دروس مادة الرياضيات للفصل الاول للشعب العلمية سنة ثالثة ثانوي

1+h - 2 + 1 4+3 4 = lim h→0 h 4 1+h -8+4+h = lim 4 + (4 + h) h→0 h = lim 4 1 + h - (4 - h) h→0 4h(4 + h) = lim 4 1+ h - (4 - h) × 4 1 + h + (4 - h) h→0 4h(4 + h) 4 1 + h(4 - h) = lim (-h + 24) h→0 4(4 + h) 4 1 + h + 4 - h = 16 24 14) = 24 = 3 (4 + 1608 32 ( )f ′ 1 = 3 ‫ ﺤﻴﺙ‬1 ‫ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬ f ‫ﺇﺫﻥ ﺍﻟﺩﺍﻟﺔ‬ 32 f (x) = x2 + 9x - 2 ; x0 = 2 : ‫( ﻟﺩﻴﻨﺎ‬7 x -1 f ( 2) = 20 ‫ ؛‬Df = \ - {1} : ‫ﻭ ﻤﻨﻪ‬ f -f (2) (2 + h)2 + 9(2 + h) - 2 - 20 hlim = lim 2+ h-1h→0 h→0 h

4 + 4h + h2 + 18 + 9h - 2 - 20 1 +h = lim h→0 h h2 + 13h + 20 -20 1+ h = lim h→0 h h2 + 13h + 20 - 20(1 + h) h2 - 7h = lim 1+ h = lim 1+ h h→0 h h→0 h = lim h (h - 7) = lim h - 7 = -7 h→0 h (1 + h) h→0 1 + h ( )f ′ 2 = -7 : ‫ ﺤﻴﺙ‬2 ‫ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬f ‫ﺇﺫﻥ ﺍﻟﺩﺍﻟﺔ‬ . 3‫ﺍﻟﺘﻤﺭﻴﻥ‬ { }D f = x ∈ \ : 9 - x2 ≥ 0 : ‫ ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ‬-1 x −∞ -3 : ‫ﻟﺩﻴﻨﺎ‬ 3 +∞ 9 − x2 - + - Df = [-3 ; 3] : ‫ﻭﻋﻠﻴﻪ‬ lim f (x) : ‫ ﺤﺴﺎﺏ‬-2 x→3 x -3lim f ( x) = lim 9 - x2 = lim 9 - x2 . 9 - x2 < x -3 < x -3 < ( x - 3) 9 - x2x→3 x→3 x→3 = lim 9 - x2 = lim (3 - x) (3 + x) < ( x - 3) 9 - x2 < ( x - 3) 9 - x2 x→3 x→3

‫∞‪= lim -(x -3) (x + 3) = lim -(3 + x) = -‬‬ ‫<‬ ‫)‪( x - 3‬‬ ‫‪9 - x2‬‬ ‫<‬ ‫‪9 - x2‬‬‫‪x→3‬‬ ‫‪x→3‬‬ ‫‪-( x + 3) → -6‬‬ ‫‪‬‬ ‫ﻷﻥ ‪:‬‬ ‫‪‬‬ ‫‪9 - x2‬‬ ‫‪→ 0‬‬‫ﻭﻨﻘﻭل ﺃﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 3‬ﻤﻥ ﺍﻟﻴﺴﺎﺭ ﻭﻋﻠﻴﻪ ‪( )C f‬‬ ‫ﻴﻘﺒل ﻨﺼﻑ ﻤﻤﺎﺱ ﻋﻨﺩ ‪ 0‬ﻤﻥ ﺍﻟﻴﺴﺎﺭ ﻴﻭﺍﺯﻱ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ‪.‬‬ ‫‪ -3‬ﺘﺒﻴﺎﻥ ﺃﻥ ‪ C f‬ﻫﻭ ﻨﺼﻑ ﺩﺍﺌﺭﺓ ‪( ):‬‬ ‫ﻟﺩﻴﻨﺎ ‪ f ( x ) = 9 - x2 :‬ﻭﻤﻨﻪ ‪y = 9 - x2‬‬ ‫‪2‬‬ ‫‪9 - x2‬‬‫‪y2 = 9 - x2‬‬ ‫= ‪( )y2‬‬‫‪y ≥ 0‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ :‬‬ ‫‪y ≥ 0‬‬ ‫‪x2 + y2 = 9‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪y‬‬ ‫≥‬ ‫‪0‬‬‫ﺇﺫﻥ ‪ C f :‬ﻤﻥ ﻨﺼﻑ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﺘﻲ ﻤﺭﻜﺯﻫﺎ ‪ O‬ﻭ ﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪ 3‬ﺤﻴﺙ ‪( ). y ≥ 0‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬ ‫‪ -1‬ﺩﺭﺍﺴﺔ ﺍﻻﺴﺘﻤﺭﺍﺭﻴﺔ ﻋﻨﺩ ‪D f = \ : 0‬‬‫‪lim‬‬ ‫)‪f (x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪sin‬‬ ‫‪x‬‬ ‫‪=1=f‬‬ ‫)‪(0‬‬ ‫‪x‬‬‫‪x→0‬‬ ‫‪x→0‬‬ ‫ﻭﻤﻨﻪ ‪ f‬ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪. 0‬‬ ‫‪ -‬ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻫﻲ ﺠﺩﺍﺀ ﺩﺍﻟﺘﻲ ﺍﻟﻤﻘﻠﻭﺏ ﻭ ‪ sin‬ﻭ ﻋﻠﻴﻪ ﻓﻬﻲ ﻤﺴﺘﻤﺭﺓ ﻋﻠﻰ‬ ‫*\ ﻟﻜﻥ ‪ f‬ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪ 0‬ﻭﻤﻨﻪ ‪ f‬ﻤﺴﺘﻤﺭﺓ ﻋﻠﻰ \‬ ‫‪ (2‬ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪: 0‬‬

‫‪f (h) - f (0) = lim‬‬ ‫‪1‬‬ ‫‪sin‬‬ ‫‪h‬‬ ‫‪1‬‬ ‫‪sin‬‬ ‫‪h‬‬ ‫‪h‬‬ ‫‪h‬‬ ‫‪h‬‬‫‪lim‬‬ ‫>‪h‬‬ ‫=‬ ‫‪lim‬‬ ‫×‬ ‫=‬ ‫∞‪+‬‬ ‫>‬ ‫‪h→0‬‬ ‫>‪h‬‬ ‫‪h→0‬‬‫‪h→0‬‬ ‫‪lim‬‬ ‫)‪f (h) - f (0‬‬ ‫‪= lim‬‬ ‫‪1‬‬ ‫‪.‬‬ ‫‪sin h‬‬ ‫∞‪= -‬‬ ‫<‬ ‫<‬ ‫‪h‬‬ ‫‪h‬‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫ﺇﺫﻥ ﺍﻟﺩﺍﻟﺔ ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪. O‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬ ‫‪( )x2 x2 + 2x + 1‬‬ ‫‪f ( x) = ( x + 1) (x2 - x + 1) ; x ≠ -1‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪(x‬‬ ‫‪x2 ( x + 1)2‬‬ ‫)‪1‬‬ ‫‪; x ≠ -1‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪+ 1) (x2 - x +‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪(x‬‬ ‫‪x.x‬‬ ‫‪+1‬‬ ‫‪+ 1) (x2‬‬ ‫)‪- x + 1‬‬ ‫‪ -1‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ ‪:‬‬ ‫‪{ }Df = x ∈ \ : (x + 1) (x2 - x + 1) ≠ 0‬‬ ‫ﻭﻋﻠﻴﻪ ‪ ( x + 1) x2 - x + 1 = 0 :‬ﺘﻜﺎﻓﺊ ‪ x + 1 = 0‬ﺃﻭ) (‬ ‫‪ x2 - x + 1 = 0‬ﻭﺃﻥ ‪ x = -1‬ﺃﻭ ‪x2 - x + 1 = 0‬‬ ‫ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪x2 - x + 1 = 0 :‬‬ ‫‪ D = −3‬ﻭﻤﻨﻪ ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭل ‪.‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪{ }D f = \ - -1 :‬‬ ‫‪ – 2‬ﺩﺭﺍﺴﺔ ﺍﻻﺴﺘﻤﺭﺍﺭﻴﺔ ﻋﻨﺩ ‪: 0‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪(x‬‬ ‫‪x.x‬‬ ‫‪+1‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪+ 1) (x2‬‬ ‫)‪- x + 1‬‬

x −∞ −1 0 : ‫ﻭﻤﻨﻪ‬ x −x −x +∞ x+1 −( x + 1) x+1 x x( x + 1) − x( x + 1) x ⋅ x+1 x+1 − x( x + 1) : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ f ( x ) = x (x + 1) x ∈ ]-∞ ; -1[ ∪ [0 ; +∞[ (x + 1) (x2 - x + 1) -x (x + 1) f ( x ) = (x + 1) (x2 - x + 1) x ∈ ]-1 ; 0] f ( -1) = 1 3 : ‫ﻭ ﻋﻠﻴﻪ‬  f ( x ) = x , x ∈ ]-∞ ; -1[ ∪ [0 ; +∞[  f ( x ) = x2 - x + 1 , x ∈ ]-1 ; 0]   −x  x2 - x + 1   1  f ( -1) = 3 lim f ( x) = lim f ( x) -x =0 : ‫ﻭ ﻟﺩﻴﻨﺎ‬ << x +1 x→0 x→0 lim f ( x) = lim f ( x) x =0 >> x +1 x→0 x→0

‫‪f‬‬ ‫)‪(0‬‬ ‫=‬ ‫‪(0‬‬ ‫‪0.0‬‬ ‫‪+1‬‬ ‫)‪1‬‬ ‫=‬ ‫‪0‬‬ ‫‪+ 1) (02‬‬ ‫‪-0+‬‬ ‫ﺇﺫﻥ ‪lim f ( x ) = lim f ( x ) = f (0) :‬‬ ‫><‬ ‫‪x→0‬‬ ‫‪x→0‬‬ ‫ﻭﻋﻠﻴﻪ ‪ f‬ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪. 0‬‬ ‫‪ -‬ﺩﺭﺍﺴﺔ ﺍﻻﺴﺘﻤﺭﺍﺭﻴﺔ ﻋﻨﺩ ‪: -1‬‬ ‫‪lim f ( x) = lim‬‬ ‫‪x2‬‬ ‫‪x‬‬ ‫‪+1‬‬ ‫=‬ ‫‪-1‬‬ ‫<<‬ ‫‪-x‬‬ ‫‪3‬‬ ‫‪x→0‬‬ ‫‪x→0‬‬ ‫ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪ – 1‬ﻤﻥ ﺍﻟﻴﺴﺎﺭ ‪( ).‬‬ ‫‪f‬‬ ‫‪f‬‬ ‫ﻭﻤﻨﻪ )‪x ≠ f (−1‬‬ ‫ﻭﻋﻠﻴﻪ‬ ‫‪lim‬‬ ‫<‬ ‫‪x → −1‬‬ ‫‪lim f ( x ) = lim‬‬ ‫‪x2‬‬ ‫‪−x‬‬ ‫=‬ ‫‪1‬‬ ‫>‬ ‫‪x→−1‬‬ ‫‪− x+1‬‬ ‫‪3‬‬ ‫‪x → −1‬‬ ‫ﻭﻤﻨﻪ )‪ lim f ( x ) = f ( − 1‬ﻭﻋﻠﻴﻪ ‪ f‬ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪ – 1‬ﻤﻥ ﺍﻟﻴﻤﻴﻥ‬ ‫‪x ;→ −1‬‬ ‫ﻟﻜﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻏﻴﺭ ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪. - 1‬‬ ‫‪ – 3‬ﺩﺭﺍﺴﺔ ﺍﻻﺸﺘﻘﺎﻗﻴﺔ ﻋﻨﺩ ‪: 0‬‬ ‫‪-h‬‬ ‫‪lim f (h) - f (0) = lim h2 - h + 1‬‬ ‫<‬ ‫‪h‬‬ ‫<‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫‪= lim‬‬ ‫‪-h 1‬‬ ‫<‬ ‫‪h2 - h + 1 × h‬‬ ‫‪h→0‬‬ ‫‪= lim‬‬ ‫‪-1‬‬ ‫=‬ ‫‪-1‬‬ ‫<‬ ‫‪h2 - h + 1‬‬ ‫‪h→ 0‬‬ ‫ﻭﻋﻠﻴﻪ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 0‬ﻤﻥ ﺍﻟﻴﺴﺎﺭ ‪.‬‬ ‫‪-h‬‬‫‪lim‬‬ ‫‪f (h) - f (0) = lim‬‬ ‫‪h2 - h + 1‬‬ ‫‪= lim‬‬ ‫‪-h 1‬‬ ‫>‬ ‫>‪h‬‬ ‫‪h‬‬ ‫>‬ ‫‪h2 - h + 1 × h‬‬ ‫‪h→0‬‬‫‪h→0‬‬ ‫‪h→0‬‬

‫‪= lim‬‬ ‫‪h2‬‬ ‫‪-1‬‬ ‫=‬ ‫‪1‬‬ ‫>‬ ‫‪-h+1‬‬ ‫‪h→0‬‬ ‫ﻭﻋﻠﻴﻪ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 0‬ﻤﻥ ﺍﻟﻴﻤﻴﻥ ﻟﻜﻨﻬﺎ ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪. 0‬‬ ‫‪ -‬ﺩﺭﺍﺴـﺔ ﺍﻻﺸﺘﻘﺎﻗﻴﺔ ﻋﻨﺩ ‪: - 1‬‬ ‫‪f (1 + h) - f (-1) = lim‬‬ ‫‪(-1‬‬ ‫‪-1 + h‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪+ h)2 - (-1 + h) +1‬‬ ‫‪3‬‬‫‪lim‬‬ ‫>‬ ‫‪h‬‬ ‫>‬ ‫‪h‬‬‫‪h→0‬‬ ‫‪h→0‬‬ ‫‪= lim‬‬ ‫‪ -1 + h‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫×‬ ‫‪1‬‬ ‫<‬ ‫‪1 - 2h + h2 + 1 - h + 1‬‬ ‫‪3 ‬‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫=‬ ‫‪lim‬‬ ‫‪ 3 (-1 + h) - (h2 - 3h + 3) ‬‬ ‫×‬ ‫‪1‬‬ ‫<‬ ‫‪‬‬ ‫‪‬‬ ‫‪h‬‬ ‫‪‬‬ ‫)‪3 (h2 - 3h + 3‬‬ ‫‪‬‬ ‫‪h→ 0‬‬ ‫‪= lim‬‬ ‫‪-3 + 3h - h2 + 3h - 3‬‬ ‫×‬ ‫‪1‬‬ ‫<‬ ‫)‪3 (h2 - 3h + 3‬‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫‪= lim‬‬ ‫‪-h2 + 6h - 6‬‬ ‫×‬ ‫‪1‬‬ ‫∞‪= +‬‬ ‫<‬ ‫)‪3 (h2 - 3h + 3‬‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫ﻭﻤﻨﻪ ‪ f‬ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 0‬ﻤﻥ ﺍﻟﻴﺴﺎﺭ ﻭﻴﻤﻜﻥ ﺍﻹﺠﺎﺒﺔ ﺃﻴﻀﺎ‬ ‫ﺒﻘﻭﻟﻨﺎ ﺒﻤﺎ ﺃﻥ ‪ f‬ﻏﻴﺭ ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪ – 1‬ﻤﻥ ﺍﻟﻴﺴﺎﺭ ﻓﻬﻲ ﻏﻴﺭ ﻗﺎﺒﻠﺔ‬ ‫ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ -1‬ﻤﻥ ﺍﻟﻴﺴﺎﺭ ‪.‬‬ ‫‪f (-1 + h) - f (-1) = lim‬‬ ‫‪(-1‬‬ ‫‪+1 - h‬‬ ‫‪+1‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫)‪+ h)2 - (-1+ h‬‬ ‫‪3‬‬‫‪lim‬‬ ‫>‬ ‫‪h‬‬ ‫>‬ ‫‪h‬‬‫‪h→ 0‬‬ ‫‪h→ 0‬‬ ‫‪1‬‬ ‫‪-‬‬ ‫‪2h‬‬ ‫‪+‬‬ ‫‪1-h‬‬ ‫‪-‬‬ ‫‪h‬‬ ‫‪+1‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪h2 + 1‬‬ ‫‪3‬‬ ‫‪= lim‬‬ ‫>‬ ‫‪h‬‬ ‫‪h→ 0‬‬

‫‪h2‬‬ ‫‪1-h‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪- 3h + 3‬‬ ‫‪3‬‬ ‫=‬ ‫‪lim‬‬ ‫>‬ ‫‪h‬‬ ‫‪h→ 0‬‬ ‫=‬ ‫‪lim‬‬ ‫‪3‬‬ ‫‪- 3h -‬‬ ‫‪h2 + 3h -‬‬ ‫‪3‬‬ ‫×‬ ‫‪1‬‬ ‫>‬ ‫‪3 (h2‬‬ ‫)‪- 3h + 3‬‬ ‫‪h‬‬ ‫‪h→ 0‬‬ ‫‪= lim‬‬ ‫‪-h2‬‬ ‫>‬ ‫)‪3h (h2 - 3h + 3‬‬ ‫‪h→ 0‬‬ ‫‪= Lim‬‬ ‫‪3(h2‬‬ ‫‪−h‬‬ ‫‪+‬‬ ‫)‪3‬‬ ‫=‬ ‫‪0‬‬ ‫‪h ;→ 0‬‬ ‫‪− 3h‬‬ ‫ﻭﻤﻨﻪ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻗﻴﺔ ﻋﻨﺩ ‪ -1‬ﻤﻥ ﺍﻟﻴﻤﻴﻥ ﻟﻜﻨﻬﺎ ﻏﻴﺭ ﻗﺎﺒﻠﺔ‬ ‫ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪. - 1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 6‬‬ ‫ﺤﺴﺎﺏ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤـﺸـﺘﻘـﺔ ‪:‬‬ ‫‪f‬‬ ‫= )‪(x‬‬ ‫‪x2 − 4x + 2‬‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪x2 + x + 1‬‬ ‫‪{ }Df = x ∈ \ : x2 + x + 1 ≠ 0‬‬ ‫ﻨﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ x2 + x + 1 = 0 :‬ﻭﻟﺩﻴﻨﺎ ‪∆ = -3‬‬ ‫ﻭﻤﻨﻪ ﻻ ﺘﻭﺠﺩ ﺤﻠﻭل ‪D f = \ .‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻫﻲ ﺩﺍﻟﺔ ﻨﺎﻁﻘﺔ ﻭ ﻤﻨﻪ ﻓﻬﻲ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪ D‬ﺤﻴﺙ ‪:‬‬‫)‪f ′( x‬‬ ‫=‬ ‫‪(2x‬‬ ‫‪-‬‬ ‫)‪4‬‬ ‫‪(x2‬‬ ‫‪+‬‬ ‫‪x + 1) - (2x +‬‬ ‫)‪1‬‬ ‫‪( x2 - 4x‬‬ ‫‪+‬‬ ‫)‪2‬‬ ‫‪( x2 + x + 1)2‬‬‫)‪f ′(x‬‬ ‫=‬ ‫‪2x3+‬‬ ‫‪2x2 +‬‬ ‫‪2x‬‬ ‫‪-‬‬ ‫‪4x2- 4x - 4-2x3‬‬ ‫‪+ 8x2-4x-x2‬‬ ‫‪+ 4x-2‬‬ ‫‪(x2 + x + 1)2‬‬

f ′( x) = 5x2 - 2x - 6 (x2 - x - 1)2 f (x) = (x - x+3 2) : ‫( ﻟﺩﻴﻨﺎ‬2 1) ( x + Df = {x ∈ \ : ( x - 1) ( x + 2) ≠ 0} Df = \ − { 1 ; − 2} : ‫ﺇﺫﻥ‬f ′(x) = 1 (x - 1) - [1 (x + 2) + 1(x - 1)] (x + 3) 1)2 (x + 2)2 (x -f ′(x) = (x - 1) (x + 2) - (2x + 1) (x + 3) (x - 1)2 (x + 2)2f ′(x) = x2 + x - (2x2 + 7x + 3) (x - 1)2 (x + 2)2 f ′(x) = - x2 - 6x - 5 : ‫ﺇﺫﻥ‬ (x - 1)2 (x + 2)2 f ( x ) = (2x + 1)2 (x2 + 3)2 : ‫( ﻟﺩﻴﻨﺎ‬3 . D f = \ : ‫ﻭ ﻤﻨﻪ‬f ′( x) = 2.2.(2x + 1) (x2 + 3)2 + 3 (2x) (x2 + 3) . (2x + 1)2f ′( x) = (x + 1) (x2 + 3) 4 (x2 + 3) + 6x (x - 1) f ′( x) = (x + 1) (x2 + 3) (16x2 + 6x + 12) 2  3  5 Df = \ : ‫ﻭ ﻤﻨﻪ‬ f (x) = x3 + 4x + 1 : ‫( ﻟﺩﻴﻨﺎ‬4  6   2  2  5   5 f ′(x) =3 × x 2 + 4 x 3 + 4x + 1

‫‪f‬‬ ‫)‪( x‬‬ ‫=‬ ‫‪- x2 + 3x -‬‬ ‫‪5‬‬ ‫‪ (5‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪x +1‬‬ ‫ﻭ ﻤﻨﻪ ‪Df = \ - {-1} :‬‬ ‫)‪f ′(x‬‬ ‫=‬ ‫‪(-2x‬‬ ‫‪+‬‬ ‫)‪3‬‬ ‫‪(x‬‬ ‫‪+ 1) - 1 (-x2‬‬ ‫‪+‬‬ ‫‪3x‬‬ ‫‪-‬‬ ‫)‪5‬‬ ‫‪(x + 1)2‬‬ ‫)‪f ′( x‬‬ ‫=‬ ‫‪-2x 2‬‬ ‫‪- 2x‬‬ ‫‪+ 3x + 3 + x2‬‬ ‫‪- 3x‬‬ ‫‪+5‬‬ ‫‪(x + 1)2‬‬ ‫)‪f ′(x‬‬ ‫=‬ ‫‪- x2 - 2x +‬‬ ‫‪8‬‬ ‫‪( x + 1)2‬‬ ‫‪ (6‬ﻟﺩﻴﻨﺎ ‪f ( x ) = -4x2 + 5 x :‬‬ ‫ﻭ ﻤﻨﻪ ‪. D f = \ :‬‬ ‫ﻭﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪ f ( x) = -4x2 + 5x ; x ≥ 0‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪f‬‬ ‫)‪( x‬‬ ‫=‬ ‫‪-4x 2‬‬ ‫‪-‬‬ ‫‪5x‬‬ ‫‪; x≤0‬‬ ‫‪ x‬ﻤﻥ ﺃﺠل ‪f ′( x ) = -8x + 5 : x > 0‬‬ ‫‪ x‬ﻤﻥ ﺃﺠل ‪f ′( x ) = -8x - 5 : x < 0‬‬ ‫ﻨﺩﺭﺱ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪( )f 0 = 0 : 0‬‬‫‪lim f (h) - f (0) = lim -4h2 -5h‬‬ ‫<‬ ‫‪h‬‬ ‫<‬ ‫‪h‬‬‫‪h→0‬‬ ‫‪h→0‬‬ ‫‪= lim h (-4h -5) = -5‬‬ ‫<‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫ﺇﺫﻥ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 0‬ﻤﻥ ﺍﻟﻴﻤﻴﻥ ﻟﻜﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻻ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪. 0‬‬

( )Df = \ : ‫ ﻭ ﻤﻨﻪ‬f x = - x2 + 8 x - 15 : ‫( ﻟﺩﻴﻨﺎ‬7  f ( x) = - x2 + 8x - 15 ; x ∈ [3 ; 5]   f ( x) = -x 2 + 8x - 15 ; x ∈ [3 ; 5] ∪ [5 ; + ∞] - x2 + 8 x - 15 : ‫ ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ‬x x2 = 3 ‫ ؛‬x1 = 5 : ‫∆ ﻭﻤﻨﻪ‬′ = 1 : ‫ﻟﺩﻴﻨﺎ‬ −∞3 5+∞→ : ‫ﻭ ﻤﻨﻪ ﺍﻹﺸﺎﺭﺓ ﻫﻲ‬ + 0- 0 + f ( x) = - x2 + 8x - 15 ; - x2 + 8 x - 15 ≥ 0 f ( x ) = - x2 + 8x - 15 ; - x2 +8x - 15 ≤ 0 f ′( x) = - 2x + 8 x ∈ ]3 ; 5[ ‫ﻟﻤﺎ‬ f ′( x) = - 2x + 8 : x ∈ ]- ∞ ; 3[ ∪ ]5 ; +∞[ ‫ﻟﻤﺎ‬ ( )f 3 = 0 : 3 ‫ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬lim f (3 + h) - f (h) = lim f (3 + h) - f (h) + 15 < h < hh→0 h→0lim f (3 +h)- f (h) = lim f (3 +h) - 8(3 +h) +15 < h < hh→0 h→0lim h2 - 2h = lim (h - 2) = - 2 < h < h→0h→0 ‫ ﻤﻥ ﺍﻟﻴﺴﺎﺭ‬3 ‫ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬ f ‫ﺇﺫﻥ‬lim f (3+ h)- f (h) = lim f (3+ h)- 8(3+ h) + 15 < h < hh→0 h→0

‫‪= lim -9 - 6h - h2 + 24 + 8h - 15 = lim -h2 + 2h‬‬ ‫>‬ ‫‪h‬‬ ‫>‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫‪= lim h (-h + 2h) = lim (-h + 2) = 2‬‬ ‫>‬ ‫‪h‬‬ ‫>‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫ﺇﺫﻥ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪. 3.......‬‬ ‫ﻟﻜﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪. 3‬‬ ‫‪ x‬ﻗﺒﻠﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪( )f 5 = 0 : 5‬‬‫‪lim f (5 + h) - f (5) = lim − (5 + h)2 - 8 (5 + h) - 15‬‬ ‫<‬ ‫‪h‬‬ ‫<‬ ‫‪h‬‬‫‪h→0‬‬ ‫‪h→0‬‬ ‫‪= lim -25 - 10h - h2 + 40 + 8h - 15 = lim -h2 - 2h‬‬ ‫<‬ ‫‪h‬‬ ‫<‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫‪= lim h (-h - 2) = lim (-h - 2) = -2‬‬ ‫<‬ ‫‪h‬‬ ‫<‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫ﺇﺫﻥ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 5‬ﻤﻥ ﺍﻟﻴﺴﺎﺭ‪.‬‬‫‪lim f (5 + h) - f (5) = lim (5 + h)2 - 8 (5 + h) + 15‬‬ ‫>‬ ‫‪h‬‬ ‫>‬ ‫‪h‬‬‫‪h→0‬‬ ‫‪h→0‬‬ ‫‪= lim 25 + 10h + h2 - 40 - 8h + 15 = lim h2 + 2h‬‬ ‫>‬ ‫‪h‬‬ ‫>‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫‪= lim h (h + 2) = lim (h + 2) = 2‬‬ ‫>‬ ‫‪h‬‬ ‫>‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫ﺇﺫﻥ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 5‬ﻤﻥ ﺍﻟﻴﻤﻴﻥ ‪.‬‬ ‫ﻟﻜﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪5‬‬ ‫‪ (8‬ﻟﺩﻴﻨﺎ ‪ f ( x ) = - x + 2 . x + 5 :‬ﻭ ﻤﻨﻪ ‪D f = \ :‬‬ ‫ﻨﻜﺘﺏ ‪ f x‬ﺩﻭﻥ ﺭﻤﺯ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ) (‬

x −∞ -5 2 +∞ -x + 2 x+5 -x + 2 -x + 2 -(-x + 2)-x + 2 x + 5 -(x + 5) x +5 x +5 -(-x+2)(x+5) (-x+2)(x+5) -(-x+2)(x+5) : ‫ﺇﺫﻥ‬ f ( x) = -(-x + 2) (x + 5) ; x ∈ ]-∞ ; -5] ∪ [2 ; +∞[ f ( x) = (-x + 2) (x + 5) ; x ∈ [-5 ; 2[ f ′( x) = -1 (x + 5) + 1 (-x + 2) : x ∈ ]-5 ; 2[ ‫* ﻟﻤﺎ‬ f ′( x) = -x - 5 - x + 2 f ′( x) = - (-2x - 3) : x ∈ ]-∞ ; -5[ ∪ ]2 ; +∞[ ‫* ﻟﻤﺎ‬ f ′( x) = 2x + 3 ( )f −5 = 0 : -5 ‫ ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬xlim f (-5 + h) - f (-5) = lim -(5 - h + 2) (-5 + h +5) < h < hh→0 h→0 = lim -h (7 - h) = lim - (7 - h) < h < h→0 h→0 = -7 .‫ ﻤﻥ ﺍﻟﻴﺴﺎﺭ‬-5 ‫ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬f ‫ﻭﻋﻠﻴﻪ‬lim f (-5 + h) - f (-5) = lim (5 - h + 2) (-5 + h +5) > h > hh→0 h→0 = lim h (- h + 7) = lim (-h + 7) = 7 > h > h→0 h→0 .-5 ‫ ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ‬f ‫ ﻤﻥ ﺍﻟﻴﻤﻴﻥ ﻟﻜﻥ ﺍﻟﺩﺍﻟﺔ‬-5 ‫ﻭﻤﻨﻪ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬

. 2‫ ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬xlim f (2 + h) - f (2) = lim (-2 - h + 2) (2 + h +5) < h < hh→0 h→0 = lim -h (7 + h) = lim -(7 + h) = -7 < h < h→0 h→0 . ‫ ﻤﻥ ﺍﻟﻴﺴﺎﺭ‬2 ‫ ﻗﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬f ‫ﺇﺫﻥ‬lim f (2 + h) - f (2) = lim -(-2 - h + 2) (2 + h +5) > h > hh→0 h→0 = lim h (7 + h) = lim (7 + h) = 7 > h > h→0 h→0 .‫ ﻤﻥ ﺍﻟﻴﻤﻴﻥ‬2 ‫ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬f ‫ﺇﺫﻥ‬ . 2 ‫ ﻻ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ‬f ‫ﻟﻜﻥ ﺍﻟﺩﺍﻟﺔ‬ . 7‫ﺍﻟﺘﻤﺭﻴﻥ‬ : ‫ﺘﻌﻴﻴﻥ ﺍﻟﺩﻭﺍل ﺍﻟﻤﺸﺘﻘﺔ‬1) f ( x ) = cos3 x , Df = \ f ′( x) = -3 sin x cos2 x2) f ( x) = sin3 x Df = {x ∈ k : cosx ≠ 0} x = π + \π , k ∈ ] : ‫ﻤﻌﻨﺎﻩ‬ cosx = 0 2 Df =\ - π + kπ / k ∈ ]     2  f ′( x) = 3 (1 + tan2 x) (tan2 x) ( )f ′( x) = 3 . tan2g 1 + tan2 x

3) f ( x) = sin x Df = {x ∈ \ : sin x ≥ 0} sin x ≥ 0 : ‫ﻨﺠﺩ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ‬ [ ]S = 2kπ ; (2k + 1) π : ‫ﺤﻠﻭﻟﻬﺎ ﻤﻥ ﺍﻟﺸﻜل‬ k ∈ ] ‫ﺤﻴﺙ‬ Df = [2kπ ; (2k + 1) π] : ‫ﻭﻤﻨﻪ‬ k ∈ ] ‫ﻤﻊ‬ ] [k ∈ ] ‫ ﻤﻊ‬2kπ ; (2k + 1) π ‫ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬f ‫ﻭ ﺍﻟﺩﺍﻟﺔ‬ . f ′(x) = cosx : ‫ﻭ ﻤﻨﻪ‬ 2 sinx4) f ′( x) = cos 2 x 1 - sin x Df = {x ∈ \ : 1 - sin x ≠ 0} sin x = 1 ‫ ﺃﻱ‬1 - sin x = 0 ‫ﻜل ﺍﻟﻤﻌﺎﺩﻟﺔ‬ x = π + 2kπ , k∈] : ‫ﻭﻤﻨﻪ‬ 2 Df =\ - π + 2kπ / k ∈ ]    2 f ′(x) = -2 sin2x (1 - sin) - (-cos x) . cos 2x (1 - sinx)2 f ′(x) = -1 sin x cosx (1 - sin x) + cosx . cos 2x (1 - sinx)2 f ′(x) = cosx [-4 sin x (1 - sin x) + cos 2x] (1 - sinx)2

f ′(x) = ( )cosx -4 sinx + 4 sin2 x + 1 - 2 sin2 x (1 - sinx)2f ′(x) = ( )cosx 2 sin2 x - 4 sinx + 1 (1 - sinx)25) f ( x) = sin x - cos2xDf = {x ∈ \ / sin x - cos2x ≥ 0} ‫ ﺃﻱ‬sin x - cos2x ≥ 0 : ‫ﻨﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ‬ sin x - (1 - 2sin2 x) ≥ 0 sin x + 1 + 2sin2 x ≥ 0 2 sin2 x + sinx - 1 ≥ 0 . . . (1) 2z2 + z - 1 ≥ 0 : ‫ ﻨﺠﺩ‬sin x = z ‫ﻨﻀﻊ‬ ∆ = 9 ‫( = ∆ ﻭﻤﻨﻪ‬1)2 - 4 (-1) (2) z2 = -1 +3 = 1 , z1 = -1 +3 = -1 : ‫ﻭﻋﻠﻴﻪ‬ 4 2 4 2z 2 + z - 1 = 2 (z + 1)  z - 1 : ‫ﺇﺫﻥ‬  2 2 sin2 x + sinx - 1 = 2 ( sinx + 1)  sinx - 1 : ‫ﺇﺫﻥ‬  2  2 ( sinx + 1)  sinx - 1 ≥ 0 : ‫( ﺘﻜﺎﻓﺊ‬1) ‫ﻭﻋﻠﻴﻪ‬  2  sin x 1 + ≥ 0 : ‫ﻷﻥ‬ sin x - 1 ≥0 : ‫ﻭﻤﻨﻪ‬ 2x∈ π + 2kπ ; 5π + 2kπ  : ‫ﺇﺫﻥ‬ sin x ≥ 1  6 6  2 : ‫ﻭﻋﻠﻴﻪ‬

‫‪Df‬‬ ‫∈‬ ‫‪π‬‬ ‫; ‪+ 2kπ‬‬ ‫‪5π‬‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫‪‬‬ ‫‪,‬‬ ‫]∈‪k‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪ 6‬‬ ‫‪6‬‬ ‫‪‬‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫;‬ ‫‪5π‬‬ ‫‪+ 2kπ‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺘﻘل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬ ‫‪ 6‬‬ ‫‪6‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫)‪f ′(x‬‬ ‫=‬ ‫‪cosx + 2 sin2x‬‬ ‫‪2 sin x - cos2x‬‬‫)‪6‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪2-‬‬ ‫‪sin x‬‬ ‫‪4-‬‬ ‫‪cosx‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪ 4 - cosx > 0‬ﻓﺈﻥ ‪D f = \ :‬‬‫)‪f ′(x‬‬ ‫=‬ ‫‪-cosx‬‬ ‫‪(4 - cos) - sin x‬‬ ‫)‪(2 -sin x‬‬ ‫‪(4 - cosx)2‬‬‫)‪f ′(x‬‬ ‫=‬ ‫‪-cosx‬‬ ‫‪(4 - cos) - sin x‬‬ ‫)‪(2 -sin x‬‬ ‫‪(4 - cosx)2‬‬‫)‪f ′(x‬‬ ‫=‬ ‫‪-4 cosx - 2 sin x‬‬ ‫‪+1‬‬ ‫‪(4 - cosx)2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 8‬‬ ‫ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ ‪: f‬‬ ‫‪ -1‬ﻓﻲ ﺍﻟﻤﺠﺎل ‪ : -∞ ; -1‬ﻟﺩﻴﻨﺎ ‪ f ′ x > 0‬ﻭﻤﻨﻪ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ‪( ) ] ].‬‬ ‫‪ -2‬ﻓﻲ ﺍﻟﻤﺠﺎل ‪ : -1 ; 0‬ﻟﺩﻴﻨﺎ ‪ f ′ x ≥ 0‬ﻭﻤﻨﻪ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ‪( ) [ ].‬‬ ‫‪ -3‬ﻓﻲ ﺍﻟﻤﺠﺎل ‪ : 0 ; 2‬ﻟﺩﻴﻨﺎ ‪ f ′ x ≤ 0‬ﻭ ﻤﻨﻪ ‪ f‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ‪( ) [ ].‬‬ ‫‪ -4‬ﻓﻲ ﺍﻟﻤﺠﺎل ‪ : 2 ; 3‬ﻟﺩﻴﻨﺎ ‪ f ′ x ≤ 0‬ﻭﻤﻨﻪ ‪ f‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ‪( ) [ ].‬‬ ‫‪ -5‬ﻓﻲ ﺍﻟﻤﺠﺎل ∞‪ f ′ x ≥ 0 : 3 ; +‬ﻭﻤﻨﻪ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ‪( ) [ [.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 9‬‬

‫ﻓﻲ ﺍﻟﻤﺠﺎل ‪ f2 x < 0 : -1 ; 0‬ﻭﻟﺩﻴﻨﺎ ‪ f1‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ‪( ) [ ].‬‬ ‫ﻭ ﻓﻲ ﺍﻟﻤﺠﺎل ‪ f2 x < 0 : 0 ; 1‬ﻭ ﻟﺩﻴﻨﺎ ‪ f1‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻓﻲ ﺍﻟﻨﻘﻁﺔ] [ ) (‬ ‫)‪ M1(-1 ; 2‬ﻟﺩﻴﻨﺎ ‪ C1‬ﻴﻘﺒل ﻨﺼﻑ ﻤﻤﺎﺱ ﻴﻭﺍﺯﻱ ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ﻭ ﻟﺩﻴﻨﺎ) (‬‫‪ f2 -1 = 0‬ﻓﻲ ﺍﻟﻨﻘﻁﺔ )‪ M1(1 ; -2‬ﻓﺈﻥ ‪ C1‬ﻴﻘﺒل ﻨﺼﻑ ﻤﻤﺎﺱ ﻴﻭﺍﺯﻱ ﻤﺤﻭﺭ) ( ) (‬‫ﺍﻟﻔﻭﺍﺼل ﻭ ﻟﺩﻴﻨﺎ ‪ f2 1 = 0 :‬ﺇﺫﻥ ‪ C1‬ﻫﻭ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ ‪ f1‬ﻭ ‪( ) ( ) ( )C2‬‬ ‫ﺍﻟﺘﻤﺜل ﺍﻟﺒﻴﺎﻨﻲ ﻟﺩﺍﻟﺘﻬﺎ ﺍﻟﻤﺸﺘﻘﺔ ‪ f1′‬ﺇﺫﻥ ‪f2 = f1′ :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 10‬‬ ‫‪ -1‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ ‪[ [D f -1 ; +∞ :‬‬ ‫ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪( )f 0 = 1 : 0‬‬ ‫‪-2‬‬‫‪lim‬‬ ‫)‪f (h) - f (0‬‬ ‫=‬ ‫‪lim‬‬ ‫‪1+h -1‬‬ ‫‪h‬‬ ‫‪h‬‬‫‪h→0‬‬ ‫‪h→0‬‬ ‫‪= lim‬‬ ‫‪( 1 + h - 1) ( )1 + h + 1‬‬ ‫‪h→0‬‬ ‫‪( )h 1 + h + 1‬‬ ‫‪= lim 1 + h - 1 = lim‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫=‬ ‫‪1‬‬ ‫‪h→0 h  1 + h + 1 h→0‬‬ ‫‪+h‬‬ ‫‪2‬‬ ‫‪( )f ′ 0‬‬ ‫=‬ ‫‪1‬‬ ‫ﺇﺫﻥ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 0‬ﺤﻴﺙ‬ ‫‪2‬‬ ‫‪y=1+‬‬ ‫‪1‬‬ ‫‪x‬‬ ‫ﺃﻱ ‪:‬‬ ‫ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ﻋﻨﺩ ‪: O‬‬ ‫‪2‬‬ ‫)‪y = f (0) + f (0) × ( x - 0‬‬ ‫‪ -3‬ﺍﻟﺘﻘﺭﻴﺏ ﺍﻟﺘﺂﻟﻔﻲ ﻟﻠﺩﺍﻟﺔ ‪: f‬‬ ‫‪( )g‬‬‫‪x‬‬ ‫‪=1+‬‬ ‫‪1‬‬ ‫‪x‬‬ ‫ﺃﺤﺴﻥ ﺘﻘﺭﻴﺏ ﺘﺂﻟﻔﻲ ﻟﻠﺩﺍﻟﺔ ﻋﻨﺩ ‪ 0‬ﻫﻭ ﺍﻟﺩﺍﻟﺔ ‪ g‬ﺤﻴﺙ ‪:‬‬ ‫‪2‬‬

‫‪ -4‬ﺤﺴﺎﺏ ﺍﻟﻘﻴﻡ ﺍﻟﻤﻘﺭﺒﺔ ‪:‬‬‫= ‪1,00007‬‬ ‫‪1 + 0,00007‬‬ ‫‬ ‫‪1+‬‬ ‫‪1‬‬ ‫)‪(0,00007‬‬ ‫‪2‬‬ ‫‪ 1 + 0,000035  1,000035‬‬ ‫ﺇﺫﻥ ‪1,00005  1,000035 :‬‬‫= ‪0,999917‬‬ ‫‪1 - 0,000083‬‬ ‫‪=1-‬‬ ‫‪1‬‬ ‫)‪(0,000083‬‬ ‫‪2‬‬ ‫‪= 0,9999585‬‬ ‫‪ -5‬ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪: f‬‬‫[∞‪Df = [-1 ; +‬‬‫‪lim f ( x) = lim‬‬ ‫‪1+x =0‬‬‫>>‬‫‪x →−1‬‬ ‫‪x →−1‬‬‫‪lim f ( x) = lim‬‬ ‫∞‪1 + x = +‬‬‫∞‪x→+‬‬ ‫∞‪x→+‬‬ ‫= )‪f ′(x‬‬ ‫‪1‬‬ ‫ﻤﻥ ﺃﺠل ‪: x > -1‬‬ ‫‪1+x‬‬ ‫ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ -1‬ﻤﻥ ﺍﻟﻴﻤﻴﻥ ‪( )f −1 = 0 :‬‬‫‪lim f (h) - f ( ) = lim‬‬ ‫‪- 0 = lim h‬‬ ‫>‬ ‫>‬ ‫‪h‬‬ ‫>‪h‬‬‫‪hh→0‬‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫∞‪= lim 1 = +‬‬ ‫>‪h‬‬ ‫‪h→0‬‬‫ﻭﻋﻠﻴﻪ ‪ f‬ﻻ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ -1‬ﻭ ﺒﻴﺎﻥ ‪ C f‬ﻴﻘﺒل ﻨﺼﻑ ﻤﻤﺎﺱ ﻴﻭﺍﺯﻱ ﺤﺎﻤل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ) (‬ ‫ﻋﻠﻰ ﻴﻤﻴﻥ ﺍﻟﻌﺩﺩ‪. -1‬‬ ‫ﺍﻟﻔﺭﻭﻉ ﺍﻟﻼﻨﻬﺎﺌﻴﺔ ‪:‬‬ ‫‪( )lim‬‬ ‫‪fx‬‬ ‫ﻓﺈﻨﻨﺎ ﻨﺤﺴﺏ ‪:‬‬ ‫‪( )lim f‬‬‫‪x‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪= +∞ :‬‬ ‫‪x‬‬ ‫∞‪x→+‬‬ ‫∞‪x→+‬‬

‫ﻟﺩﻴﻨﺎ ‪:‬‬‫‪lim f ( x) = lim 1 + x‬‬‫∞‪x→+‬‬ ‫‪x‬‬ ‫∞‪x→+‬‬ ‫‪x‬‬‫‪= lim‬‬ ‫‪1+x . 1+x‬‬ ‫‪= lim‬‬ ‫‪1+x‬‬ ‫×‬ ‫‪1 =0‬‬ ‫∞‪x→+‬‬ ‫‪x 1+x‬‬ ‫∞‪x→+‬‬ ‫‪x‬‬ ‫‪1+x‬‬ ‫ﺇﺫﻥ ‪ C f‬ﻴﻘﺒل ﻓﺭﻉ ﺘﻘﺎﻁﻊ ﻤﻜﺎﻓﺊ ﺒﺎﺘﺠﺎﻩ ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ﻋﻨﺩ ∞‪( ). +‬‬ ‫ﻨﻘﻁ ﺍﻟﺘﻘﺎﻁﻊ ‪ :‬ﻤﻥ ﺃﺠل ‪ x = 0 :‬ﻓﺈﻥ ‪ y = 1 :‬ﻤﻥ ﺃﺠل ‪ y = 0‬ﻓﺈﻥ ‪. x = 1 :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 11‬‬ ‫ﺤﺴﺎﺏ ﺍﻟﺩﻭﺍل ﺍﻟﻤﺸﺘﻘﺔ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪ f x = xn :‬ﻭ ﻋﻠﻴﻪ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ \ ﺤﻴﺙ ‪( ):‬‬ ‫‪f ′( x ) = nxn-1‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f ′‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ \ ﺤﻴﺙ ‪( )f ′′ x = n (n -1) xn-2 :‬‬‫ﺍﻟﺩﺍﻟﺔ ‪ f ′′‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ \ ﺤﻴﺙ ‪( )f ′′′ x = n (n -1) (n - 2) xn-3 :‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f ′′′‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ \ ﺤﻴﺙ ‪:‬‬ ‫‪f (4) ( x ) = n (n -1) (n - 2) (n - 3) xn-4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 12‬‬ ‫ﺤﺴﺎﺏ ﺍﻟﺩﻭﺍل ﺍﻟﻤﺸﺘﻘﺔ ‪:‬‬

f ( x ) = cos (ax + b) ‫ﻟﺩﻴﻨﺎ‬ ( )f ′ x = a sin (ax + b) :‫ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ \ ﺤﻴﺙ‬f ‫ﺍﻟﺩﺍﻟﺔ‬ f ′(x) = a cos  ax +b + π : ‫ﻭﻤﻨﻪ‬  2 ( )f ′′ x = a2 sin  ax + b + π : ‫\ ﺤﻴﺙ‬ ‫ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬ f ′ ‫ﺍﻟﺩﺍﻟﺔ‬  2  f ′′( x ) = a2 cos (ax + b + π) : ‫ﻭﻤﻨﻪ‬ f ′′′( x ) = -a3 sin (ax + b + π) ‫ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬f ′′ ‫ﺍﻟﺩﺍﻟﺔ‬ f ′′′( x) = a3 sin  ax +b+ 3π   2  . 13‫ﺍﻟﺘﻤﺭﻴﻥ‬ y′′ = -  4π 2 y : ‫ﻭﻤﻨﻪ‬ y′′ + 16π2 y = 0 : ‫ﻟﺩﻴﻨﺎ‬  5  25 y = a cos 4π x + sin 4π x : ‫ﻭﻋﻠﻴﻪ ﺤﻠﻬﺎ ﻫﻭ‬ 5 5 1 = a cos0 + b sin0 : ‫ ﻭﻋﻠﻴﻪ‬y = 1 : x = 0 ‫ﻭﻟﺩﻴﻨﺎ ﻤﻥ ﺃﺠل‬ y′ = 0 : x= 5 : ‫ﻭ ﻤﻥ ﺃﺠل‬ a = 1 : ‫ﻭﻤﻨﻪ‬ 4 4πa 4π 4πb 4πx y′ = 5 sin 5 x + 5 cos 5 : ‫ﻟﻜﻥ‬ 0 = 4πa sinπ + 4πb cosπ : ‫ﻭﻤﻨﻪ‬ 5 5 4πb b = 0 ‫ﻭﻤﻨﻪ‬ 5 cosπ = 0 : ‫ﻭﻋﻠﻴﻪ‬

‫‪y = cos‬‬ ‫‪4πx‬‬ ‫‪x‬‬ ‫ﻭﻋﻠﻴﻪ ﺍﻟﺤل ﻫﻭ ‪:‬‬ ‫‪5‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 14‬‬ ‫‪ -1‬ﻨﺒﻴﻥ ﺃﻥ ‪f ′′( x ) + 4 f ( x ) - 2 = 0 :‬‬‫ﻟﺩﻴﻨﺎ ‪ f ′( x ) = -2sin x cosx‬ﻭﻤﻨﻪ ‪f ′( x ) = -sin2x :‬‬ ‫ﺇﺫﻥ ‪f ′′( x ) = -2 cos2x :‬‬‫ﻭﻋﻠﻴﻪ ‪f ′′( x) + 4 f ( x) - 2 = -2 cos2x + 4 cos2 x - 2 :‬‬ ‫‪= -2 (2cos2 x - 1) + 4 cos2 x - 2‬‬ ‫‪= -4 cos2 x + 2 + 4 cos2 x - 2‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪f ′′( x) + 4 f ( x) - 2 = 0‬‬ ‫‪ -2‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ‪:‬‬ ‫‪y′′ = -4y + 2‬‬ ‫ﻨﻔﺭﺽ ) ‪ y = f ( x‬ﻓﻨﺠﺩ ‪f ′′( x) + 4 f ( x ) - 2 = 0 :‬‬ ‫ﻭﻤﻥ ﺍﻟﺴﺅﺍل )‪( )f x = cos2 x : (1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 15‬‬ ‫‪ -1‬ﺤﺴﺎﺏ ﺍﻟﻤﺸﺘﻘﺎﺕ ﺍﻟﻤﺘﺘﺎﺒﻌﺔ ‪:‬‬‫‪f ′( x) = 4x3 - 18x2 + 24x + 24‬‬‫‪f ′′( x) = 12x2 - 36x + 24‬‬‫‪f ((3) x ) = 24x - 36‬‬‫‪f (4) ( x ) = 24‬‬‫‪f ((5) x ) = 0‬‬ ‫ﻨﻼﺤﻅ ﺃﻨﻪ ﻤﻥ ﺃﺠل ‪( )f (n) x = 0 : n ≥ 5‬‬ ‫‪ -2‬ﺩﺭﺍﺴﺔ ﺇﺸﺎﺭﺓ ‪f ′′( x ) = 12x2 - 36x + 24 : ( )f ′′ x‬‬

‫ﻭﻤﻨﻪ ‪( )f ′′( x) = 12 x2 - 3x + 2 :‬‬ ‫‪ f ′′( x ) = 0‬ﺘﻜﺎﻓﺊ ‪x2 - 3x + 2 = 0 :‬‬‫‪x2‬‬ ‫=‬ ‫‪3+1‬‬ ‫‪=2‬‬ ‫‪,‬‬ ‫‪x1‬‬ ‫=‬ ‫‪3-1‬‬ ‫‪=1‬‬ ‫ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪:‬‬ ‫‪∆ =1‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪−∞ 1‬‬ ‫∞‪2 +‬‬ ‫‪x‬‬ ‫)‪f ′′( x‬‬ ‫‪+‬‬ ‫‪-+‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f ′′‬ﺘﻨﻌﺩﻡ ﻋﻨﺩ ‪ 1‬ﻭ ‪ 2‬ﻤﻐﻴﺭﺓ ﻓﻲ ﻜل ﻤﺭﺓ ﺇﺸﺎﺭﺘﻬﺎ ﻭﻤﻥ ﻫﻨﺎﻙ ﻨﻘﻁﺘﻲ ﺍﻨﻌﻁﺎﻑ ‪:‬‬ ‫))‪M2 ( 2 ; f (2)) , M1 (1 ; f (1‬‬ ‫ﺤﻴﺙ ‪f ( 2) = 40 , f (7) :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 16‬‬ ‫‪ -1‬ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪: f‬‬‫[∞‪Df = [0 ; +‬‬‫∞‪lim f ( x) = lim x x = 0 ; lim f ( x) = lim x x = +‬‬ ‫>‬ ‫>‬ ‫∞‪x→+‬‬ ‫∞‪x→+‬‬‫‪x→0‬‬ ‫‪x→0‬‬ ‫‪x+ x‬‬ ‫‪2x‬‬ ‫= ‪] [( )f ′ x‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫∞‪0 ; +‬‬ ‫‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬‫)‪f ′(x‬‬ ‫=‬ ‫‪2x‬‬ ‫‪+x‬‬ ‫‪+‬‬ ‫‪3x‬‬ ‫=‬ ‫‪3‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪x‬‬ ‫‪2x‬‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ ‪ f ′ x > 0‬ﻭﻋﻠﻴﻪ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل) (‬ ‫[∞‪[0 ; +‬‬

‫ﺍﻹﻨﺸﺎﺀ ﺒﺂﻟﺔ ﺒﻴﺎﻨﻴﺔ ‪:‬‬ ‫‪ -2‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ‪2y′ = 3 x :‬‬‫‪y′‬‬ ‫=‬ ‫‪3x‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪2y′ = 3 x‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2‬‬‫ﻭﺒﺎﻟﺘﺎﻟﻲ ﺤﺴﺏ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺴﺎﺒﻘﺔ ﻓﺈﻥ ‪ f x‬ﺤل ﻟﻬﺫﻩ ﺍﻟﻤﻌﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ﻭﻤﻨﻪ ‪( ) ( )y = f x :‬‬ ‫ﺃﻱ ‪y = x x‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 17‬‬ ‫ﺘﻌﻴﻴﻥ ‪ a‬ﻭ ‪ b‬ﻭ‪c‬‬‫‪ f‬ﺤل ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ‪y′′ = 12x2 - 24x + 1 :‬‬ ‫ﺃﻱ ‪f ′′( x ) = x2 - 4x + 1 :‬‬ ‫ﻟﻜﻥ ‪f ′( x ) = 4ax3 + 3bx2 + 2cx :‬‬‫‪f ′′( x ) = 12ax2 + 6bx2 + 2cx‬‬‫‪a = 1‬‬ ‫‪12a = 12‬‬‫‪ 6b = -24‬ﺃﻱ ﺃﻥ ‪b = -4 :‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬‫‪c = 5‬‬ ‫‪2c = 10‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 18‬‬ ‫‪ (1‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ ‪{ }D f = \ - -1 :‬‬

‫∞‪] [ ] [Df = -∞ ; -1 ∪ -1 ; +‬‬ ‫‪ (2‬ﺩﺭﺍﺴﺔ ﺍﻟﺘﻐﻴﺭﺍﺕ ‪:‬‬‫‪lim f ( x) = lim‬‬ ‫‪1‬‬ ‫‪=0‬‬ ‫;‬ ‫∞‪lim f ( x) = -‬‬‫∞‪x→−‬‬ ‫∞‪x→−‬‬ ‫‪(1 + x)3‬‬ ‫<‬ ‫‪x →−1‬‬‫‪lim f ( x) = lim‬‬ ‫‪1‬‬ ‫‪=0‬‬ ‫;‬ ‫∞‪lim f ( x) = +‬‬‫∞‪x→+‬‬ ‫∞‪x→+‬‬ ‫‪(1 + x)3‬‬ ‫>‬ ‫‪x →−1‬‬‫)‪f (x‬‬ ‫=‬ ‫‪-3 (1 + x)2‬‬ ‫=‬ ‫‪-3‬‬ ‫‪(1 + x)6‬‬ ‫‪(1 + x)4‬‬‫ﻭﻋﻠﻴﻪ ‪ f ′ x < 0 :‬ﻭﻤﻨﻪ ‪ f‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ ‪ -∞ ; -1‬ﻭ) ( [ ]‬ ‫[∞‪]-1 ; +‬‬ ‫‪x‬‬ ‫‪−∞ -1‬‬ ‫∞‪+‬‬ ‫)‪f ′(x‬‬ ‫‪--‬‬ ‫)‪f (x‬‬ ‫∞‪0 +‬‬ ‫‪0‬‬ ‫∞‪−‬‬ ‫‪ -3‬ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ‪y = f (0) × ( x - 0) + f (0) :‬‬ ‫ﺤﻴﺙ ‪f ′(0) = -3 , f (0) = 1 :‬‬ ‫ﻭﻋﻠﻴﻪ ‪ y = -3x +1 :‬ﻭﻫﻲ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ∆ ‪( ).‬‬ ‫‪ -4‬ﺇﻨﺸﺎﺀ ) ∆ ( ﻭ ) ‪: ( γ‬‬ ‫‪y‬‬ ‫‪0,5‬‬ ‫‪x‬‬ ‫‪0 0,5‬‬

‫‪ -5‬ﺃﺤﺴﻥ ﺘﻘﺭﺏ ﺘﺂﻟﻔﻲ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻭﻫﻭ ﺍﻟﺩﺍﻟﺔ ‪x 6 -3x + 1 :‬‬ ‫‪ -6‬ﺤﺴﺎﺏ ﺍﻟﻘﻴﻡ ﺍﻟﻤﻘﺭﺒﺔ ‪:‬‬ ‫‪1‬‬ ‫=‬ ‫‪1‬‬ ‫‪ - 3 (0,001) + 1  0,997‬‬ ‫‪(1,001)3‬‬ ‫‪(1 + 0,001)3‬‬ ‫‪1‬‬ ‫‪ 0,997‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪(1,001)3‬‬ ‫‪1‬‬ ‫=‬ ‫‪1‬‬ ‫‪ - 3 (-0,001) + 1‬‬ ‫‪(0,999)3‬‬ ‫‪(1 - 0,001)3‬‬ ‫‪ 1,003‬‬ ‫‪1‬‬ ‫‪ 1,003‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪(0,999)3‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 19‬‬ ‫‪ -1‬ﺘﻌﻴﻴﻥ ‪: D f‬‬‫‪{ }Df = x ∈ \ : x2 - 3x + 2 ≠ 0‬‬ ‫ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ x2 - 3x + 2 = 0 :‬ﻫﻲ ‪. 2 ، 1 :‬‬ ‫}‪Df = \ - {1 ; 2‬‬ ‫‪ -2‬ﺘﻌﻴﻴﻥ ﺍﻷﻋﺩﺍﺩ ‪ a‬ﻭ ‪ b‬ﻭ ‪: c‬‬‫)‪f (x‬‬ ‫‪=a+‬‬ ‫‪b‬‬ ‫‪+‬‬ ‫‪c‬‬ ‫‪x -1‬‬ ‫‪x -2‬‬ ‫‪a‬‬ ‫‪(x‬‬ ‫‪-‬‬ ‫)‪1‬‬ ‫‪(x‬‬ ‫)‪- 2) + b (x - 2‬‬ ‫‪+‬‬ ‫‪c‬‬ ‫‪(x‬‬ ‫‪-‬‬ ‫)‪1‬‬‫‪f‬‬ ‫(‬ ‫)‪x‬‬ ‫=‬ ‫)‪(x - 1) (x - 2‬‬‫‪f‬‬ ‫)‪( x‬‬ ‫=‬ ‫‪ax 2‬‬ ‫‪-‬‬ ‫‪3ax‬‬ ‫‪+ 2a + bx +‬‬ ‫‪2b‬‬ ‫‪+ cx‬‬ ‫‪-‬‬ ‫‪c‬‬ ‫‪x2 - 3x - 2‬‬‫‪f‬‬ ‫)‪( x‬‬ ‫=‬ ‫‪a‬‬ ‫‪x2‬‬ ‫‪(b - 3a+c) x +‬‬ ‫‪2b‬‬ ‫‪-c‬‬ ‫‪x2 - 3x + 2‬‬

f (x) = x2 + 2x + 1 a = 1 x2 - 3x + 2 : ‫ ﻷﻥ‬b - 3a + c = 2 : ‫ﻭﻤﻨﻪ‬ 2a - 2b -c = 1 a = 1 a = 1 b = -4 : ‫ ﺃﻱ‬b + c = 5 : ‫ﻭﻋﻠﻴﻪ‬ b = 9 2b - c = -1 f (x) =1- 4 + 9 : ‫ﻭﻋﻠﻴﻪ‬ x -1 x -2 : f ‫ ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ‬-2• Df = ]-∞ ; 1[ ∪ ]1 ; 2[ ∪lim f ( x) lim 1- 4 + 9 =1 x -1 x -2x→−∞ x→−∞lim f ( x) lim 1- 4 + 9 = +∞ < < x -1 x -2x →1 x→1lim f ( x) lim 1- 4 + 9 = -∞ > > x -1 x -2x →1 x→1lim f ( x) lim 1- 4 + 9 = -∞ < < x -1 x -2x→2 x→2lim f ( x) lim 1- 4 + 9 = +∞ > > x -1 x -2x→2 x→2lim f ( x) lim 1- 4 + 9 =1 x -1 x -2x→+∞ x→+∞ f ′(x) = 4 - 9 ( x - 1)2 ( x - 2)2

‫)‪f ′(x‬‬ ‫=‬ ‫‪4‬‬ ‫‪( x - 2)2 - 9 ( x - 1)2‬‬ ‫‪( x - 1)2 ( x - 2)2‬‬‫)‪f ′(x‬‬ ‫‪[2 (x‬‬ ‫])‪- 2) - 3 ( x - 1‬‬ ‫‪[2 ( x - 2) + 3 ( x‬‬ ‫])‪- 1‬‬ ‫‪( x - 1)2‬‬ ‫‪( x - 2)2‬‬‫)‪f ′(x‬‬ ‫=‬ ‫)‪(− x - 1) - (5x - 7‬‬ ‫‪( x - 1)2 ( x - 2)2‬‬ ‫ﺇﺸﺎﺭﺓ ﺍﻟﻤﺸﺘﻕ ‪:‬‬ ‫‪x‬‬ ‫‪−∞ -1‬‬ ‫‪1‬‬ ‫‪7‬‬ ‫∞‪2 +‬‬ ‫‪+‬‬ ‫‪5‬‬ ‫)‪f ′(x‬‬ ‫‪-‬‬ ‫‪-‬‬ ‫‪--‬‬ ‫; ‪[ [1‬‬‫‪7‬‬ ‫‪5 ‬‬ ‫ﻭ‬ ‫‪-1 ; 1‬‬ ‫ﺇﺫﻥ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ‬ ‫∞‪] [ ] ]2 ; +‬‬‫‪7‬‬ ‫‪2‬‬ ‫ﻭ‬ ‫‪ 5‬‬ ‫;‬ ‫ﻭ‬ ‫‪-∞ ; -1‬‬ ‫ﻭﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻻﺕ‬ ‫ﺠﺩﻭل ﺍﻟﺘﻐﻴﺭﺍﺕ ‪:‬‬ ‫‪x −∞ -1‬‬ ‫‪1‬‬ ‫‪7‬‬ ‫∞‪2 +‬‬ ‫‪5‬‬‫‪f ′(x) - - + - -‬‬‫‪f (x) 1‬‬ ‫∞‪+‬‬ ‫‪f‬‬ ‫‪‬‬ ‫‪7‬‬ ‫‪‬‬ ‫∞‪+‬‬ ‫‪‬‬ ‫‪5‬‬ ‫‪‬‬ ‫)‪f (-1‬‬ ‫∞‪−‬‬ ‫∞‪−‬‬ ‫‪1‬‬‫‪f‬‬ ‫)‪(1‬‬ ‫‪=1‬‬ ‫‪-‬‬ ‫‪4‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪9‬‬ ‫‪2‬‬ ‫=‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪2‬‬ ‫=‪-3‬‬ ‫‪0‬‬ ‫‪-1 -‬‬ ‫‪-1 -‬‬

‫‪f‬‬ ‫‪‬‬ ‫‪7‬‬ ‫‪‬‬ ‫‪=1 -‬‬ ‫‪4‬‬ ‫‪+‬‬ ‫‪9‬‬ ‫‪=1-‬‬ ‫‪4‬‬ ‫‪9‬‬ ‫‪‬‬ ‫‪5‬‬ ‫‪‬‬ ‫‪2+‬‬ ‫‪-3‬‬ ‫‪7‬‬ ‫‪-1‬‬ ‫‪7‬‬ ‫‪-‬‬ ‫‪2‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬‫‪f‬‬ ‫‪‬‬ ‫‪7‬‬ ‫‪=1‬‬ ‫‪-‬‬ ‫‪4‬‬ ‫×‬ ‫‪5‬‬ ‫‪+9‬‬ ‫×‬ ‫‪5‬‬ ‫‪= 1 - 10 -15 = -24‬‬ ‫‪‬‬ ‫‪5 ‬‬ ‫‪2‬‬ ‫‪-3‬‬ ‫‪ – 3‬ﺘﻌﻴﻥ ﻤﻌﺎﺩﻻﺕ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ ﺍﻟﻤﻘﺎﺭﺒﺔ ‪:‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪ lim f (x) = 1 :‬ﻓﺈﻥ ‪ y = 1‬ﻤﻌﺎﺩﻟﺔ ﻤﺴﺘﻘﻴﻡ ﻤﻘﺎﺭﺏ ﻴﻭﺍﺯﻱ ‪( ). y′y‬‬ ‫∞‪x→ +‬‬‫‪( ) ( ).‬‬ ‫‪x′x‬‬ ‫ﻓﺈﻥ ‪ x = 1 :‬ﻤﻌﺎﺩﻟﺔ ﻤﺴﺘﻘﻴﻡ ﻤﻘﺎﺭﺏ ﻴﻭﺍﺯﻱ‬ ‫‪lim f‬‬ ‫‪x‬‬ ‫ﻭﺒﻤﺎ ﺃﻥ ∞‪= +‬‬ ‫‪x→1‬‬ ‫‪ -‬ﺩﺭﺍﺴﺔ ﺍﻟﻭﻀﻌﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ)‪ (C‬ﻭ ﺍﻟﻤﺴﺘﻘﻴﻡ ∆ ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪( ).y = 1 :‬‬‫)‪f (x‬‬ ‫=‪-1‬‬ ‫‪-4‬‬ ‫‪+‬‬ ‫‪9‬‬ ‫=‬ ‫)‪-4(x - 2) + 9 (x - 1‬‬ ‫‪x -1‬‬ ‫‪x -2‬‬ ‫)‪(x - 1) (x - 2‬‬‫‪f‬‬ ‫)‪( x‬‬ ‫=‬ ‫‪(x‬‬ ‫‪-‬‬ ‫‪x -1‬‬ ‫‪-‬‬ ‫)‪2‬‬ ‫‪1) (x‬‬ ‫‪x‬‬ ‫∞‪−‬‬ ‫‪1‬‬ ‫∞‪1 2 +‬‬ ‫‪5‬‬ ‫‪++‬‬ ‫‪5x - 1‬‬‫)‪( x - 1) (x - 2‬‬ ‫‪-+‬‬ ‫‪f (x) - y‬‬ ‫‪++ - +‬‬ ‫‪-+-+‬‬ ‫‪( )M1‬‬‫‪‬‬ ‫‪1‬‬ ‫;‬ ‫‪1‬‬ ‫‪‬‬ ‫∆‬ ‫ﻭﻋﻠﻴﻪ )‪ (C‬ﻭ‬ ‫‪‬‬ ‫‪5‬‬ ‫‪‬‬ ‫ﺘﺘﻘﺎﻁﻌﺎﻥ ﻓﻲ ﺍﻟﻨﻘﻁﺔ‬ ‫∞‪] [ ( )2 ; +‬‬ ‫ﻭ‬ ‫‪1‬‬ ‫‪; 1‬‬ ‫∆‬ ‫)‪ (C‬ﻓﻭﻕ‬ ‫‪ 5‬‬ ‫ﻓﻲ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ‬

‫‪] [ ( )1 ; 2‬‬‫‪‬‬ ‫∞‪-‬‬ ‫;‬ ‫‪1‬‬ ‫∆‬ ‫)‪ (C‬ﻴﻘﻊ ﺘﺤﺕ‬ ‫ﻭ‬ ‫‪‬‬ ‫‪5 ‬‬ ‫ﻓﻲ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ‬ ‫‪ -4‬ﺍﻟﺘﻘﺭﻴﺏ ﺍﻟﺘﺂﻟﻔﻲ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻋﻨﺩ ‪: 0‬‬ ‫ﻟﻨﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ﻋﻨﺩ ‪y = f ( 0) . ( x - 0) + f (0) :‬‬‫‪y‬‬ ‫=‬ ‫‪7‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫)‪f ′(0‬‬ ‫=‬ ‫‪7‬‬ ‫‪,‬‬ ‫‪f‬‬ ‫)‪(0‬‬ ‫=‬ ‫‪1‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪4‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪2‬‬ ‫‪7‬‬ ‫‪1‬‬ ‫‪x‬‬ ‫‪6‬‬ ‫‪4‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪2‬‬ ‫ﻭﻋﻠﻴﻪ ﺃﺤﺴﺏ ﺘﻘﺭﻴﺏ ﺘﺂﻟﻔﻲ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻫﻭ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﺂﻟﻔﻴﺔ ‪:‬‬ ‫‪ -5‬ﺇﻨﺸﺎﺀ )‪(C‬‬ ‫‪y‬‬ ‫‪25‬‬ ‫‪20‬‬ ‫‪15‬‬ ‫‪10‬‬ ‫‪5‬‬‫‪-5 -4 -3 -2 -1 0‬‬ ‫‪1 2 3 4 5x‬‬ ‫‪-5‬‬ ‫‪-10‬‬ ‫‪-15‬‬ ‫‪-20‬‬ ‫‪-25‬‬ ‫‪-30‬‬ ‫‪ (1 (II‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ ‪: g‬‬‫‪{ }Dg = x ∈ \ : x2 - 3 x + 2= 0‬‬

‫ﻨﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ x2 - 3 x + 2 = 0 :‬ﺃﻱ ‪x 2 - 3 x + 2 = 0 :‬‬ ‫ﻨﻭﻀﻊ ‪ x = z :‬ﻨﺠﺩ ‪ z2 -3z + 2 =0 :‬ﻭ ﺤﻠﻴﻬﺎ‬ ‫‪ z = 1‬ﺃﻭ ‪ z = 2‬ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪ x = 1 :‬ﺃﻭ ‪x = 2‬‬ ‫ﻭﻤﻨﻪ ‪ x = 1 :‬ﺃﻭ ‪ x = -1‬ﺃﻭ ‪ x = -2‬ﺃﻭ ‪x = 2‬‬ ‫}‪Df = \ - {-2 ; -1 ; 1 ; 2‬‬ ‫‪ -2‬ﻜﺘﺎﺒﺔ ‪ g x‬ﺩﻭﻥ ﺭﻤﺯ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ) (‬‫‪‬‬ ‫‪f‬‬ ‫(‬ ‫)‪x‬‬ ‫=‬ ‫‪x2+ 2 x + 1‬‬ ‫;‬ ‫[∞‪x ∈ ]0 ; +1[ ∪ ]1 ; 2[ ∪ ].... ; +‬‬‫‪‬‬ ‫‪x2- 3x + 2‬‬‫‪‬‬‫‪‬‬ ‫‪x2‬‬ ‫‪-2x +1‬‬‫‪‬‬ ‫‪f‬‬ ‫(‬ ‫)‪x‬‬ ‫=‬ ‫‪x2‬‬ ‫‪+ 3x + 2‬‬ ‫;‬ ‫[‪x ∈ ]-∞ ; -2[ ∪ ]-2 ; -1[ ∪ ]-1 ; 0‬‬ ‫‪ -3‬ﺩﺭﺍﺴﺔ ﺍﻻﺴﺘﻤﺭﺍﺭﻴﺔ ﻋﻨﺩ ‪: 0‬‬‫)‪g(0‬‬ ‫=‬ ‫‪1‬‬ ‫‪2‬‬‫‪lim g ( x) = lim‬‬ ‫‪x2‬‬ ‫‪+ 2x‬‬ ‫‪+1‬‬ ‫=‬ ‫‪1‬‬ ‫=‬ ‫)‪g(0‬‬ ‫>>‬ ‫‪x2‬‬ ‫‪+ 3x‬‬ ‫‪+2‬‬ ‫‪2‬‬‫‪x→0‬‬ ‫‪x→0‬‬‫‪lim g ( x) = lim‬‬ ‫‪x2‬‬ ‫‪+ 2x‬‬ ‫‪+1‬‬ ‫=‬ ‫‪1‬‬ ‫)‪= g(0‬‬ ‫<<‬ ‫‪x2‬‬ ‫‪+ 3x‬‬ ‫‪+2‬‬ ‫‪2‬‬‫‪x→0‬‬ ‫‪x→0‬‬ ‫ﺇﺫﻥ ‪ g‬ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪.0‬‬ ‫‪( )g 0 = 0‬‬ ‫‪ -‬ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪: 0‬‬‫‪lim‬‬ ‫‪g(h) - g(0) = lim‬‬ ‫‪ h2‬‬ ‫‪+ 2h‬‬ ‫‪+1‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫×‬ ‫‪1‬‬ ‫>‬ ‫‪‬‬ ‫‪+ 3h‬‬ ‫‪+2‬‬ ‫‪‬‬ ‫‪h‬‬ ‫>‪h‬‬ ‫‪‬‬ ‫‪h‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪‬‬‫‪x→0‬‬ ‫‪x→0‬‬ ‫‪2h2 + 4h + 2 - h2 + 3h - 2‬‬ ‫‪1‬‬ ‫‪2 h2 - 3h + 2‬‬ ‫‪h‬‬ ‫‪( )= lim‬‬ ‫×‬ ‫>‬ ‫‪x→0‬‬

‫‪= lim‬‬ ‫‪h2 + 7h‬‬ ‫×‬ ‫‪1 = lim‬‬ ‫‪h+7‬‬ ‫=‬ ‫‪7‬‬ ‫>‪h‬‬ ‫‪h2 - 3h + 2‬‬ ‫‪4‬‬ ‫‪x→0‬‬ ‫>‪( ) ( )2‬‬‫‪x→0‬‬ ‫‪h2 - 3h + 2‬‬ ‫‪2‬‬ ‫ﺇﺫﻥ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 0‬ﻤﻥ ﺍﻟﻴﻤﻴﻥ ‪.‬‬‫‪lim‬‬ ‫‪g(h) - g(0) = lim‬‬ ‫‪ h2‬‬ ‫‪+‬‬ ‫‪2h‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫×‬ ‫‪1‬‬ ‫<‬ ‫‪‬‬ ‫‪+‬‬ ‫‪3h‬‬ ‫‪+‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪h‬‬ ‫<‪h‬‬ ‫‪‬‬ ‫‪h‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪‬‬‫‪x→0‬‬ ‫‪x→0‬‬ ‫‪ 2h2 - 4h + 2 - h2 - 2h - 2 ‬‬ ‫‪1‬‬ ‫‪‬‬‫‪( )= lim‬‬ ‫‪‬‬ ‫×‬ ‫‪h‬‬ ‫‪‬‬ ‫<‬ ‫‪‬‬ ‫‪x→0‬‬ ‫‪2 h2 + 3h + 2‬‬ ‫‪h2 - 6h‬‬ ‫‪1‬‬ ‫‪( )= lim‬‬ ‫×‬ ‫‪h‬‬ ‫<‬ ‫‪2‬‬ ‫‪h2 + 3h + 2‬‬ ‫‪x→0‬‬ ‫‪h-6‬‬ ‫‪3‬‬ ‫‪( )= lim‬‬ ‫×‬ ‫‪-‬‬ ‫‪2‬‬ ‫<‬ ‫‪2‬‬ ‫‪h2 + 3h + 2‬‬ ‫‪x→0‬‬ ‫ﺇﺫﻥ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 0‬ﻤﻥ ﺍﻟﻴﺴﺎﺭ ﻟﻜﻥ ‪ f‬ﻏﻴﺭ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ‪.0‬‬ ‫‪ -4‬ﺩﺭﺍﺴﺔ ﺸﻔﻌﻴﺔ ﺍﻟﺩﺍﻟﺔ ‪: g‬‬ ‫ﻟﺩﻴﻨﺎ ﻤﻥ ﺃﺠل ﻜل ‪ x‬ﻤﻥ ‪− x ∈ \ : D f‬‬ ‫)‪g(−x‬‬ ‫=‬ ‫‪( − x )2‬‬ ‫‪+‬‬ ‫‪2‬‬ ‫‪-x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫)‪= g(x‬‬ ‫ﻭﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪3‬‬ ‫‪-‬‬ ‫‪x+‬‬ ‫‪2‬‬ ‫‪x2 -‬‬ ‫ﻭﻤﻨﻪ ‪ g‬ﺯﻭﺠﻴﺔ ‪.‬‬ ‫‪ -5‬ﺍﺴﺘﻨﺘﺎﺝ ℘ ﻤﻥ )‪( ): (C‬‬ ‫ﻟﺩﻴﻨﺎ ﻤﻥ ﺃﺠل ‪x ∈[0 ; 1[ ∪ ]1 ; 2[ ∪ ]2 ; +∞[ :‬‬ ‫)‪g(x) = f (x‬‬‫ﻭﻤﻨﻪ ℘ ﻴﻨﻁﺒﻕ ﻋﻠﻰ )‪. (C‬ﻭ ﺒﺎﻗﻲ ﺍﻟﺒﻴﺎﻥ ﻴﺴﺘﻨﺘﺞ ﺒﺎﻟﺘﻨﺎﻅﺭ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻷﻥ ﺍﻟﺩﺍﻟﺔ ‪( )g‬‬ ‫ﺯﻭﺠﻴﺔ‪.‬‬ ‫‪ -6‬ﺍﻟﻤﻨﺎﻗﺸﺔ ﺍﻟﺒﻴﺎﻨﻴﺔ ‪:‬‬

‫‪(m - 1) x2 - (3m + 2) x + 2m - 1 = 0‬‬‫‪mx2 - x2 - 3mx - 2x + 2m - 1 = 0‬‬‫‪( )m x2 - 3x + 2 = x2 + 2x + 1‬‬‫=‪m‬‬ ‫‪x2 + 2x + 1‬‬ ‫‪x2 - 3x + 2‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫=‪m‬‬ ‫‪( x + 1)2‬‬ ‫‪x2 - 3x + 2‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫)‪m = f (x‬‬ ‫ﻟﻤﺎ ‪ m ∈ -∞ ; 0 :‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪] [.‬‬ ‫ﻟﻤﺎ ‪ : m = 0‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺜﻼﺙ ﺤﻠﻭل ﺃﺤﺩﻫﻤﺎ ﻤﻀﺎﻋﻑ ‪.‬‬ ‫ﻟﻤﺎ ‪ : m ∈ 0 ; 1‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺃﺭﺒﻌﺔ ﺤﻠﻭل‪] [.‬‬ ‫∈ ‪ : m‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺃﺭﺒﻌﺔ ﺤﻠﻭل‪.‬‬ ‫; ‪1‬‬ ‫‪27 ‬‬ ‫ﻟﻤﺎ ‪2 ‬‬ ‫‪ :‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺜﻼﺙ ﺤﻠﻭل ﺃﺤﺩﻫﻤﺎ ﻤﻀﺎﻋﻑ‪.‬‬ ‫=‪m‬‬ ‫‪27‬‬ ‫ﻟﻤﺎ‬ ‫‪2‬‬ ‫∈ ‪ : m‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ‪.‬‬ ‫‪ 27‬‬ ‫;‬ ‫∞‪+‬‬ ‫‪‬‬ ‫ﻟﻤﺎ‬ ‫‪ 2‬‬ ‫‪‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 20‬‬ ‫‪ -1‬ﺩﺭﺍﺴﺔ ﺇﺸﺎﺭﺓ ﺍﻟﺩﺍﻟﺔ ‪: f‬‬ ‫[∞‪Df = ]-∞ ; -1[ ∪ ]-1 ; 1[ ∪ ]1 ; +‬‬

( )• lim f x = lim x2 - 4x + 4 = lim x2 =1 x→−∞ x→−∞ x2 - 1 x2 x→−∞( )• lim f x = lim x2 - 4x + 4 = lim x2 =1 x→+∞ x→+∞ x2 - 1 x2 x→+∞x −∞ -1 1 +∞x2 -1 + - + : ‫ﻷﻥ‬ : ‫ﻷﻥ‬( )lim fx = lim ( x - 2)2 = +∞ : ‫ﻷﻥ‬ < < x2 - 1x →−1 x →−1 ( x - 2)2 → 9 ( x2 - 1 >→ 0( )lim fx = lim ( x - 2)2 = -∞ > > x2 - 1x →−1 x →−1 ( x - 2)2 → 9 ( x2 - 1 <→ 0( )lim fx = lim ( x - 2)2 = -∞ < < x2 - 1x →1 x→1 ( x - 2)2 → 1 ( x2 - 1 <→ 0lim f ( x) = lim ( x - 2)2 = +∞ > > x2 - 1x →1 x→1

‫‪( x - 2)2 → 1‬‬ ‫ﻷﻥ ‪:‬‬ ‫‪( x2 - 1 >→ 0‬‬‫•‬ ‫)‪f ′(x‬‬ ‫=‬ ‫‪2 (x‬‬ ‫‪- 2) (x2 - 1) - 2x‬‬ ‫‪(x‬‬ ‫‪- 2)2‬‬ ‫‪(x2 - 1)2‬‬ ‫)‪f ′(x‬‬ ‫=‬ ‫‪2 (x‬‬ ‫)‪- 2‬‬ ‫‪ x2‬‬ ‫‪-1-x‬‬ ‫‪(x‬‬ ‫‪- 2)‬‬ ‫‪(x 2‬‬ ‫‪- 1)2‬‬ ‫)‪f ′(x‬‬ ‫=‬ ‫‪2 (x - 2) (2x‬‬ ‫)‪- 1‬‬ ‫‪(x2 - 1)2‬‬ ‫ﺇﺸﺎﺭﺓ ﺍﻟﻤﺸﺘﻕ ‪:‬‬ ‫‪x −∞ -1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫∞‪2 +‬‬ ‫‪2‬‬‫‪f ′(x) + + - - +‬‬‫∞‪[ [ ] [2 ; +‬‬ ‫‪ -1‬‬ ‫‪1‬‬ ‫ﻭ‬ ‫;‬ ‫‪2 ‬‬ ‫ﻭ‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻻﺕ ‪-∞ ; -1‬‬ ‫‪] ].‬‬ ‫‪1‬‬ ‫‪; 1‬‬ ‫‪1;2‬‬ ‫ﻭ‬ ‫‪ 2‬‬ ‫ﻭﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ‬

‫‪x‬‬ ‫‪−∞ -1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫ﺠﺩﻭل ﺍﻟﺘﻐﻴﺭﺍﺕ ‪:‬‬ ‫‪2‬‬ ‫)‪f ′( x‬‬ ‫‪-‬‬ ‫∞‪2 +‬‬ ‫‪++-‬‬ ‫∞‪+‬‬ ‫)‪f (x‬‬ ‫‪+‬‬ ‫∞‪+‬‬ ‫‪f‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪‬‬ ‫=‬ ‫‪−3‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫∞‪1 −‬‬ ‫‪−∞ f (0) = 0‬‬ ‫‪ -‬ﺍﻟﻔﺭﻭﻉ ﺍﻟﻼﻨﻬﺎﺌﻴﺔ ﻭ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ ﺍﻟﻤﻘﺎﺭﺒﺔ‬ ‫ﻫﻨﺎﻙ ‪ 6‬ﻓﺭﻭﻉ ﻻ ﻨﻬﺎﺌﻴﺔ ﻭ ﺜﻼﺙ ﻤﺴﺘﻘﻴﻤﺎﺕ ﻤﻘﺎﺭﺒﺔ ﻤﻌﺎﺩﻟﺘﻬﺎ ‪y = 1 , x = 1 , x = -1‬‬ ‫‪ -‬ﺩﺭﺍﺴﺔ ﺍﻟﻭﻀﻌﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ )‪ (C‬ﻭ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﻤﻘﺎﺭﺏ ﺍﻷﻓﻘﻲ ‪y = 1 :‬‬‫)‪f (x‬‬ ‫=‪-1‬‬ ‫‪(x - 2)2‬‬ ‫=‪-1‬‬ ‫‪x2‬‬ ‫‪- 4x + 4 - x2‬‬ ‫‪+1‬‬ ‫‪x2 - 1‬‬ ‫‪x2 - 1‬‬ ‫‪-4x‬‬ ‫‪+5‬‬‫‪f‬‬ ‫)‪(x‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫=‬ ‫‪x2‬‬ ‫‪-1‬‬ ‫‪x‬‬ ‫‪−∞ -1‬‬ ‫‪1‬‬ ‫‪5‬‬ ‫∞‪+‬‬ ‫‪4‬‬ ‫)‪f ′( x‬‬ ‫‪+‬‬ ‫‪-‬‬ ‫‪+‬‬ ‫‪+‬‬‫‪−4x + 5‬‬ ‫‪+++-‬‬‫‪f (x)−1‬‬ ‫‪+-+-‬‬ ‫‪( )M‬‬ ‫‪5‬‬ ‫‪; 1‬‬ ‫∆‬ ‫)‪ (C‬ﻴﻘﻁﻊ‬ ‫‪ 4‬‬ ‫ﻓﻲ ﺍﻟﻨﻘﻁﺔ‬ ‫; ‪] [ ( )1‬‬‫‪5‬‬ ‫‪4 ‬‬ ‫ﻭ‬ ‫‪-∞ ; -1‬‬ ‫ﻓﻲ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ‬ ‫∆‬ ‫)‪ (C‬ﻴﻘﻁﻊ ﻓﻭﻕ‬

‫‪] [ ( ) 5‬‬‫‪‬‬ ‫‪‬‬‫‪ 4‬‬‫;‬ ‫∞‪+‬‬ ‫ﻭ‬ ‫‪-1 ; 1‬‬ ‫ﻓﻲ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ‬ ‫∆‬ ‫)‪ (C‬ﻴﻘﻁﻊ ﺘﺤﺕ‬ ‫‪ -‬ﺃﻨﺸﺄ )‪ (C‬ﺒﺂﻟﺔ ﺒﻴﺎﻨﻴﺔ ‪:‬‬ ‫‪y‬‬ ‫‪5‬‬ ‫‪4‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪1‬‬‫‪-6 -5 -4 -3 -2 -1 0‬‬ ‫‪1 2 3 4 5x‬‬ ‫‪-1‬‬ ‫‪-2‬‬ ‫‪-3‬‬ ‫‪-4‬‬ ‫‪-5‬‬ ‫‪ -‬ﺍﻟﻤﻨﺎﻗﺸﺔ ﺍﻟﺒﻴﺎﻨﻴﺔ ‪:‬‬‫‪(m - 1) x2 + 4x - m - 4 = 0‬‬‫‪m x2 - x2 + 4x - m - 4 = 0‬‬‫‪m (x2 - 1) = x2 - 4x + 4‬‬‫‪m (x2 - 1) = (x - 2)2‬‬

‫‪m‬‬ ‫‪(x 2‬‬ ‫)‪- 1‬‬ ‫‪x2‬‬ ‫‪-1‬‬ ‫ﻭﻤﻨﻪ ) ‪. m = f ( x‬‬ ‫‪ x‬ﻟﻤﺎ ‪ : m ∈ -∞ ; -3‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ‪] [.‬‬ ‫‪ x‬ﻟﻤﺎ ‪ : m = -3‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻤﻀﺎﻋﻑ‪.‬‬ ‫‪ x‬ﻟﻤﺎ ‪ : m ∈ -3 ; 0‬ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭل‪] [.‬‬ ‫‪ x‬ﻟﻤﺎ ‪ : m = 0‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻤﻀﺎﻋﻑ‪.‬‬ ‫‪ x‬ﻟﻤﺎ ‪ : m ∈ 0 ; 1‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪] [.‬‬ ‫‪ x‬ﻟﻤﺎ ‪ : m = 1‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻭﺤﻴﺩ ‪.‬‬ ‫‪ x‬ﻟﻤﺎ ∞‪ : m ∈ 1 ; +‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ ‪] [.‬‬ ‫‪ (2‬ﻜﺘﺎﺒﺔ ‪ g x‬ﺩﻭﻥ ﺭﻤﺯ ﺍﻟﻘﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ) (‬ ‫‪‬‬ ‫‪g‬‬ ‫(‬ ‫‪x‬‬ ‫)‬ ‫=‬ ‫‪(x - 2)2‬‬ ‫‪;x‬‬ ‫∈‬ ‫[‪[0 ; 1‬‬ ‫∪‬ ‫[∞‪]1 ; +‬‬ ‫‪‬‬ ‫‪x2 - 1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪(-x - 2)2‬‬ ‫‪‬‬ ‫‪g‬‬ ‫(‬ ‫)‪g‬‬ ‫=‬ ‫‪x2 - 1‬‬ ‫‪;x‬‬ ‫∈‬ ‫[‪[-∞ ; -1‬‬ ‫∪‬ ‫[‪]-1 ; 0‬‬ ‫‪ -‬ﺩﺭﺍﺴﺔ ﺍﻻﺴﺘﻤﺭﺍﺭﻴﺔ ﻋﻨﺩ ‪( )g 0 = -4 : 0‬‬‫‪lim g ( x) = lim‬‬ ‫‪(x - 2)2‬‬ ‫)‪= - 4 = g(0‬‬‫<<‬ ‫‪x2 - 1‬‬‫‪x→0‬‬ ‫‪x→0‬‬‫‪lim g ( x) = lim‬‬ ‫‪(x - 2)2‬‬ ‫)‪= - 4 = g(0‬‬‫>>‬ ‫‪x2 - 1‬‬‫‪x→0‬‬ ‫‪x→0‬‬ ‫ﻭﻤﻨﻪ ‪ g‬ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪. 0‬‬ ‫‪ -‬ﺩﺭﺍﺴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪. 0‬‬

‫)‪lim g(h)- g(0‬‬ ‫‪(-h‬‬ ‫‪- 2)2‬‬ ‫‪+‬‬ ‫‪4‬‬ ‫‪h2‬‬ ‫‪+4h+ 4+ 4h2‬‬ ‫‪-‬‬ ‫‪4‬‬ ‫‪h2‬‬ ‫‪-4‬‬ ‫)‪h (4h2 - 1‬‬ ‫=‬ ‫‪lim‬‬ ‫=‬ ‫‪lim‬‬ ‫<‬ ‫‪h‬‬ ‫<‬ ‫‪h‬‬ ‫<‬‫‪x→0‬‬ ‫‪x→0‬‬ ‫‪x→0‬‬ ‫‪= lim‬‬ ‫‪5h2 - 4h‬‬ ‫‪= lim‬‬ ‫‪5h + 4‬‬ ‫‪= -4‬‬ ‫<‬ ‫)‪h (h2 - 1‬‬ ‫<‬ ‫‪h2 - 1‬‬ ‫‪x→0‬‬ ‫‪x→0‬‬ ‫ﻭﻤﻨﻪ ‪ g‬ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ﻴﺴﺎﺭ ‪0‬‬‫)‪lim g(h)- g(0‬‬ ‫‪(h -‬‬ ‫‪2)2‬‬ ‫‪+‬‬ ‫‪4‬‬ ‫‪h2‬‬ ‫‪+‬‬ ‫‪4h + 4‬‬ ‫‪+‬‬ ‫‪4h2‬‬ ‫‪-4‬‬ ‫>‬ ‫‪h2‬‬ ‫‪-4‬‬ ‫‪h (4h2‬‬ ‫‪-‬‬ ‫)‪1‬‬ ‫=‬ ‫‪lim‬‬ ‫=‬ ‫‪lim‬‬‫‪hx → 0‬‬ ‫>‬ ‫‪h‬‬ ‫>‬ ‫‪x→0‬‬ ‫‪x→0‬‬ ‫‪= lim‬‬ ‫‪5h2 - 4h‬‬ ‫‪= lim‬‬ ‫‪5h + 4‬‬ ‫‪=4‬‬ ‫>‬ ‫)‪h (h2 - 1‬‬ ‫>‬ ‫‪h2 - 1‬‬ ‫‪x→0‬‬ ‫‪x→0‬‬ ‫ﻭﻤﻨﻪ ‪ g‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ﻴﻤﻴﻥ ‪. 0‬‬ ‫ﻟﻜﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻻ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪. 0‬‬ ‫‪ -‬ﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ ‪g ( x) = f ( x ) : [0 , 1[ ∪ ] [1 ; +∞ :‬‬ ‫‪ -‬ﺩﺭﺍﺴﺔ ﺸﻔﻌﻴﺔ ﺍﻟﺩﺍﻟﺔ ‪: g‬‬ ‫ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪ x‬ﻤﻥ ‪ Dg‬ﻟﺩﻴﻨﺎ ‪- x ∈ Dg :‬‬ ‫‪( )( ) x - 2 2‬‬ ‫ﻭﻟﺩﻴﻨﺎ ‪ g − x = (- x)2 - 1 = g( x) :‬ﻭﻤﻨﻪ ‪ g‬ﺯﻭﺠﻴﺔ ‪.‬‬ ‫‪ -‬ﺍﺴﺘﻨﺘﺎﺝ ‪( ): C′‬‬ ‫ﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ ∞‪ C′ : 0 ; 1 ∪ 1 ; +‬ﻴﻨﻁﺒﻕ ﻋﻠﻰ )‪ (C‬ﻓﻲ ﺍﻟﻤﺠﺎل[ ] [ [ ) (‬ ‫‪ ℘ : -1 ; 1‬ﻴﻨﺎﻅﺭ )‪ (C‬ﺒﺎﻟﻨﺴﺒﺔ ﻟﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل[ ] ) (‬ ‫‪ -4‬ﺘﻌﻴﻴﻥ ‪: D f‬‬‫‪{ }Df = x ∈ \ : sin2 x - 1 ≠ 0‬‬ ‫ﻨﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ sin2 x - 1 = 0 :‬ﺃﻱ )‪(sin x - 1) ( sin x + 1‬‬

sin x = -1 ‫ ﺃﻭ‬sin x = 1 ‫ﻭﻤﻨﻪ‬ k ∈ ... ‫ﻤﻊ‬ x = - π + 2kπ ‫ﺃﻭ‬ x= π + 2kπ : ‫ﻭﻋﻴﻪ‬ 2 2 Dα =\ - π + 2kπ ; - π + 2kπ / k ∈ ]   2   2  : ‫ ﻤﺭﻜﺏ ﺩﺍﻟﺘﻴﻥ‬α ‫ ﺘﺒﻴﺎﻥ ﺃﻥ‬- ( )f : x → f x : ‫ ﻩ ﻤﺭﻜﺏ ﺍﻟﺩﺍﻟﺘﻴﻥ‬α g : x → sinx α( x) = ( fog) ( x) = f ( g ( x)) = f (sin x) : ‫ﺃﻥ‬ α( x) = (sin x - 2 )2 sin2 x - 1 α = fog : ‫ﺇﺫﻥ‬ : α ‫ ﺤﺴﺎﺏ ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬-α′( x) = g′( x) . f ′ g( x)α′( x) = 2 (sin x - 2 ) (2sin x -2) (sin2 x - 1)2α′( x) = 2 cosx (sin x - 2 ) (2sin x -1) (sin2 x - 1)2 0=α+ . 21‫ﺍﻟﺘﻤﺭﻴﻥ‬ : γ ‫ ﻭ‬β ‫ ﻭ‬α ‫ ﺘﻌﻴﻴﻥ‬-1 γ : ‫ ﻭﻋﻠﻴﻪ‬f (0) = 0 : ‫ ﻓﺈﻥ‬O ∈ (Γ) ‫ﺒﻤﺎ ﺃﻥ‬ f (1) = -1 ( ): ‫ ﻓﺈﻥ‬A ∈ Γ ‫ﻭﺒﻤﺎ ﺃﻥ‬ -1 = α + β + γ :‫ﻭﻤﻨﻪ‬

‫‪α + γ = 0‬‬ ‫‪‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪α‬‬ ‫‪+‬‬ ‫‪β + γ = -1‬‬‫‪( )f ′ 0‬‬ ‫=‬ ‫‪-‬‬ ‫‪3‬‬ ‫ﻭﻋﻠﻴﻪ‬ ‫‪-‬‬ ‫‪3‬‬ ‫ﻭﻟﺩﻴﻨﺎ ﻤﻌﺎﻤل ﺘﻭﺠﻴﻪ ﺍﻟﻤﻤﺎﺱ ﻋﻨﺩ ‪ 0‬ﻴﺴﺎﻭﻱ‬ ‫‪4‬‬ ‫‪4‬‬‫)‪f ′(0‬‬ ‫=‬ ‫‪β‬‬ ‫‪ f ′( x ) = 2‬ﻭﻤﻨﻪ‬ ‫‪β‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫‪2γ‬‬ ‫‪βx + γ‬‬ ‫ﻭﻋﻠﻴﻪ ‪-6 γ = 4β :‬‬ ‫‪β‬‬ ‫=‬ ‫‪-‬‬ ‫‪3‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪2γ‬‬ ‫‪4‬‬ ‫ﺃﻱ ‪4β + 6 γ = 0 :‬‬ ‫)‪α + γ = 0 . . . (1‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫)‪α + β + γ = -1 . . . (2‬‬ ‫)‪2β + 3 γ = 0 . . . (3‬‬‫ﻤﻥ )‪ γ = -α (1‬ﻭﻋﻠﻴﻪ ‪ γ = α2 :‬ﻤﻊ ‪α < 0‬‬‫ﻤﻥ )‪ β + γ = -α - 1 (2‬ﺃﻱ ‪β + γ + 1= -α‬‬ ‫ﻭﻤﻨﻪ‪ β + γ + 1 2 = α2 :‬ﻷﻥ ‪( )α < 0 :‬‬ ‫ﻭﻋﻠﻴﻪ ‪β + γ + 2 β + γ + 1 = α2 :‬‬ ‫ﻭﻤﻨﻪ ‪β + α2 + 2 β + γ + 1 = α2 :‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪2 β + γ = -β - 1 :‬‬‫ﻭﺒﺎﻟﺘﺭﺒﻴﻊ ‪ 4(β + γ ) = β2 + 2β + 1 :‬ﻤﻊ ‪β < -1‬‬ ‫ﺇﺫﻥ ‪β2 - 2β + 1 = 4γ . . .(4) :‬‬‫ﻤﻥ )‪ 3 γ = -2β (3‬ﻭﻤﻨﻪ ‪(β < 0) 9γ = 4β2 :‬‬

‫ﺇﺫﻥ ‪:‬‬ ‫‪β2‬‬ ‫‪-‬‬ ‫‪2β‬‬ ‫‪+1‬‬ ‫=‬ ‫‪16‬‬ ‫‪β2‬‬ ‫ﻭﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ )‪ (4‬ﻨﺠﺩ ‪:‬‬ ‫‪γ‬‬ ‫=‬ ‫‪4‬‬ ‫‪β2‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪9β2 - 18β + 9 = 162β2‬‬ ‫ﻭﻋﻠﻴﻪ ‪ 7β2 + 18β - 9 = 0 :‬ﻟﺩﻴﻨﺎ ‪∆′ = 144 :‬‬ ‫= ‪β2‬‬ ‫‪-9 - 12‬‬ ‫‪= -3‬‬ ‫= ‪ β1‬؛‬ ‫‪-9 - 12‬‬ ‫‪= -3‬‬ ‫ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪γ‬‬ ‫=‬ ‫‪4 (-3)2‬‬ ‫‪=4‬‬ ‫ﻭﻋﻠﻴﻪ‪:‬‬ ‫‪) β = -3‬ﻷﻥ ‪(β < - 1‬‬ ‫ﻭﻤﻨﻪ‪:‬‬ ‫‪9‬‬ ‫ﻭ ﻋﻠﻴﻪ‪ α = - 4 = -2 :‬ﺇﺫﻥ ‪. f ( x ) = -2 + -3x + 4 :‬‬ ‫‪ -2‬ﺃ‪ -‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ‪: g‬‬‫‪• g ( x) = -2 + 4 - 3x , x < 1‬‬ ‫}‪ x < 1‬و ‪D1 = {x ∈ \ : 4 - 3x ≥ 0‬‬ ‫‪D1‬‬ ‫=‬ ‫‪‬‬ ‫‪x‬‬ ‫∈‬ ‫\‬ ‫‪:x‬‬ ‫≤‬ ‫‪3‬‬ ‫و‬ ‫‪x < 1‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪‬‬ ‫[‪D1 = ]-∞ ; 1‬‬‫‪• g(x) = x - 3 + x - 1‬‬ ‫}‪ x ≥ 1‬و ‪D2 = {x ∈ \ : x - 1 ≥ 0‬‬ ‫ﻭﻋﻠﻴﻪ ‪ D2 = 1 ; +∞ :‬ﻭﻤﻨﻪ ‪[ [Dg = D1 ∪ D2 = \ :‬‬ ‫ﺏ‪ -‬ﺩﺭﺍﺴﺔ ﺍﻻﺴﺘﻤﺭﺍﺭﻴﺔ ﻋﻨﺩ ‪: 1‬‬‫‪g(1) = 1 - 3 + 1 - 1 = -2‬‬‫)‪lim g ( x) = lim x - 3 + x - 1 = -2 = g(1‬‬‫>>‬‫‪x →1‬‬ ‫‪x→1‬‬ ‫ﺇﺫﻥ ‪ g‬ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪ 1‬ﻤﻥ ﺍﻟﻴﻤﻴﻥ‬‫‪lim g ( x) = lim - 2 + 4 - 3x = -1‬‬‫<<‬‫‪x →1‬‬ ‫‪x→1‬‬

‫ﻭﻋﻠﻴﻪ ‪ lim g x ≠ g 1 :‬ﻭ ﻋﻠﻴﻪ ‪ g‬ﻏﻴﺭ ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪ 1‬ﻤﻥ ﺍﻟﻴﺴﺎﺭ ؛ ﻭ ﻋﻠﻴﻪ ‪( ) ( )g‬‬ ‫<‬ ‫‪x →1‬‬ ‫ﻏﻴﺭ ﻤﺴﺘﻤﺭﺓ ﻋﻨﺩ ‪. 1‬‬ ‫‪ -‬ﺩﺭﺍﺴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪: 1‬‬‫•‬ ‫‪lim‬‬ ‫‪g(1 + h) - g(1) = lim‬‬ ‫‪1+h-3+‬‬ ‫‪1+h-1 +2‬‬ ‫>‬ ‫>‪h‬‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫‪= lim‬‬ ‫‪h + h = lim‬‬ ‫‪1‬‬ ‫∞‪+ 1 = +‬‬ ‫>‬ ‫>‪h‬‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫ﻭﻋﻠﻴﻪ ‪ g‬ﻻ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 1‬ﻤﻥ ﺍﻟﻴﻤﻴﻥ‬‫‪• lim‬‬ ‫‪g(1 + h) - g(1) = lim‬‬ ‫‪-2 +‬‬ ‫‪4 - 3 (1 + h) + 2‬‬ ‫<‬ ‫<‪h‬‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫‪h→0‬‬ ‫‪= lim‬‬ ‫‪1 - 3h‬‬ ‫∞‪= -‬‬ ‫<‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫ﻭﻤﻨﻪ ‪ f‬ﻻ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 1‬ﻤﻥ ﺍﻟﻴﺴﺎﺭ ‪.‬‬ ‫ﺠـ ‪ -‬ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ ‪: g‬‬‫‪lim‬‬ ‫= )‪g(x‬‬ ‫‪lim‬‬ ‫‪-2 +‬‬ ‫∞‪4 - 3x  = +‬‬‫∞‪h→−‬‬ ‫∞‪h→−‬‬‫‪lim g( x) = lim x - 3 +‬‬ ‫∞‪x - 1 = +‬‬‫∞‪h→+‬‬ ‫∞‪h→+‬‬ ‫‪g′( x) = 2‬‬ ‫‪-3‬‬ ‫ﻤﻥ ﺃﺠل ‪: x < 1‬‬ ‫‪1 - 3x‬‬ ‫ﺇﺫﻥ ‪ g′ x < 0‬ﻭﻋﻠﻴﻪ ‪ g‬ﻤﺘﻨﺎﻗﺹ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ‪] ] ( ). -∞ ; 1‬‬ ‫‪g′( x) = 1 + 2‬‬ ‫‪-3‬‬ ‫‪ x‬ﻤﻥ ﺃﺠل ‪: x > 1‬‬ ‫‪x -1‬‬ ‫ﻭﻤﻨﻪ ‪ g′ x > 0‬ﻭﻋﻠﻴﻪ ‪ g‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ∞‪[ [ ( )1 ; +‬‬

‫‪x‬‬ ‫‪−∞ 1‬‬ ‫∞‪+‬‬‫)‪g′( x‬‬ ‫‪-‬‬ ‫∞‪−∞ +‬‬ ‫‪+‬‬‫∞‪g( x) +‬‬ ‫‪2-‬‬ ‫∞‪+‬‬ ‫‪ -5‬ﺩﺭﺍﺴﺔ ﺍﻟﻔﺭﻭﻉ ﺍﻟﻼﻨﻬﺎﺌﻴﺔ ﻭ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ ﺍﻟﻤﻘﺎﺭﺒﺔ ‪:‬‬ ‫ﻫﻨﺎﻙ ﻓﺭﻋﻴﻥ ﻻﻨﻬﺎﺌﻴﻴﻥ ‪.‬‬‫‪( )lim‬‬‫‪g‬‬ ‫‪x‬‬ ‫‪= lim‬‬ ‫‪-2 +‬‬ ‫‪4 - 3x‬‬ ‫‪= lim‬‬ ‫‪-2‬‬ ‫‪+‬‬ ‫‪4 - 3x‬‬ ‫‪x‬‬ ‫∞‪x→−‬‬ ‫‪x‬‬ ‫∞‪x→−‬‬ ‫‪x‬‬ ‫‪x‬‬‫∞‪x→−‬‬ ‫‪= lim‬‬ ‫‪-2‬‬ ‫‪+‬‬ ‫‪4 - 3x . 4 - 3x‬‬ ‫∞‪x→−‬‬ ‫‪x‬‬ ‫‪x 4 - 3x‬‬ ‫‪= lim‬‬ ‫‪-2‬‬ ‫‪+‬‬ ‫‪4‬‬ ‫‪- 3x‬‬ ‫×‬ ‫‪1 =0‬‬ ‫∞‪x→−‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫‪4 - 3x‬‬ ‫‪1‬‬ ‫ﻭ‪→0‬‬ ‫‪4 - 3x‬‬ ‫‪→ -3‬‬ ‫ﻭ‬ ‫‪−2‬‬ ‫ﻷﻥ ‪→ 0 :‬‬ ‫‪4 - 3x‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫ﺇﺫﻥ )‪ (C‬ﻴﻘﺒل ﻓﺭﻉ ﻗﻁﻊ ﻤﻜﺎﻓﺊ ﻓﻲ ﺍﺘﺠﺎﻩ ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ﻋﻨﺩ ∞‪−‬‬‫‪( )lim‬‬‫‪g‬‬ ‫‪x‬‬ ‫‪= lim‬‬ ‫‪x -3+‬‬ ‫‪x -1‬‬ ‫‪x‬‬ ‫∞‪x→+‬‬ ‫‪x‬‬‫∞‪x→+‬‬ ‫‪= lim‬‬ ‫‪x-‬‬ ‫‪3‬‬ ‫‪+‬‬ ‫‪x -1‬‬ ‫‪x -1‬‬ ‫∞‪x→+‬‬ ‫‪x‬‬ ‫×‪x‬‬ ‫‪x -1‬‬ ‫‪= lim‬‬ ‫‪x -3‬‬ ‫‪+‬‬ ‫‪x -1‬‬ ‫×‬ ‫‪1 =1‬‬ ‫∞‪x→+‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫‪x -1‬‬ ‫‪1‬‬ ‫→‬ ‫‪0‬‬ ‫ﻭ‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫→‬ ‫‪1‬‬ ‫ﻭ‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪3‬‬ ‫→‬ ‫‪1‬‬ ‫ﻷﻥ ‪:‬‬ ‫‪x-‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫‪1‬‬‫‪lim‬‬ ‫)‪g( x‬‬ ‫‪-‬‬ ‫‪x ‬‬ ‫=‬ ‫‪lim‬‬ ‫‪-3 +‬‬ ‫∞‪x - 1 = +‬‬‫∞‪x→+‬‬ ‫∞‪x→+‬‬

‫ﻭﻤﻨﻪ )‪ (C‬ﻴﻘﺒل ﻓﺭﻉ ﻗﻁﻊ ﻤﻜﺎﻓﺊ ﺒﺎﺘﺠﺎﻩ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﺫﻱ‬ ‫ﻤﻌﺎﺩﻟﺘﻪ‪ y = x :‬ﻋﻨﺩ ∞‪. +‬‬ ‫ﻫـ ‪ -‬ﺘﻌﻴﻴﻥ ﻨﻘﻁ ﺘﻘﺎﻁﻊ )‪ (C‬ﻭ ∆ ‪( ):‬‬ ‫‪ x‬ﻤﻥ ﺃﺠل ‪ g ( x ) = x : x < 1‬ﺇﺫﻥ ‪-2 + 4 - 3x = x :‬‬ ‫‪4 - 3x = (x +2)2‬‬ ‫ﻭﻋﻠﻴﻪ ‪4 - 3x = x + 2 :‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪x ≥ -2‬‬ ‫ﻭﻋﻠﻴﻪ ‪ 4 - 3x = x2 + 4x + 4 :‬ﻭ ‪x ≥ -2‬‬ ‫ﺇﺫﻥ ‪ x2 + 7x = 0 :‬ﻭﻤﻨﻪ ‪ x = 0‬ﺃﻭ ‪x = -7‬‬ ‫ﺇﺫﻥ ‪ x = 0 :‬ﻷﻥ ‪ x ≥ -2‬ﻭ ‪x < 1‬‬ ‫ﻭﻤﻨﻪ ﻨﻘﻁ ﺍﻟﺘﻘﺎﻁﻊ ﻫﻲ )‪O (0 ; 0‬‬ ‫‪ x‬ﻤﻥ ﺃﺠل ‪g ( x ) = x : x ≥ 1‬‬ ‫ﺇﺫﻥ ‪ x - 3 + x - 1 = x :‬ﻭﻤﻨﻪ ‪x - 1 = 3 :‬‬ ‫ﺇﺫﻥ ‪ x - 1 = 9 :‬ﻭﻤﻨﻪ ‪x = 10 :‬‬ ‫ﻭﻋﻠﻴﻪ ‪ . g(10) = 10 :‬ﺇﺫﻥ ﻨﻘﻁ ﺍﻟﺘﻘﺎﻁﻊ ﻫﻲ )‪B (10 ; 10‬‬ ‫ﻭﻤﻨﻪ ‪. (C) ∩ ( D) = {O ; B} :‬‬ ‫ﻭ‪ -‬ﻜﺘﺎﺒﺔ ﻤﻌﺎﺩﻻﺕ ﺍﻟﻤﻤﺎﺴﺎﺕ ‪:‬‬ ‫‪y‬‬ ‫=‬ ‫‪-‬‬ ‫‪3‬‬ ‫‪x‬‬ ‫ﻋﻨﺩ ‪ y = g′(0) . (x - 0) + g(0) : O‬ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫) ‪g′ ( 0‬‬ ‫=‬ ‫‪-‬‬ ‫‪2‬‬ ‫ﻷﻥ‬ ‫ﻋﻨﺩ ‪y = g′( x1 ) ( x - x1 ) + g( x1 ) : B‬‬‫=‪y‬‬ ‫‪7‬‬ ‫)‪( x - 10‬‬ ‫‪+ 10‬‬ ‫)‪ y = g′(10) ( x - 10) + g(10‬؛‬ ‫‪6‬‬ ‫=‪y‬‬ ‫‪7‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪35‬‬ ‫‪+ 10‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪g′‬‬ ‫)‪(10‬‬ ‫=‬ ‫‪7‬‬ ‫ﻷن‬ ‫‪‬‬ ‫‪6‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪6‬‬ ‫‪‬‬

‫‪y‬‬ ‫=‬ ‫‪7‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪5‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪6‬‬ ‫‪3‬‬ ‫ﻱ‪ -‬ﺇﻨﺸﺎﺀ )‪: (C‬‬‫‪y‬‬‫‪14‬‬‫‪13‬‬‫‪12‬‬‫‪11‬‬‫‪10‬‬‫‪9‬‬‫‪8‬‬‫‪7‬‬‫‪6‬‬‫‪5‬‬‫‪4‬‬‫‪3‬‬‫‪2‬‬‫‪1‬‬‫‪-4 -3 -2 -1 0‬‬ ‫‪1 2 3 4 5 6 7 8 9 10 11 12 x‬‬ ‫‪-1‬‬ ‫‪-2‬‬

‫‪ – 3‬ﺍﻟﺩﻭﺍل ﺍﻷﺼﻠﻴﺔ‬ ‫ﺍﻟﻜﻔﺎﺀﺓ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ‬ ‫‪ -1‬ﺘﻌﻴﻴﻥ ﺩﺍﻟﺔ ﺃﺼﻠﻴﺔ ﻟﺩﺍﻟﺔ ﻤﺴﺘﻤﺭﺓ ﻋﻠﻰ ﻤﺠﺎل ‪.‬‬ ‫‪ - 2‬ﺘﻌﻴﻴﻥ ﺩﺍﻟﺔ ﺃﺼﻠﻴﺔ ﻟﺩﺍﻟﺔ ﻤﺄﻟﻭﻓﺔ ‪.‬‬ ‫‪ -3‬ﺘﻌﻴﻴﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻷﺼﻠﻴﺔ ﺍﻟﺘﻲ ﺘﺄﺨﺫ ﻗﻴﻤﺔ ‪ γ 0‬ﻤﻥ ﺃﺠل ﺍﻟﻘﻴﻤﺔ ‪ x0‬ﻟﻠﻤﺘﻐﻴﺭ‬ ‫ﺗﺼﻤﻴﻢ اﻟﺪرس‬ ‫ﺃﻨﺸﻁﺔ‬ ‫‪ – 1‬ﺘﻌﺭﻴﻑ‬ ‫‪ – 2‬ﻭﺠﻭﺩ ﺍﻟﺩﻭﺍل ﺍﻷﺼﻠﻴﺔ‬ ‫‪ – 3‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﺩﻭﺍل ﺍﻷﺼﻠﻴﺔ ﻟﺩﺍﻟﺔ ﻋﻠﻰ ﻤﺠﺎل‬‫‪ – 4‬ﺍﻟﺩﺍﻟﺔ ﺍﻷﺼﻠﻴﺔ ﺍﻟﺘﻲ ﺘﺄﺨﺫ ﺍﻟﻘﻴﻤﺔ ‪ y0‬ﻤﻥ ﺃﺠل ﺍﻟﻘﻴﻤﺔ ‪ x0‬ﻟﻠﻤﺘﻐﻴﺭ‬ ‫‪ – 5‬ﺠﺩﻭل ﺍﻟﺩﻭﺍل ﺍﻷﺼﻠﻴﺔ ﺍﻟﻤﺄﻟﻭﻓﺔ‬ ‫‪ – 6‬ﺍﻟﺒﺤﺙ ﻋﻥ ﺍﻟﺩﻭﺍل ﺍﻷﺼﻠﻴﺔ‬ ‫ﺘﻤـﺎﺭﻴﻥ ﻭ ﻤﺸﻜﻼﺕ‬ ‫ﺍﻟﺤـﻠــــــﻭل‬

‫ﺃﻨﺸﻁﺔ‬ ‫‪x2 + 2x‬‬ ‫ﺍﻟﻨﺸﺎﻁ ‪: 1‬‬ ‫ﻜﻤﺎ ﻴﻠﻲ ‪x2 + x + 1 2 :‬‬‫‪f‬‬‫‪( )( )،‬‬‫=‪x‬‬ ‫‪ α‬ﻭ ‪ β‬ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ‬ ‫‪ f‬ﻭ ‪ g‬ﺩﺍﻟﺘﺎﻥ ﻋﺩﺩﻴﺘﺎﻥ ﻤﻌﺭﻓﺘﺎﻥ ﻋﻠﻰ‬ ‫)‪g( x‬‬ ‫=‬ ‫‪αx‬‬ ‫‪+β‬‬ ‫‪1‬‬ ‫‪x2 +‬‬ ‫‪x+‬‬ ‫ﻋﻴﻥ ‪ α‬ﻭ ‪ β‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪: x‬‬ ‫) ‪ g′( x ) = f ( x‬؛ ﻤﺎﺫﺍ ﻨﺴﺘﻨﺘﺞ ؟‬ ‫ﺍﻟﺤل ‪:‬‬ ‫ﺘﻌﻴﻴﻥ ‪ α‬ﻭ ‪: β‬‬ ‫(‪g′‬‬ ‫)‪x‬‬ ‫=‬ ‫‪α‬‬ ‫(‬ ‫‪x2‬‬ ‫‪+‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫)‪1‬‬ ‫‪− (α x + β‬‬ ‫)‬ ‫(‬ ‫‪2x‬‬ ‫‪+‬‬ ‫)‪1‬‬ ‫(‬ ‫‪x2‬‬ ‫‪)+ x + 1 2‬‬ ‫)‬ ‫‪α‬‬ ‫‪x2‬‬ ‫‪+‬‬ ‫‪α‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪α‬‬ ‫‪− 2α x2 − α‬‬ ‫‪x‬‬ ‫‪−‬‬ ‫‪2β‬‬ ‫‪x‬‬ ‫‪−‬‬ ‫‪β‬‬ ‫‪x2 + x + 1 2‬‬ ‫(‪( )g′‬‬‫‪x‬‬ ‫=‬ ‫)‪x‬‬ ‫‪−α‬‬ ‫‪x2 −‬‬ ‫‪2β x + α‬‬ ‫‪−‬‬ ‫‪β‬‬ ‫‪x2‬‬ ‫‪+ x+1 2‬‬ ‫(‪( )g′‬‬ ‫=‬ ‫‪ −α = 1‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ ) ‪ g′( x ) = f ( x‬ﻓﺈﻥ ‪−2β = 2 :‬‬ ‫‪α − β = 0‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪ α = −1 :‬ﻭ ‪β = −1‬‬

‫= )‪g(x‬‬ ‫‪−x −1‬‬ ‫‪:‬‬ ‫ﻤﻨﻪ‬ ‫ﻭ‬ ‫‪x2 + x + 1‬‬ ‫ﻨﺴﺘﻨﺘﺞ ﺃﻥ ‪ g :‬ﺩﺍﻟﺔ ﺃﺼﻠﻴﺔ ﻟﻠﺩﺍﻟﺔ ‪. f‬‬ ‫ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫ﺍﻟﻨﺸﺎﻁ ‪: 2‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺘﺎﻥ ‪ g‬ﻭ ‪ h‬ﺍﻟﻤﻌﺭﻓﺘﺎﻥ ﻋﻠﻰ‬ ‫)‪h( x‬‬ ‫=‬ ‫‪4x2 − 5x + 10‬‬ ‫‪،‬‬ ‫)‪g( x‬‬ ‫=‬ ‫‪2x2‬‬ ‫‪x‬‬ ‫‪5‬‬ ‫‪2x2 − 3x + 5‬‬ ‫‪− 3x +‬‬ ‫ﻋﻴﻥ ﻜل ﻤﻥ ‪ g′‬ﻭ ‪ h′‬؛ ﻤﺎﺫﺍ ﻨﻼﺤﻅ ؟‬ ‫ﺤﻴﺙ ‪:‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ g‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬ ‫×‪( ( ) )g′( x) = 1‬‬ ‫‪2x2 − 3x + 5 − (4x − 3)× x‬‬ ‫‪2x2 − 3x + 5 2‬‬ ‫)‪x‬‬ ‫‪2x2‬‬ ‫‪− 3x‬‬ ‫‪+‬‬ ‫‪5 − 4x2 +‬‬ ‫‪3x‬‬ ‫‪2x2‬‬ ‫‪−‬‬ ‫‪3x + 5 2‬‬ ‫(‪( )g′‬‬ ‫=‬ ‫‪−2x2 + 5‬‬ ‫‪2x2 − 3x + 5 2‬‬ ‫= )‪( )g′( x‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ h‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ﺤﻴﺙ ‪:‬‬‫( ‪h′‬‬ ‫‪x‬‬ ‫)‬ ‫=‬ ‫)‪( 8 x-5‬‬ ‫(‬ ‫()‪2x2-3x + 5)-(4x-3‬‬ ‫‪4 x 2 -5‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫)‪10‬‬ ‫‪( )2x2-3x + 5 2‬‬ ‫‪16 x3 -24 x2‬‬ ‫‪+‬‬ ‫‪40x-10x2 +‬‬ ‫‪15 x-25-16 x 3‬‬ ‫‪2x2-3x + 5‬‬‫(‪( )h′‬‬‫)‪x‬‬ ‫=‬ ‫‪2‬‬ ‫‪+‬‬

‫‪20x2 − 40x + 12 x2 − 15x + 30‬‬ ‫‪( )2x2-3x + 5 2‬‬ ‫‪−2x2 + 5‬‬‫‪2x2 − 3x + 5 2‬‬‫= )‪( )h′( x‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫ﺍﻻﺴﺘﻨﺘﺎﺝ ﻟﺩﻴﻨﺎ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪: x‬‬‫)‪g′( x) = h′( x) = f ( x‬‬ ‫‪−2x2 + 5‬‬‫ﺤﻴﺙ ‪2 x2 − 3 x + 5 2 :‬‬‫= )‪( )f ( x‬‬‫ﻭ ﻨﻘﻭل ﺃﻥ ‪ g‬ﻭ ‪ h‬ﺩﺍﻟﺘﺎﻥ ﺃﺼﻠﻴﺘﺎﻥ ﻟﻨﻔﺱ ﺍﻟﺩﺍﻟﺔ ‪. f‬‬