﻤﻥ‬f ‫ ﻭ ﻤﻨﻪ‬f ′ x < 0 ‫ﻭ ﻤﻨﻪ‬

‫ﺍﻟﻤﺠﺎﻟﻴﻥ ‪]0;+∞[ , ]−∞;0[ :‬‬‫∞‪x −∞ 0 +‬‬‫)‪f ′( x‬‬ ‫‪-‬‬ ‫‪-‬‬‫‪f (x) 0‬‬ ‫∞‪+‬‬ ‫‪2‬‬ ‫∞‪−‬‬ ‫ﻓﺈﻥ ‪ y = 0‬ﻤﻌﺎﺩﻟﺔ ﻤﺴﺘﻘﻴﻡ ﻤﻘﺎﺭﺏ) (‬‫‪x‬‬ ‫‪lim f‬‬ ‫‪ -2‬ﺒﻤﺎ ﺃﻥ ‪= 0‬‬ ‫∞‪x→−‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ ‪ lim f x = 2‬ﻓﺈﻥ ‪ y = 2‬ﻤﻌﺎﺩﻟﺔ ﻤﺴﺘﻘﻴﻡ ﻤﻘﺎﺭﺏ) (‬ ‫∞‪x→+‬‬‫ﻭ ﺒﻤﺎ ﺃﻥ ∞‪ lim f x = +‬ﻓﺈﻥ ‪ x = 0‬ﻤﻌﺎﺩﻟﺔ ﻤﺴﺘﻘﻴﻡ ﻤﻘﺎﺭﺏ) (‬ ‫‪x→0‬‬ ‫‪ -3‬ﻨﺒﻴﻥ ﺃﻥ ‪ w 0;1‬ﻤﺭﻜﺯ ﺘﻨﺎﻅﺭ ‪( ):‬‬ ‫ﻤﻥ ﺃﺠل ‪: x ∈ D f‬‬‫ﻟﺩﻴﻨﺎ ‪ 2α − x ∈ Df :‬ﻭ ‪f ( 2α − x ) + f ( x ) = 2β‬‬ ‫ﺃﻱ ‪f ( 2 × 0 − x) + f ( x) = 2 × 1 :‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪f ( − x) + f ( x) = 2 :‬‬ ‫ﻟﺩﻴﻨﺎ ﻤﻥ ﺃﺠل ﻜل ‪ x‬ﻤﻥ ‪− x ∈ D f : D f‬‬

2e− x 2e x 2 2e x e−x − 1 ex −1 ex −1f (−x)+ f ( x) = + = ex + : ‫ﺇﺫﻥ‬ 1 ex −1 2 2e x 2e x e x -1 e x -1 = ex + = 2 + 1 1-e x ex -1 = −2 1 + 2e x ex − ex −1 ( )= −2 + 2e x = 2 ex −1 =2 ex −1 ex −1 ( ). c ‫ ﻤﺭﻜﺯ ﺘﻨﺎﻅﺭ‬w ‫ﻭ ﻤﻨﻪ‬ ( ): ‫ ﺩﻭﻥ ﺭﻤﺯ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ‬g x ‫( ﻜﺘﺎﺒﺔ‬4( )g ( x) = − 2e x , e x -1<0 ‫ أي‬x < 0 ex −1 e x − 1 > 0 ‫ أي‬x > 0 2e x g( x) = ex −1 ,  g ( x) = − f ( x) , x ∈ ]−∞;0]  : ‫ﺇﺫﻥ‬  g ( x ) = f ( x) , x ∈ ]0; + ∞[] ] ] [( ) ( ) ( )‫ ﻨﻅﻴﺭ‬δ : −∞;0 ‫ ﻭ ﻓﻲ ﺍﻟﻤﺠﺎل‬c ‫ ﻴﻨﻁﺒﻕ ﻋﻠﻰ‬δ : 0;+∞ ‫ﻓﻲ ﺍﻟﻤﺠﺎل‬ ( ). ‫ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل‬c

‫‪ -‬ﺇﻨﺸﺎﺀ )‪ (c‬ﻭ ) ‪: (δ‬‬ ‫‪y‬‬ ‫‪4‬‬ ‫‪3,5‬‬ ‫‪3‬‬ ‫‪2,5‬‬ ‫‪2‬‬ ‫‪1,5‬‬ ‫‪1‬‬ ‫‪0,5‬‬‫‪-3,5 -3 -2,5 -2 -1,5 -1 -0,5 0‬‬ ‫‪0,5 1 1,5 2 2,5 3 3,5 4‬‬ ‫‪x‬‬ ‫‪-0,5‬‬ ‫‪-1‬‬ ‫‪-1,5‬‬ ‫‪-2‬‬ ‫‪ -‬ﺍﻟﻤﻨﺎﻗﺸﺔ ﺍﻟﺒﻴﺎﻨﻴﺔ ‪( )m − 3 e x − 1 = 2e x :‬‬ ‫)‪m − 3= g(x‬‬ ‫ﺃﻱ ‪:‬‬ ‫=‪m−3‬‬ ‫‪2e x‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪ex −1‬‬ ‫ﻭ ﻤﻨﻪ ﺒﻭﻀﻊ ‪ m − 3 = α‬ﻨﺠﺩ ‪( )g x = α :‬‬ ‫ﻟﻤﺎ ‪ α ≤ 2‬ﺃﻱ ‪ m − 3 ≤ 2 :‬ﺃﻱ ‪m ≤ 5 :‬‬ ‫‪ -‬ﻭﻤﻨﻪ ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭل ‪.‬‬ ‫‪ -‬ﻟﻤﺎ ‪ α > 2‬ﺃﻱ ‪ : m > 5‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺨﺘﻠﻔﻴﻥ ﻓﻲ ﺍﻹﺸﺎﺭﺓ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪16‬‬ ‫‪ -1 (I‬ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ ‪: f‬‬ ‫[∞‪Df = ]−∞;+‬‬‫‪( )lim f ( x ) = lim 2x2 − 3x e x = lim 2xe2x − 3xe x = 0‬‬‫∞‪x→−‬‬ ‫∞‪x→−‬‬ ‫∞‪x→−‬‬

‫∞‪( )lim f ( x) = lim 2x2 − 3x ex = lim x(2x − 3)ex = +‬‬‫∞‪x→+‬‬ ‫∞‪x→+‬‬ ‫∞‪x→+‬‬ ‫‪( )f ′( x) = (4x − 3) e x + 2x2 − 3x e x‬‬ ‫ﺇﺫﻥ ‪( )f ′( x) = 4x − 3 + 2x2 − 3x e x :‬‬ ‫ﻭ ﻤﻨﻪ ‪( )f ′( x ) = 2x2 + x − 3 e x :‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪ f ′ x :‬ﻟﻪ ﻨﻔﺱ ﺍﻹﺸﺎﺭﺓ ‪( )2 x2 + x − 3‬‬ ‫)‪ ∆ = (1)2 − 4( −3)( 2‬ﺃﻱ ‪ ∆ = 25 :‬ﻴﻭﺠﺩ ﺠﺫﺭﺍﻥ ‪:‬‬ ‫‪x2‬‬ ‫=‬ ‫‪−1 +‬‬ ‫‪5‬‬ ‫=‬ ‫‪1‬‬ ‫‪,‬‬ ‫= ‪x1‬‬ ‫‪−1 −‬‬ ‫‪5‬‬ ‫=‬ ‫‪−3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪2‬‬ ‫‪x‬‬ ‫∞‪−‬‬ ‫‪−3‬‬ ‫∞‪1 +‬‬ ‫‪2‬‬ ‫)‪f ′( x‬‬ ‫‪+‬‬ ‫‪-‬‬ ‫‪+‬‬‫ﻭ ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ[ [‬ ‫‪‬‬ ‫‪−∞;−‬‬ ‫‪3‬‬ ‫‪‬‬ ‫∞‪1;+‬‬ ‫ﻭ‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ‬ ‫‪f‬‬ ‫‪.‬‬ ‫‪‬‬ ‫‪−‬‬ ‫‪3‬‬ ‫‪;1‬‬ ‫ﻋﻠﻰ‬ ‫‪‬‬ ‫‪2‬‬

‫‪x‬‬ ‫∞‪−‬‬ ‫‪−3‬‬ ‫‪1‬‬ ‫∞‪+‬‬ ‫‪+‬‬ ‫‪2‬‬ ‫‪−e‬‬ ‫‪+‬‬ ‫)‪f ′( x‬‬ ‫‪0‬‬ ‫‪-‬‬ ‫∞‪+‬‬ ‫)‪f (x‬‬ ‫‪−3‬‬ ‫‪9e 2‬‬‫‪f‬‬ ‫‪‬‬ ‫‪−‬‬ ‫‪3‬‬ ‫‪‬‬ ‫=‬ ‫‪2‬‬ ‫‪−3‬‬ ‫‪2‬‬ ‫‪−‬‬ ‫‪3‬‬ ‫‪−3‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪e‬‬ ‫‪−3‬‬ ‫=‬ ‫‪‬‬ ‫‪9‬‬ ‫‪+‬‬ ‫‪9‬‬ ‫‪‬‬ ‫‪e‬‬ ‫‪−3‬‬ ‫=‬ ‫‪−3‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪9e 2‬‬ ‫‪( )f (1) = 2(1)2 − 3(1) e1 = −e‬‬ ‫‪ -2‬ﺩﺭﺍﺴﺔ ﺍﻟﻔﺭﻭﻉ ﺍﻟﻼﻨﻬﺎﺌﻴﺔ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪ y = 0‬ﻤﻌﺎﺩﻟﺔ ﻤﺴﺘﻘﻴﻡ ﻤﻘﺎﺭﺏ ﻋﻨﺩ ∞‪−‬‬‫‪lim‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪‬‬ ‫‪2x2 −‬‬ ‫‪3x‬‬ ‫‪‬‬ ‫‪e‬‬ ‫‪x‬‬ ‫=‬ ‫‪lim (2x‬‬ ‫‪−‬‬ ‫‪3)ex‬‬ ‫=‬ ‫∞‪+‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪‬‬‫∞‪x→+‬‬ ‫‪x‬‬ ‫∞‪x→+‬‬ ‫‪‬‬ ‫‪‬‬ ‫∞‪x→+‬‬ ‫ﺇﺫﻥ ‪ c‬ﻴﻘﺒل ﻓﺭﻉ ﻗﻁﻊ ﻤﻜﺎﻓﺊ ﺒﺎﺘﺠﺎﻩ ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻋﻨﺩ ∞‪( )+‬‬ ‫‪ -3‬ﻨﺒﻴﻥ ﺃﻥ ‪ c‬ﻴﻘﺒل ﻨﻘﻁﺘﻲ ﺍﻨﻌﻁﺎﻑ ‪( ):‬‬ ‫‪( )f ′′( x) = (4x + 1)e x + 2x2 + x − 3 e x‬‬ ‫‪( )f ′′( x) = 4x + 1 + 2x2 + x − 3 e x‬‬ ‫‪( )f ′′( x) = 2x2 + 5x − 2 e x‬‬ ‫‪ f ′′( x ) = 0‬ﻴﻜﺎﻓﺊ ‪∆ = 41 , 2x2 + 5x − 2 = 0‬‬ ‫‪x2‬‬ ‫=‬ ‫‪−5 41‬‬ ‫‪,‬‬ ‫= ‪x1‬‬ ‫‪−5 −‬‬ ‫‪41‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪:‬‬

‫∞‪x −∞ x1 x2 +‬‬ ‫‪f ′′( x) + - +‬‬ ‫ﺇﺫﻥ ‪ f ′′ x‬ﻴﻨﻌﺩﻡ ﻋﻨﺩ ‪ x1‬ﻤﻐﻴﺭﺍ ﺇﺸﺎﺭﺘﻪ ﻭ ﻴﻨﻌﺩﻡ ﻜﺫﻟﻙ ﻋﻨﺩ ‪ x2‬ﻤﻐﻴﺭﺍ ﺇﺸﺎﺭﺘﻪ ﻭ ﻋﻠﻴﻪ) (‬‫ﺍﻟﻨﻘﻁﺘﺎﻥ ﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺘﻴﻥ ‪ x1‬ﻭ ‪ x2‬ﻫﻤﺎ ﻨﻘﻁﺘﻲ ﺍﻨﻌﻁﺎﻑ ‪x1 0, 35 , x2 −2, 85‬‬ ‫‪ -4‬ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ∆ ‪( ):‬‬ ‫ﻟﺩﻴﻨﺎ ‪y = f ′(0) × ( x − 0) + f (0) + f (0) :‬‬ ‫ﺤﻴﺙ ‪f ′(0) = −3 , f (0) = 0 :‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪ y = −3 x :‬ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ∆) (‬ ‫‪ -5‬ﺭﺴﻡ ) ∆ ( ﻭ )‪: (c‬‬ ‫‪y‬‬ ‫‪6‬‬ ‫‪5‬‬ ‫‪4‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪1‬‬‫‪-6 -5 -4 -3 -2 -1 0‬‬ ‫‪1 2x‬‬ ‫‪-1‬‬ ‫‪-2‬‬ ‫‪-3‬‬

‫‪ -1 (II‬ﺘﻌﻴﻴﻥ ‪ a‬ﻭ ‪: b‬‬ ‫‪( )g ( x) = (4x + a)e x + 2x2 + ax + b e x‬‬ ‫‪( )g ( x) = 4x + a + 2x2 + ax + b e x‬‬ ‫‪g( x) = (2x2 + (a + 4) x + a + b)ex‬‬ ‫ﺘﻜﻭﻥ ‪ g‬ﺩﺍﻟﺔ ﺃﺼﻠﻴﺔ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﺇﺫﺍ ﻭ ﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ‬ ‫ﺤﻘﻴﻘﻲ ‪. x‬‬ ‫‪a = −7‬‬ ‫‪a + 4 = −3‬‬ ‫)‪g′( x) = f ( x‬‬‫ﻭ ﻴﻜﻭﻥ ‪:‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪:‬‬ ‫‪‬‬ ‫‪b‬‬ ‫=‬ ‫‪7‬‬ ‫‪‬‬ ‫‪a‬‬ ‫‪+‬‬ ‫‪b‬‬ ‫=‬ ‫‪0‬‬ ‫‪( )g ( x) = 2x2 − 7 x + 7 e x‬‬ ‫‪ -2‬ﺘﻌﻴﻴﻥ ‪ a‬ﻭ ‪ b‬ﺤﻴﺙ ﺘﻘﺒل ‪ g‬ﻗﻴﻤﺔ ﺤﺩﻴﺔ ﻜﺒﺭﻯ ﻭ ﺼﻐﺭﻯ ﻭ ﻫﻲ ﺃﻥ‬ ‫ﻴﻨﻌﺩﻡ ﺍﻟﻤﺸﺘﻕ ﻤﺭﺘﻴﻥ ﻤﻐﻴﺭﺍ ﺇﺸﺎﺭﺘﻪ ﻤﺭﺘﻴﻥ ‪.‬‬ ‫‪ g′( x ) = 0‬ﻴﻜﺎﻓﺊ ‪2x2 + (a + 4) x + a + b = 0 :‬‬ ‫)‪ ∆ = (a + 4)2 − 4(a + b‬ﻭ ﻤﻨﻪ ‪∆ = a2 + 8a + 16 − 4a − 4b :‬‬ ‫ﺇﺫﻥ ‪∆ = a2 + 4a − 4b + 16 :‬‬‫ﻭ ﻋﻠﻴﻪ ﺇﺫﺍ ﻜﺎﻥ ‪ ∆ > 0‬ﻓﺈﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ ﻭ ﺘﻜﻭﻥ ﺇﺸﺎﺭﺓ ‪ g′ x‬ﻤﺘﻐﻴﺭﺓ ﻤﺭﺘﻴﻥ ﺃﻱ) (‬ ‫ﻤﻥ ﺃﺠل ‪a2 + 4a − 4b + 16 > 0 :‬‬ ‫ﻴﻜﻭﻥ ﻟﻠﺩﺍﻟﺘﻴﻥ ﻗﻴﻤﺘﻴﻥ ﺤﺩﻴﺘﻴﻥ ﻭﺍﺤﺩﺓ ﻜﺒﺭﻯ ﻭ ﺍﻷﺨﺭﻯ ﺼﻐﺭﻯ ‪.‬‬

‫‪ - 5‬ﺍﻟﺩﺍﻟﺔ ﺍﻟﻠﻭﻏﺎﺭﺘﻤﻴﺔ‬ ‫ﺍﻟﻜﻔﺎﺀﺓ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ‬ ‫‪ -1‬ﺤل ﻤﺸﻜﻼﺕ ﺒﺘﻭﻅﻴﻑ ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻡ‪.‬‬ ‫‪ - 2‬ﻤﻌﺭﻓﺔ ﻭﺘﻔﺴﻴﺭ ﻨﻬﺎﻴﺎﺕ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻤﻴﺔ ﺍﻟﻨﻴﺒﻴﺭﻴﺔ‪.‬‬ ‫‪ -3‬ﺇﻨﺸﺎﺀ ﺒﻴﺎﻨﺎﺕ ﺩﻭﺍل ﻟﻭﻏﺎﺭﻴﺘﻤﻴﺔ ‪.‬‬ ‫ﺗﺼﻤﻴﻢ اﻟﺪرس‬ ‫ﺃﻨﺸﻁﺔ‬ ‫‪ - I‬ﺍﻟﺩﺍﻟﺔ ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻤﻴﺔ ﺍﻟﻨﻴﺒﺭﻴﺔ‬ ‫‪ –II‬ﺍﻟﺩﺍﻟﺔ ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻤﻴﺔ ﺍﻟﻌﺸﺭﻴﺔ‬ ‫ﺘﻜﻨﻭﻟﻭﺠﻴﺎ ﺍﻹﻋﻼﻡ ﻭ ﺍﻻﺘﺼﺎل‬ ‫ﺘﻤـﺎﺭﻴﻥ ﻭ ﻤﺸﻜﻼﺕ‬ ‫ﺍﻟﺤـﻠــــــﻭل‬

‫ﺃﻨﺸﻁﺔ‬ ‫ﺍﻟﻨﺸﺎﻁ ‪:‬‬‫ﻟﻠﺩﺍﻟﺔ ‪( ): f‬‬‫‪c‬‬ ‫ﺍﻟﺒﻴﺎﻨﻲ‬ ‫ﺍﻟﺘﻤﺜﻴل‬ ‫‪‬‬ ‫;‪O‬‬ ‫→‬ ‫‪,‬‬ ‫→‬ ‫‪‬‬ ‫ﻤﺘﺠﺎﻨﺱ‬ ‫ﺃﻨﺸﺊ ﻓﻲ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻭ‬ ‫‪-1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪i‬‬ ‫‪j‬‬ ‫‪x ex‬‬ ‫‪ -2‬ﺃﺭﺴﻡ ﺍﻟﻤﺴﺘﻘﻴﻡ ) ∆ ( ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪y = x‬‬ ‫‪ -3‬ﺃﻨﺸﺊ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ‪ Γ‬ﺼﻭﺭﺓ ‪ c‬ﺒﺎﻟﺘﻨﺎﻅﺭ ﺍﻟﻤﺤﻭﺭﻱ ﺒﺎﻟﻨﺴﺒﺔ) ( ) (‬ ‫ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻡ ) ∆ ( ‪.‬‬ ‫‪ -4‬ﻟﺘﻜﻥ ‪ g‬ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﻲ ﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ‪( ). Γ‬‬ ‫‪ -‬ﻤﺎ ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ ‪ - . g‬ﻤﺎ ﻫﻲ ﺼﻭﺭﺓ ‪ 1‬ﺒﺎﻟﺩﺍﻟﺔ ‪. g‬‬ ‫‪ -‬ﺍﺩﺭﺱ ﺇﺸﺎﺭﺓ ﺍﻟﺩﺍﻟﺔ ‪ g‬ﺒﻴﺎﻨﻴﺎ ‪.‬‬ ‫)‪lim g ( x‬‬ ‫)‪, lim g ( x‬‬ ‫‪,‬‬ ‫‪lim‬‬ ‫‪g‬‬ ‫‪(x‬‬ ‫)‬ ‫‪ -‬ﻤﺎ ﻫﻭ ﺘﺨﻤﻴﻨﻙ ﺤﻭل ‪:‬‬ ‫∞‪x → +‬‬ ‫‪x→0‬‬ ‫∞‪x→+‬‬ ‫‪x‬‬ ‫‪x>0‬‬ ‫‪ -‬ﻤﺎ ﻫﻲ ﺼﻭﺭﺓ ﺍﻟﻌﺩﺩ ‪ e‬ﺒﺎﻟﺩﺍﻟﺔ ‪g‬‬ ‫‪y‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪2‬‬ ‫‪ -1‬ﺇﻨﺸﺎﺀ ) ‪(c‬‬ ‫‪ -2‬ﺇﻨﺸﺎﺀ ∆ ‪( )1,5‬‬ ‫‪ -3‬ﺇﻨﺸﺎﺀ ) ‪1 ( r‬‬ ‫‪ -4‬ﻤﺠﻤﻭﻉ ﺍﻟﺩﺍﻟﺔ ‪0,5 g‬‬‫‪-1,5‬‬ ‫‪-1 -0,5 0 0,5 1 1,5 2 x‬‬ ‫ﻫﻲ ‪]0;+∞[ :‬‬ ‫‪-0,5‬‬ ‫‪-1‬‬ ‫‪ -‬ﺘﻌﻴﻴﻥ ﺼﻭﺭﺓ ‪: 1‬‬

‫ﻟﺩﻴﻨﺎ ‪ ، f (0) = 1‬ﻨﻅﻴﺭﺓ ﺍﻟﻨﻘﻁﺔ )‪ B (0;1‬ﺒﺎﻟﺘﻨﺎﻅﺭ ﺍﻟﻤﺤﻭﺭﻱ ﻫﻲ ﺍﻟﻨﻘﻁﺔ )‪ A(1;0‬ﻭ ﻤﻨﻪ‬ ‫‪g (1) = 0‬‬ ‫‪ -‬ﺩﺭﺍﺴﺔ ﺇﺸﺎﺭﺓ ‪ g‬ﺒﻴﺎﻨﻴﺎ ‪:‬‬‫∞‪x 0 1 +‬‬‫‪g(x) -‬‬ ‫‪+‬‬ ‫‪ -‬ﺍﻟﻤﺨﻤﻨﺎﺕ ‪:‬‬‫∞‪lim g ( x) =-‬‬ ‫‪,‬‬ ‫∞‪lim g ( x) =+‬‬ ‫‪,‬‬ ‫‪lim‬‬ ‫‪g‬‬ ‫‪(x‬‬ ‫)‬ ‫‪=0‬‬ ‫‪-‬‬‫‪x→0‬‬ ‫∞‪x → +‬‬ ‫∞‪x→+‬‬ ‫‪x‬‬‫‪x>0‬‬ ‫‪ -‬ﺘﻌﻴﻴﻥ ﺼﻭﺭﺓ ‪ e‬ﺒﺎﻟﺩﺍﻟﺔ ‪ : g‬ﻟﺩﻴﻨﺎ ‪ e1 = e :‬ﻭ ﻋﻠﻴﻪ ‪f (1) = e‬‬ ‫ﻨﻅﻴﺭﺓ ﺍﻟﻨﻘﻁﺔ ) ‪ c (1;e‬ﺒﺎﻟﺘﻨﺎﻅﺭ ﺍﻟﻤﺤﻭﺭﻱ ﻫﻭ )‪D (e;1‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪. g (e) = 1 :‬‬

‫‪ - I‬ﺍﻟﺩﺍﻟﺔ ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻤﻴﺔ ﺍﻟﻨﻴﺒﺭﻴﺔ‬ ‫ﻭ ﺘﺄﺨﺫ‬ ‫‪ -1‬ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻡ ﺍﻟﻨﻴﺒﺭﻱ ﻟﻌﺩﺩ ‪:‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ x e x‬ﻤﻌﺭﻓﺔ ﻭ ﻤﺴﺘﻤﺭﺓ ﻭ ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ‬ ‫ﻗﻴﻤﻬﺎ ﻓﻲ ‪ . +‬ﺤﺴﺏ ﻨﻅﺭﻴﺔ ﺍﻟﻘﻴﻡ ﺍﻟﻤﺘﻭﺴﻁﺔ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ‬ ‫ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ﻓﺈﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ e x = a‬ﺤل ﻭﺤﻴﺩ ﻫﺫﺍ ﺍﻟﺤل ﻴﺴﻤﻰ‬ ‫ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻡ ﺍﻟﻨﻴﺒﺭﻱ ﻟﻠﻌﺩﺩ ‪ a‬ﻭ ﻨﺭﻤﺯ ﻟﻪ ﺒﺎﻟﺭﻤﺯ ‪. lna‬‬ ‫‪y‬‬ ‫‪a = eα‬‬‫‪y = ex‬‬ ‫‪j‬‬ ‫‪o i lna = α x‬‬ ‫‪eln2 = 2‬‬ ‫ﺃﻤﺜﻠﺔ ‪:‬‬ ‫‪eln10 = 10‬‬ ‫‪ -‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ e x = 2‬ﻫﻭ ‪ ln2‬ﺃﻱ ﺃﻥ ‪:‬‬ ‫‪ -‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ e x = 10‬ﻫﻭ ‪ ln10‬ﺃﻱ ﺃﻥ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪ e0 = 1‬ﻭ ﻋﻠﻴﻪ ‪ln1 = 0 :‬‬ ‫ﻨﺘﺎﺌﺞ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪ e1 = e‬ﻭ ﻋﻠﻴﻪ ‪lne = 1 :‬‬ ‫ﺃ(‬ ‫ﺏ(‬

‫ﺝ( ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ e x = a‬ﻫﻭ ‪ x = lna‬ﻤﻥ ﺃﺠل ‪a > 0‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪. elna = a :‬‬ ‫ﺩ( ﻟﺩﻴﻨﺎ ‪ lnea = a‬ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪a‬‬ ‫ﻤﺒﺭﻫﻨﺔ ‪:‬‬ ‫‪ a‬ﻭ ‪ b‬ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ﻤﻭﺠﺒﺎﻥ ﺘﻤﺎﻤﺎ ‪.‬‬ ‫ﻟﺩﻴﻨﺎ ‪ln (a.b) = lna + lnb :‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬‫ﻟﺩﻴﻨﺎ ‪ ) elna+lnb = elna .elnb :‬ﻭ ﺫﻟﻙ ﺤﺴﺏ ﺨﻭﺍﺹ ﺍﻟﺩﺍﻟﺔ ﺍﻷﺴﻴﺔ ( ﻟﻜﻥ ‪ elna = a‬ﻭ‬ ‫‪ elnb‬ﻭ ﻋﻠﻴﻪ ‪elna+lnb = a.b :‬‬ ‫ﻭ ﻟﺩﻴﻨﺎ ﺃﻴﻀﺎ ‪ eln(a.b) = a.b :‬ﻭ ﻋﻠﻴﻪ ‪eln(a.b) = elna+lnb :‬‬ ‫ﻭ ﻤﻨﻪ ﻨﺴﺘﻨﺘﺞ ﺃﻥ‪. ln (a.b) = lna + lnb :‬‬ ‫ﻨﺘﺎﺌﺞ ‪:‬‬ ‫ﺘﻤﺎﻤﺎ ‪.‬‬ ‫ﻤﻭﺠﺏ‬ ‫ﻋﺩﺩ ﺤﻘﻴﻘﻲ‬ ‫‪a‬‬ ‫ﺤﻴﺙ‬ ‫‪ln‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪‬‬ ‫=‬ ‫‪−lna‬‬ ‫ﺃ(‬ ‫‪‬‬ ‫‪a‬‬ ‫‪‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬‫‪ln (a.c) = ln1‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪a.c = 1‬‬ ‫ﻨﺠﺩ ‪:‬‬ ‫‪1‬‬ ‫‪=c‬‬ ‫ﺒﻭﻀﻊ‬ ‫‪a‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪ lna + lnc = 0 :‬ﺇﺫﻥ ‪lnc = −lna :‬‬ ‫‪ln‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪‬‬ ‫=‬ ‫‪−lna‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪‬‬ ‫‪a‬‬ ‫‪‬‬‫ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ﻤﻭﺠﺒﺎﻥ ﺘﻤﺎﻤﺎ ‪.‬‬ ‫ﻭ‪b‬‬ ‫‪a‬‬ ‫‪ ln‬ﺤﻴﺙ‬ ‫‪‬‬ ‫‪a‬‬ ‫‪‬‬ ‫=‬ ‫‪lna‬‬ ‫‪−‬‬ ‫‪lnb‬‬ ‫ﺏ(‬ ‫‪‬‬ ‫‪b‬‬ ‫‪‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬‫‪ln‬‬ ‫‪‬‬ ‫‪a‬‬ ‫‪‬‬ ‫=‬ ‫‪ln‬‬ ‫‪‬‬ ‫‪a.‬‬ ‫‪1‬‬ ‫‪‬‬ ‫=‬ ‫‪lna‬‬ ‫‪+‬‬ ‫‪ln‬‬ ‫‪1‬‬ ‫=‬ ‫‪lna‬‬ ‫‪−‬‬ ‫‪lnb‬‬ ‫‪‬‬ ‫‪b‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪b‬‬ ‫‪‬‬ ‫‪b‬‬‫ﺝ( ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ‪ a‬ﻭ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻨﺎﻁﻕ ‪n‬‬ ‫ﻟﺩﻴﻨﺎ ‪lnan = nlna :‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬

‫ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ‪ a‬ﻭ ﻤﻥ ﺃﺠل ﻋﺩﺩ ﻨﺎﻁﻕ ‪ : n‬ﻟﺩﻴﻨﺎ ‪ elnan = an :‬ﻭ‬ ‫ﻟﺩﻴﻨﺎ ‪( )enlna = elna n = an :‬‬ ‫ﻭ ﻤﻨﻪ ‪ elnan = enlna :‬ﻭ ﻋﻠﻴﻪ ‪lnan = nlna :‬‬ ‫‪ -2‬ﺍﻟﺩﺍﻟﺔ ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻤﻴﺔ ﺍﻟﻨﻴﺒﻴﺭﻴﺔ ‪:‬‬ ‫ﺃ( ﺘﻌﺭﻴﻑ ‪:‬‬‫ﻨﺴﻤﻲ ﺩﺍﻟﺔ ﻟﻭﻏﺎﺭﻴﺘﻤﻴﺔ ﻨﻴﺒﻴﺭﻴﺔ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﻲ ﻨﺭﻤﺯ ﻟﻬﺎ ﺒﺎﻟﺭﻤﺯ ‪ ، ln‬ﻭﺍﻟﺘﻲ ﺘﺭﻓﻕ ﺒﻜل ﻋﺩﺩ ‪ x‬ﻤﻥ‬ ‫ﺍﻟﻤﺠﺎل [∞‪ ]0;+‬ﺍﻟﻌﺩﺩ ‪lnx‬‬ ‫ﺃﻱ ‪x lnx :‬‬ ‫ﺏ( ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺩﺍﻟﺔ ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻤﻴﺔ ﺍﻟﻨﻴﺒﺭﻴﺔ ‪:‬‬ ‫ﻤﺒﺭﻫﻨﺔ ‪:‬‬ ‫‪ x‬ﻋﻠﻰ‬ ‫‪] [1‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ ln‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬ ‫ﻭ ﺩﺍﻟﺘﻬﺎ ﺍﻟﻤﺸﺘﻘﺔ ﻫﻲ ﺍﻟﺩﺍﻟﺔ ‪x‬‬ ‫∞‪0; +‬‬ ‫[∞‪]0; +‬‬‫‪ x‬ﺃﻱ‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪ elnx = x‬ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ‪x‬‬ ‫ﻟﻨﺤﺴﺏ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺩﺍﻟﺔ ‪x elnx‬‬ ‫ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺩﺍﻟﺔ ‪ x elnx‬ﻋﻠﻰ ﺍﻟﻤﺠﺎل ∞‪ 0; +‬ﻫﻲ ﺍﻟﺩﺍﻟﺔ ‪] [ln′x .elnx‬‬‫‪ x‬ﻭ ﻋﻠﻴﻪ ‪ xln′ ( x ) = 1 :‬ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪ x‬ﻫﻲ ﺍﻟﺩﺍﻟﺔ ‪1‬‬ ‫‪x ln′( x) :‬‬ ‫ﻭ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺩﺍﻟﺔ ‪x‬‬‫‪ x‬ﻋﻠﻰ ﺍﻟﻤﺠﺎل‬ ‫‪1‬‬ ‫ﻫﻲ ﺍﻟﺩﺍﻟﺔ‬ ‫‪x‬‬ ‫‪lnx‬‬ ‫ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺩﺍﻟﺔ‬ ‫ﻭ ﻤﻨﻪ‬ ‫= )‪ln′ ( x‬‬ ‫‪1‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫[∞‪]0; +‬‬‫‪ x‬ﻫﻲ ﺍﻟﺩﺍﻟﺔ ﺍﻷﺼﻠﻴﺔ ﺍﻟﺘﻲ ﺘﻨﻌﺩﻡ ﻋﻨﺩ ‪1‬‬ ‫ﻭ ﻋﻠﻴﻪ ﺒﻤﺎ ﺃﻥ ‪ ln1 = 0 :‬ﻓﺈﻥ ﺍﻟﺩﺍﻟﺔ ‪lnx :‬‬ ‫‪ x‬ﻋﻠﻰ ﺍﻟﻤﺠﺎل [∞‪. ]0;+‬‬ ‫‪1‬‬ ‫ﻟﻠﺩﺍﻟﺔ ‪x‬‬

‫ﺝ( ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺩﺍﻟﺔ ) ‪h : x ln g ( x‬‬ ‫ﺤﻴﺙ ‪ g‬ﺩﺍﻟﺔ ﻏﻴﺭ ﻤﻌﺩﻭﻤﺔ ﻭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻠﻰ ﻤﺠﺎل ‪. I‬‬‫ﺍﻟﺩﺍﻟﺔ ‪ x ln g x‬ﻫﻲ ﻤﺭﻜﺏ ﺍﻟﺩﺍﻟﺘﺎﻥ ‪ g‬ﻭ ‪ ln‬ﺤﻴﺙ ‪ g‬ﻤﻭﺠﺒﺔ ﻭﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ) (‬ ‫ﻋﻠﻰ ‪ I‬ﻭ ﺍﻟﺩﺍﻟﺔ ‪ ln‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪+‬‬‫ﻭ ﻋﻠﻴﻪ ‪ h‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪ I‬ﻭ ﺩﺍﻟﺘﻬﺎ ﺍﻟﻤﺸﺘﻘﺔ ﻫﻲ ﺍﻟﺩﺍﻟﺔ‬‫‪x‬‬ ‫)‪g′( x‬‬ ‫‪h′ : x‬‬ ‫‪g′‬‬ ‫(‬ ‫)‪x‬‬ ‫×‬ ‫‪g‬‬ ‫‪1‬‬ ‫ﺃﻱ ‪g ( x) :‬‬ ‫)‪(x‬‬ ‫ﺩ( ﻨﻬﺎﻴﺎﺕ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻤﻴﺔ ‪:‬‬ ‫• ∞‪lim lnx = +‬‬ ‫∞‪x→+‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻤﺠﺎل ∞‪ A;+‬ﺤﻴﺙ ‪[ [10n ≤ A < 10n+1 :‬‬ ‫ﺤﺴﺏ ﺘﻌﺭﻴﻑ ‪ e‬ﻟﺩﻴﻨﺎ ‪ 2 < e < 3 :‬ﻭ ﻋﻠﻴﻪ ‪e2 < 10 :‬‬‫ﻭ ﻤﻨﻪ ‪ e2 5 < 105 :‬ﺇﺫﻥ ‪ e10 < 105 :‬ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪( )lne10 < ln105 :‬‬‫ﺃﻱ ‪ 10 < ln105 :‬ﺇﺫﺍ ﻜﺎﻥ ‪ x ≥ 105 10n :‬ﻓﺈﻥ ‪( )lnx ≥ ln (10)5 :‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪ lnx ≥ 10n.ln105 :‬ﻟﻜﻥ ‪ln105 > 10 :‬‬ ‫ﻭ ﻤﻨﻪ ‪ lnx ≥ 10n+1 > A :‬ﺇﺫﻥ ‪lnx > A :‬‬‫[∞‪( )lnx ∈[ A;+‬‬‫‪:‬‬ ‫ﻓﺈﻥ‬ ‫‪x‬‬ ‫∈‬ ‫‪‬‬ ‫‪105‬‬ ‫‪10‬‬ ‫;‬ ‫∞‪+‬‬ ‫‪‬‬ ‫ﻭ ﻋﻠﻴﻪ ﻤﻥ ﺃﺠل‬ ‫‪‬‬ ‫‪‬‬ ‫∞‪lim lnx = +‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫∞‪x→+‬‬ ‫∞‪lim lnx = −‬‬ ‫•‬ ‫‪x→0‬‬ ‫‪x>0‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬‫∞‪t → +‬‬ ‫ﻓﺈﻥ ‪:‬‬ ‫‪x →0‬‬ ‫ﻟﻤﺎ‬ ‫=‪t‬‬ ‫‪1‬‬ ‫ﻨﺠﺩ ‪:‬‬ ‫‪x‬‬ ‫=‬ ‫‪1‬‬ ‫ﺒﻭﻀﻊ ‪:‬‬ ‫‪x‬‬ ‫‪t‬‬ ‫‪x>0‬‬‫‪( )lim‬‬‫‪lnx‬‬ ‫‪1‬‬ ‫‪lim‬‬ ‫‪t‬‬‫‪x→0‬‬ ‫∞‪t →+‬‬ ‫=‬ ‫‪lim‬‬ ‫‪ln‬‬ ‫=‬ ‫‪−lnt‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪= −∞ :‬‬‫‪x>0‬‬ ‫∞‪t →+‬‬

‫‪lim‬‬ ‫‪lnx‬‬ ‫=‬ ‫‪0‬‬ ‫•‬ ‫‪x‬‬ ‫∞‪x → +‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬ ‫ﺭﺃﻴﻨﺎ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪ e x ≥ x : x‬ﻭ ﻋﻠﻴﻪ‬ ‫ﻤﻥ ﺃﺠل ‪ lne x ≥ lnx : x > 0‬ﺃﻱ ﺃﻥ ‪x ≥ lnx :‬‬ ‫ﻭ ﻋﻠﻴﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ‪x ≥ ln x :‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪ 2 x ≥ lnx :‬ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪x‬‬ ‫≥‬ ‫‪1‬‬ ‫‪lnx‬‬ ‫‪:‬‬ ‫ﺇﺫﻥ‬ ‫‪2‬‬‫‪lim‬‬ ‫‪2‬‬ ‫‪=0‬‬ ‫‪2 lnx‬‬ ‫‪2 x lnx‬‬ ‫‪x‬‬ ‫‪:‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ‬ ‫≥‬ ‫ﻭ ﻋﻠﻴﻪ ‪:‬‬ ‫‪x ≥x‬‬‫‪x→0‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫‪x>0‬‬ ‫‪lim‬‬ ‫‪lnx‬‬ ‫=‬ ‫‪0‬‬ ‫ﻓﺤﺴﺏ ﻤﺒﺭﻫﻨﺔ ﺍﻟﺤﺩ ﻤﻥ ﺍﻷﺩﻨﻰ ﻓﺈﻥ ‪:‬‬ ‫‪x‬‬ ‫∞‪x→+‬‬ ‫• ‪lim xlnx = 0‬‬ ‫‪x→0‬‬ ‫‪x>0‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬‫∞‪t → +‬‬ ‫ﻓﺈﻥ ‪:‬‬ ‫‪x →0‬‬ ‫ﻟﻤﺎ‬ ‫ﻓﺈﻨﻪ ‪:‬‬ ‫=‪t‬‬ ‫‪1‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪x‬‬ ‫=‬ ‫‪1‬‬ ‫ﺒﻭﻀﻊ ‪:‬‬ ‫‪x‬‬ ‫‪t‬‬ ‫‪x>0‬‬ ‫‪lim‬‬ ‫‪xlnx‬‬ ‫=‬ ‫‪lim 1 ln 1‬‬ ‫=‬ ‫‪lim‬‬ ‫‪−lnt‬‬ ‫=‬ ‫‪0‬‬ ‫‪tt→+∞ t‬‬ ‫‪t‬‬ ‫‪x→0‬‬ ‫∞‪t → +‬‬ ‫‪x>0‬‬ ‫‪lim‬‬ ‫‪ln (1 +‬‬ ‫)‪x‬‬ ‫=‬ ‫‪1‬‬ ‫•‬ ‫‪x→0‬‬ ‫‪x‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ ‪ t → lnt : f‬؛ ﻫﺫﻩ ﺍﻟﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻭ ﻗﺎﺒﻠﺔ ﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬ ‫∞‪ 0;+‬ﻭ ﻋﻠﻴﻪ ﻓﻬﻲ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪] [. 1‬‬ ‫‪f ′ (1) = lim‬‬ ‫‪f‬‬ ‫‪(1+ h) −‬‬ ‫‪f‬‬ ‫)‪(1‬‬ ‫ﻤﻥ ﺠﻬﺔ ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪h→0‬‬ ‫‪h‬‬‫ﻭ ﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯ ‪:‬‬ ‫‪(1) ...‬‬ ‫‪f‬‬ ‫‪′‬‬ ‫)‪(1‬‬ ‫=‬ ‫‪lim‬‬ ‫‪ln‬‬ ‫‪(1 +‬‬ ‫)‪h‬‬ ‫ﺃﻱ ﺃﻥ ‪:‬‬ ‫‪h→0‬‬ ‫‪h‬‬