c6-2sh-exp

‫ﻣﺬﻛﺮة رﻗﻢ ‪6 :‬‬ ‫ﻣﺬﻛﺮة رﻗﻢ‪ 6‬ﻓﻲ درس اﻟﺪوال اﻷﺳﻴﺔ‬

‫اﻷﺳﺘﺎذ ‪ :‬ﻋﺜﻤﺎﻧﻲ ﻧﺠﯿﺐ‬ ‫اﻷھﺪاف اﻟﻘﺪرات اﻟﻤﻨﺘﻈﺮة ﻣﻦ اﻟﺪرس ‪:‬‬

‫‪ (1‬ﺗﻌﺮﯾﻒ اﻟﺪاﻟﺔ اﻷﺳﯿﺔ اﻟﻨﺒﯿﺮﯾﺔ‬
‫اﻟﺪاﻟﺔ اﻷﺳﯿﺔ اﻟﻨﺒﯿﺮﯾﺔ ﯾﺮﻣﺰ ﻟﮭﺎ ب ‪ exp‬وھﻲ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ¡ ب ‪:‬‬

‫‪("x Î ¡);exp x = e x‬‬

‫‪ (2‬ﺧﺎﺻﯿﺔ ﻣﻘﺒﻮﻟﺔ‬

‫) ‪("y Ρ)("x Î]0;+¥[) , (x = ey Û ln(x ) = y‬‬

‫اﻟﺪاﻟﺔ ‪ exp‬ﺗﺰاﯾﺪﯾﺔ ﻗﻄﻌﺎ ﻋﻠﻰ ¡ ﯾﻌﻨﻲ ‪ex > ey Û x > y‬‬
‫‪ ex = ey Û x = y‬ﻟﻜﻞ ‪ x‬و ‪ y‬ﻣﻦ ¡‬

‫‪ex = 2 (3 ex-2 = e (2‬‬ ‫وﻟﺪﯾﻨﺎ‪ e0 = 1:‬و ‪e1 = e‬‬

‫ﻣﺜﺎل ‪:‬ﺣﻞ ﻓﻲ ¡ اﻟﻤﻌﺎدﻻت اﻟﺘﺎﻟﯿﺔ‪ex = 1 (1:‬‬
‫اﻟﺤﻞ‪ex = 1 Û ex = e0 (1 :‬‬
‫‪ x = 0 Û‬وﻣﻨﮫ ‪S = {0} :‬‬

‫‪ex-2 = e1 Û ex-2 = e (2‬‬
‫‪ x = 3 Û x - 2 = 1 Û‬وﻣﻨﮫ ‪S = {3} :‬‬

‫‪ x = ln 2 Û ex = 2 (3‬وﻣﻨﮫ ‪S = {ln 2} :‬‬

‫ﺧﺎﺻﯿﺎت ﺟﺒﺮﯾﺔ‬

‫‪( )("x Ρ)("y Ρ) ("x Î ¡ ); ln ex = x‬‬

‫و ‪( )erx = ex r‬‬ ‫‪ex‬و‬ ‫‪= ex-y‬‬ ‫‪ ("x Î ¡) ex f 0‬و ‪ e x ´ e y = e x+y‬و ‪e-x = 1‬‬

‫‪ey‬‬ ‫‪ex‬‬

‫‪C‬‬ ‫=‬ ‫‪e7‬‬ ‫و‬ ‫‪( )B = e-4 ´ e6 3‬‬ ‫و‬ ‫ﻣﺜﺎل ‪ :‬ﻟﯿﻜﻦ ‪ a‬و ‪ b‬ﻋﺪدﯾﻦ ﺣﻘﯿﻘﯿﯿﻦ‪ ,‬أﺣﺴﺐ وﺑﺴﻂ ﻣﺎ ﯾﻠﻲ ‪A = e3 ´e5 :‬‬
‫‪e4‬‬

‫‪C‬و‬ ‫‪e7‬‬ ‫= ‪B‬اﻟﺤﻞ‪ :‬و ) ( ) ( ) (‬
‫=‬ ‫‪e4‬‬ ‫‪= e7-4‬‬ ‫‪= e3‬‬ ‫‪e-4 ´ e6‬‬ ‫‪3‬‬ ‫‪e-4+6‬‬ ‫‪3‬‬ ‫‪e2‬‬ ‫‪3 = e2´3 = e6‬‬ ‫‪A = e3 ´e5 = e3+5 = e8‬‬

‫=‬ ‫=‬

‫=‬ ‫‪e‬‬ ‫‪2x ´ e 3x‬‬ ‫ﺗﻤﺮﯾﻦ‪ :1‬ﺑﺴﻂ ﻣﺎ ﯾﻠﻲ‪( )B = e2-x 2 ´ e3x-4 , A = e-x ´ e2x :‬‬
‫‪ex 4‬‬
‫‪( )C‬‬

‫اﻟﺤﻞ‪B = e2-x 2 ´ e3x-4 = e2´(2-x) ´ e3x-4 = e2´(2-x)+3x-4 = e4-2x+3x-4 = ex A = e- x ´ e2x = e- x+2x = ex :‬و ) (‬

‫=‬ ‫‪e2x ´ e3x‬‬ ‫=‬ ‫‪e2 x +3x‬‬ ‫=‬ ‫‪e5x‬‬ ‫‪= e5x-4x‬‬ ‫‪= ex‬‬
‫‪ex 4‬‬ ‫‪e4x‬‬ ‫‪e4x‬‬
‫‪( )C‬‬

‫ﺗﻤﺮﯾﻦ‪ :2‬ﺣﻞ ﻓﻲ ¡ اﻟﻤﻌﺎدﻻت اﻟﺘﺎﻟﯿﺔ‪ex+1 = 4 (1 :‬‬

‫‪e2x - 5ex + 6 = 0 (6‬‬ ‫‪e 2 x +1‬‬ ‫‪=e‬‬ ‫‪(5‬‬ ‫‪e2- x‬‬ ‫‪= ex-1‬‬ ‫‪(4‬‬ ‫‪e1+ x‬‬ ‫=‬ ‫‪1‬‬ ‫‪(3‬‬ ‫‪e1-x ´ e2x = e (2‬‬
‫‪e x-3‬‬ ‫‪e1+2 x‬‬ ‫‪e 2 x -3‬‬

‫اﻟﺤﻞ‪e1-x+2x = e1 Û e1-x ´ e2x = e (1 :‬‬

‫‪ x = 0 Û 1+ x =1 Û e1+x = e1 Û‬وﻣﻨﮫ ‪S = {0} :‬‬

‫‪ex-2 = e1 Û ex-2 = e (2‬‬
‫‪ x = 3 Û x - 2 = 1 Û‬وﻣﻨﮫ ‪S = {3} :‬‬

‫‪(3e1+x‬‬ ‫‪1‬‬
‫=‬ ‫)‪e-( 2 x -3‬‬ ‫‪Û‬‬ ‫‪e1+ x‬‬ ‫=‬ ‫‪e2 x -3‬‬

‫‪ 1 http:// xyzmath.e-monsite.com‬اﻷﺳﺘﺎذ ‪ :‬ﻋﺜﻤﺎﻧﻲ ﻧﺠﯿﺐ‬

S = ì 3 ü : ‫ وﻣﻨﮫ‬x = 2 Û e1+x = e-2x+3 Û 1+ x = -2x + 3 Û 3x = 2 Û
íî 2 ýþ 3

e( 2- x )-(1+ 2 x ) = ex-1 Û e2-x = ex-1 (4
e1+2 x

2 - x -1- 2x = x -1 Û (2 - x) - (1+ 2x) = x -1 Û

S = ì 1 ü : ‫وﻣﻨﮫ‬ x= 1 Û x = -2 Û -4x = -2 Û -x - 2x - x = -1+1- 2 Û
í 2 ý 2 -4
î þ

e = e Û (52x+1-x+31 e2 x+1
e(2 x+1)- (x-3) = e1 Û e x-3 =e

S ={-3} : ‫ وﻣﻨﮫ‬x = -3 Û x+4 =1Û 2x+1-x+3=1Û

: ‫ ﻓﻲ اﻟﺤﺎﻻت اﻵﺗﯿﺔ‬f ‫ ﺣﺪد ﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﯾﻒ اﻟﺪاﻟﺔ‬:3‫ﺗﻤﺮﯾﻦ‬

f (x ) = xe x + 2x .1

f (x) = ex +1 .2
ex -1

Df = ¡ (1 :‫اﻟﺤﻞ‬

{ }Df = x Î ¡ / ex -1 ¹ 0 (2

Df =¡-{0} : ‫ وﻣﻨﮫ‬x = 0 Û ex = e0 Û ex =1Ûex -1=0

:‫ ﺣﻞ ﻓﻲ ¡ اﻟﻤﺘﺮاﺟﺤﺎت اﻟﺘﺎﻟﯿﺔ‬:4‫ﺗﻤﺮﯾﻦ‬

e7 x-1 ³ e2x-3 ´ e x-2 (2 e2x-1 ³ 1 (1

2x -1 ³ 0 Û e2x-1 ³ e0 Û e2x-1 ³ 1(1 :‫اﻟﺤﻞ‬

S = é 1 ; +¥ ëêé : ‫وﻣﻨﮫ‬ x ³ 1 Û 2x ³1Û
ëê 2
2

e7 x-1 ³ e2 x-3+x-2 Û e7 x-1 ³ e2 x-3 ´ ex-2 (2

4x ³ -4 Û 7x -1 ³ 2x - 3 + x - 2 Û

S = [-1; +¥[ : ‫ وﻣﻨﮫ‬x ³ -1 Û

: ‫( اﻟﻨﮭﺎﯾﺎت‬3
‫ﻧﻘﺒﻞ اﻟﻨﮭﺎﯾﺘﯿﻦ اﻟﺘﺎﻟﯿﺘﯿﻦ‬

lim ex = +¥ : 2‫ و ﺧﺎﺻﯿﺔ‬lim ex = 0 : 1‫ﺧﺎﺻﯿﺔ‬
x®+¥ x®-¥

:‫ أﺣﺴﺐ اﻟﻨﮭﺎﯾﺎت اﻵﺗﯿﺔ‬:5‫ﺗﻤﺮﯾﻦ‬

lim 2ex -1 (4 lim e x (3 lim 2ex -1 (2 lim æ 2x -1 ö (1
ex +1 çè ex ø÷
x®-¥ x ®+¥ e x + 3 x®+¥ ex +2 x ®+¥

lim æ 2x - 1 ö = +¥ : ‫اذن‬ lim ex = +¥:‫ﻷن‬ lim 1 =0:‫وﻟﺪﯾﻨﺎ‬ lim 2x = +¥ ‫(ﻟﺪﯾﻨﺎ‬1 :‫اﻟﺤﻞ‬
èç ex ø÷
x®+¥ x®+¥ ex®+¥ x x®+¥

‫ﺷﻜﻞ ﻏﯿﺮ ﻣﺤﺪد‬ lim 2ex -1 = +¥ (2
ex + 2 +¥
x®+¥

lim ex = +¥:‫ﻷن‬ 2ex -1 = ex æ 2 - 1 ö 2 - 1 = 2-0 =2
çè ex ÷ø ex
x ®+¥ lim lim = lim

x® +¥ e x + 2 x ® +¥ ex æ 1 + 2 ö x ® +¥ 1+ 2 1+0
èç ex ø÷ ex

‫ﺷﻜﻞ ﻏﯿﺮ ﻣﺤﺪد‬ (3lim ex = +¥
ex +3 +¥
x®-¥

:‫ﻷن‬lim ex = +¥ lim ex = lim ex = lim 1 = 1 =1
ex + èæç1 + 1+ 0
x ®+¥ x® +¥ x ® +¥

3 ex 3 ö x® +¥ 1+ 3
ex ø÷ ex

lim 2ex -1 = 2´0 -1 = -1 (4
ex +1 0 +1
x®-¥

x a e x .‫( ﻣﺸﺘﻘﺔ اﻟﺪاﻟﺔ‬4
¡ ‫ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ‬exp ‫ﻧﻘﺒﻞ أن اﻟﺪاﻟﺔ‬

( )("x Î ¡) : e x ¢ = e x :‫و ﻟﺪﯾﻨﺎ‬

‫ ﻋﺜﻤﺎﻧﻲ ﻧﺠﯿﺐ‬: ‫ اﻷﺳﺘﺎذ‬2 http:// xyzmath.e-monsite.com

‫‪ (5‬ﺟﺪول ﺗﻐﯿﺮات اﻟﺪاﻟﺔ ‪: x ® e x‬‬
‫‪ (6‬ﻣﻨﺤﻨﻰ اﻟﺪاﻟﺔ ‪: exp‬‬

‫ﺗﻤﺮﯾﻦ‪ :6‬أﺣﺴﺐ )‪ f ¢( x‬ﻓﻲ اﻟﺤﺎﻻت اﻵﺗﯿﺔ ‪:‬‬

‫‪f (x ) = e x -1 (3 f (x ) = xe x + 3x (2 f (x ) = e x + 2 (1‬‬

‫‪ex +1‬‬

‫اﻟﺤﻞ‪f ¢( x) = (ex + 2)¢ = (ex )¢ + (2)¢ = ex + 0 = ex (1 :‬‬

‫‪f ¢(x) =( xex +3x)¢ = (xex )¢ +(3x)¢ = (x)¢ ex + x(ex )¢ +3=ex + xex +3(2‬‬

‫)‪¢(x‬‬ ‫‪æ‬‬ ‫‪ex‬‬ ‫‪- 1 ö¢‬‬ ‫‪(e x‬‬ ‫‪- 1)¢‬‬ ‫´‬ ‫‪(ex‬‬ ‫‪+ 1)- (ex‬‬ ‫´)‪- 1‬‬ ‫‪(e x‬‬ ‫‪+ 1)¢‬‬ ‫‪(3‬‬
‫‪ç‬‬ ‫‪ex‬‬ ‫‪( )e x + 1 2‬‬
‫‪f‬‬ ‫=‬ ‫‪è‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫÷‬ ‫=‬
‫‪ø‬‬

‫´ ‪( ( ) ( ) ) ( ) ( )f ¢(x) = ex‬‬
‫‪ex‬‬ ‫‪+1 - ex‬‬ ‫‪-1‬‬ ‫‪´ex‬‬ ‫=‬ ‫‪ex ´ex‬‬ ‫‪+ ex‬‬ ‫‪-ex ´ex‬‬ ‫‪+ ex‬‬ ‫=‬ ‫‪2ex‬‬
‫‪ex +1 2‬‬ ‫‪ex‬‬ ‫‪ex +1 2‬‬
‫‪2‬‬

‫‪+1‬‬

‫ﺗﻤﺮﯾﻦ‪ :7‬ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﯾﺔ ‪ f‬اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﯾﻠﻲ‪f (x ) = e x + 3x :‬‬

‫‪(1‬ﺣﺪد ‪ Df‬ﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﯾﻒ اﻟﺪاﻟﺔ ‪f‬‬

‫‪(2‬أﺣﺴﺐ )‪ f (0‬و )‪ ) f (1‬أﻋﻂ ﻗﯿﻤﺔ ﻣﻘﺮﺑﺔ ﻟﻠﻨﺘﺎﺋﺞ(‬

‫‪ (3‬أﺣﺴﺐ )‪ f ¢(x‬و وﺑﯿﻦ أن اﻟﺪاﻟﺔ ‪ f‬ﺗﺰاﯾﺪﯾﺔ ﻗﻄﻌﺎ ﻋﻠﻰ ‪Df‬‬

‫‪(4‬أﺣﺴﺐ ) ‪ lim f (x‬و) ‪lim f (x‬‬
‫‪x ®+¥‬‬ ‫‪x ® -¥‬‬

‫‪ (5‬ﺣﺪدﺟﺪول ﺗﻐﯿﺮات اﻟﺪاﻟﺔ ‪f‬‬

‫اﻟﺤﻞ‪Df = ¡ (1 :‬‬

‫‪f (0) = e 0 + 3´ 0 = 1- 0 = 1(2‬‬

‫‪f (1) = e1 + 3´1 = e + 3 ; 2, 7 + 3 ; 5, 7‬‬

‫‪f ¢(x ) = (ex +3x )¢ = (ex )¢ +(3x )¢ =ex +3> 0(3‬‬

‫ﻷن‪ ("x Î ¡) ex > 0 :‬وﻣﻨﮫ ‪ f‬ﺗﺰاﯾﺪﯾﺔ ﻗﻄﻌﺎ ﻋﻠﻰ ¡‬

‫‪(4‬أﺣﺴﺐ ‪lim f (x ) = lim ex +3x = 0+3(-¥) = -¥‬‬
‫‪x ®-¥‬‬ ‫‪x ®-¥‬‬

‫‪lim f (x ) = lim ex +3x = +¥+3(+¥) = +¥‬‬
‫‪x ®+¥‬‬ ‫‪x ®+¥‬‬

‫‪(5‬ﺟﺪول ﺗﻐﯿﺮات اﻟﺪاﻟﺔ ‪f‬‬

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