www.9alami.com ﺍﻟﻔﻬﺮﺱ
ﺗﻤﺮﻳﻦ 1 ﺗﻤﺮﻳﻦ 2 ﺗﻤﺮﻳﻦ 3 ﺗﻤﺮﻳﻦ 4 ﺗﻤﺮﻳﻦ 5 ﺗﻤﺮﻳﻦ 6 ﺗﻤﺮﻳﻦ 7 ﺗﻤﺮﻳﻦ 8 ﺗﻤﺮﻳﻦ9 ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺗﻤﺮﻳﻦ 10 ﺗﻤﺮﻳﻦ 11 ﺗﻤﺮﻳﻦ 12 ﺗﻤﺮﻳﻦ 13 ﺗﻤﺮﻳﻦ 14 ﺗﻤﺮﻳﻦ 15 ﺗﻤﺮﻳﻦ 16 ﺗﻤﺮﻳﻦ 17 ﺗﻤﺮﻳﻦ18 ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ ﺍﻟﺤﻞ
ﺗﻤﺮﻳﻦ:1
ﺗﻤﺮﻳﻦ:2
ﺗﻤﺮﻳﻦ:3
ﺗﻤﺮﻳﻦ:4
ﺗﻤﺮﻳﻦ:5
ﺗﻤﺮﻳﻦ:6
ﺗﻤﺮﻳﻦ:7
ﺗﻤﺮﻳﻦ:8
ﺗﻤﺮﻳﻦ:9
ﺗﻤﺮﻳﻦ: 10 GG وﻟﻴﻜﻦ ) (Cﻣﻨﺤﻨﺎهﺎ ﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ) (o, i, j -a-1ﺣﺪد Dﺣﻴﺰ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ . f -bاﺣﺴﺐ آﻼ ﻣﻦ اﻟﻨﻬﺎﻳﺎت اﻟﺘﺎﻟﻴﺔ: ) lim f ( xو ) lim f ( xو ) lim+ f ( xو ) lim− f ( x ∞x →+
∞x →−
x →0
x →0
– a-2ﺗﺤﻘﻖ ﻣﻦ أن : x +1 x +1 27 =) ( ) ( ∀x ∈ D : f ( x) − 2x 2x x 2 + 27 + x x +1 = ( Δ1 ) : yﻣﻘﺎرب ﻣﺎﺋﻞ ل ) (Cﺑﺠﻮار ∞+ -bاﺳﺘﻨﺘﺞ أن اﻟﻤﺴﺘﻘﻴﻢ 2 x +1 – Cﺑﻴﻦ أن ( Δ 2 ) : y = −ﻣﻘﺎرب ﻣﺎﺋﻞ ل ) ( Cﺑﺠﻮار ∞− 2 x 3 − 27 = ) f '(xﻟﻜﻞ xﻣﻦ . D -a-3ﺑﻴﻦ أن 2x 2 x 2 + 27 -bﺗﺤﻘﻖ ﻣﻦ أن fﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ اﻟﻤﺠﺎل [∞ [3, +وﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ آﻞ ﻣﻦ اﻟﻤﺠﺎﻟﻴﻦ [ ]−∞, 0و ]. ]0,3 -cاﻋﻂ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ . f – a-4ﺣﺪد ﺗﻘﺎﻃﻊ ) (Cﻣﻊ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ. -bﻧﻘﺒﻞ أن ) A( x0 , y0ﺣﻴﺚ x0 ≈ −5, 2و y0 ≈ 2,9هﻲ ﻧﻘﻄﺔ اﻻﻧﻌﻄﺎف اﻟﻮﺣﻴﺪة ﻟﻠﻤﻨﺤﻨﻰ ) (Cوأن ) f’(xﺳﺎﻟﺒﺔ G G ﻋﻠﻰ اﻟﻤﺠﺎل [ [ x0 , 0وﻣﻮﺟﺒﺔ ﻋﻠﻰ آﻞ ﻣﻦ اﻟﻤﺠﺎﻟﻴﻦ ] ]−∞, x0و [∞ . ]0, +ﻧﺄﺧﺬ i = j = 1cm أﻧﺸﺊ . C
ﺗﻤﺮﻳﻦ: 11 ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fﻟﻠﻤﺘﻐﻴﺮ اﻟﺤﻘﻴﻘﻲ xاﻟﻤﻌﺮﻓﺔ آﺎﻟﺘﺎﻟﻲ: 2 x2 −1 . 1 + x2 = )f ( x 3 x GG وﻟﻴﻜﻦ ) (ζاﻟﻤﻨﺤﻨﻰ اﻟﻤﻤﺜﻞ ﻟﻠﺪاﻟﺔ fﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ) (o, i, j -1ﺣﺪد Dﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ ،fوﺗﺤﻘﻖ ﻣﻦ أن fﻓﺮدﻳﺔ. -2اﺣﺴﺐ ) lim f ( xو ) lim f (xوأول هﻨﺪﺳﻴﺎ اﻟﻨﺘﻴﺠﺘﻴﻦ اﻟﻤﺤﺼﻞ ﻋﻠﻴﻬﻤﺎ . ∞x →+
-3ﺑﻴﻦ أﻧﻪ ﻟﻜﻞ xﻣﻦ Dﻟﺪﻳﻨﺎ . :
-4أﻧﺸﺊ ) . (ζ
x →0 x ;0
3 x . 1 + x2 4
= ) f '( xواﺳﺘﻨﺘﺞ ﺗﻐﻴﺮات . f
-5ﻟﻴﻜﻦ gﻗﺼﻮر اﻟﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل [∞. ]0, + أ -ﺑﻴﻦ أن gﺗﻘﺎﺑﻞ ﻣﻦ [∞ ]0, +ﻧﺤﻮ ﻣﺠﺎل Jﻳﻨﺒﻐﻲ ﺗﺤﺪﻳﺪﻩ. ب -ﺑﻴﻦ أن g −1ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ g −1 ) . Jهﻲ اﻟﺪاﻟﺔ اﻟﻌﻜﺴﻴﺔ ﻟﻠﺪاﻟﺔ ( g ج -أﻧﺸﺊ ) ' (ζاﻟﻤﻨﺤﻨﻰ اﻟﻤﻤﺜﻞ ﻟﻠﺪاﻟﺔ . g −1
ﺗﻤﺮﻳﻦ: 12
2 3
ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fاﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ [∞ [ 0, +ب f ( x) = 6 x − 4 x : ) f (x -1اﺣﺴﺐ x
. limأول هﻨﺪﺳﻴﺎ اﻟﻨﺘﻴﺠﺔ . x →0 x ;0
-2ﺣﺪد اﻟﻔﺮع اﻟﻼﻧﻬﺎﺋﻲ ل ) (ζﻣﻨﺤﻨﻰ . f -3اﺣﺴﺐ ) f’(xﻟﻜﻞ xﻣﻦ [∞ ]0, +ﺛﻢ ﺣﺪد ﺟﺪول ﺗﻐﻴﺮات . f -4أ -ﺣﺪد ﻧﻘﻄﺘﻲ ﺗﻘﺎﻃﻊ ) (ζﻣﻊ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ. 27 ب -ﺣﺪد ﻣﻌﺎدﻟﺔ ) ( Δﻣﻤﺎس ) (ζﻓﻲ اﻟﻨﻘﻄﺔ ذات اﻷﻓﺼﻮل 8 GG ج -أﻧﺸﺊ ) ( Δو ) (ζﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ) (o, i, j
) وﺣﺪة اﻟﻘﻴﺎس .( 1 cm : g -5ﻗﺼﻮر fﻋﻠﻰ اﻟﻤﺠﺎل [∞I = [1, +
ﺑﻴﻦ أن gﺗﻘﺎﺑﻞ ﻣﻦ Iﻧﺤﻮ ﻣﺠﺎل ﻳﺘﻢ ﺗﺤﺪﻳﺪﻩ .اﺣﺴﺐ ) (.g ) ' ( 0 −1
ﺗﻤﺮﻳﻦ: 13 ﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ ﻟﻠﻤﺘﻐﻴﺮ اﻟﺤﻘﻴﻘﻲ xاﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ IRﺑﻤﺎ ﻳﻠﻲ: f ( x) = ( 1 + x 2 − x) 2
GG ) (ζهﻮ ﻣﻨﺤﻨﻰ اﻟﺪاﻟﺔ fﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ) (o, i, j
-1أ -اﺣﺴﺐ اﻟﻨﻬﺎﻳﺔ ) lim f (x
∞x →−
ب -ﺑﻴﻦ أن lim f ( x ) = 0 :
∞x →+
)−2 f ( x
= )f '( x
-2أ -ﺑﻴﻦ أن ﻟﻜﻞ xﻣﻦ : IR 1 + x2 ب -أﺛﺒﺖ أن f '( x ) ≠ 0ﻟﻜﻞ xﻣﻦ IRﺛﻢ ﺿﻊ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ . f )f ( x limﺛﻢ اﺳﺘﻨﺘﺞ اﻟﻔﺮع اﻟﻼﻧﻬﺎﺋﻲ ل ) (ζﺟﻮار ∞− -3ﺑﻴﻦ أن ∞= − ∞n →− x -4أ -اآﺘﺐ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻘﻴﻢ ) (Tﻣﻤﺎس اﻟﻤﻨﺤﻨﻰ ) (ζﻓﻲ اﻟﻨﻘﻄﺔ ذات اﻷﻓﺼﻮل .0 ب -أﻧﺸﺊ اﻟﻤﺴﺘﻘﻴﻢ ) (Tواﻟﻤﻨﺤﻨﻰ ) ) (ζاﻟﻮﺣﺪة ( 2cm -5أ -ﺑﻴﻦ أن fﺗﻘﺎﺑﻞ ﻣﻦ IRﻧﺤﻮ ﻣﺠﺎل Jﻳﺘﻢ ﺗﺤﺪﻳﺪﻩ. ب -اﺣﺴﺐ )( f −1 ) ' (1 ج -اﺣﺴﺐ ) f’(xﻟﻜﻞ xﻣﻦ . J
ﺗﻤﺮﻳﻦ: 14
ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fاﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ [∞ [ 0, +ﺑﻤﺎ ﻳﻠﻲ: f ( x) = x − 2 + 3 x 2 + 1
وﻟﻴﻜﻦ ) (ζﺗﻤﺜﻴﻠﻬﺎ اﻟﻤﺒﻴﺎﻧﻲ ﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ) (o, i, j -1أ -اﺣﺴﺐ ) lim f (x
∞x →+
ب -ادرس اﻟﻔﺮع اﻟﻼﻧﻬﺎﺋﻲ ﻟﻠﻤﻨﺤﻨﻰ 3 3 ( x 2 + 1) 2 + 2 x
) (ζ = )(∀x ∈ [ 0, +∞[) , f '( x
-2أ -ﺑﻴﻦ أن : 3 3 ( x 2 + 1) 2 ب -اﻋﻂ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ . f -3ﺑﻴﻦ أن fﺗﻘﺎﺑﻞ ﻣﻦ [∞ [ 0, +ﻧﺤﻮ ﻣﺠﺎل Iﻳﺠﺐ ﺗﺤﺪﻳﺪﻩ
1 -4أ -ﺑﻴﻦ أن اﻟﻤﻌﺎدﻟﺔ f(x) =0ﺗﻘﺒﻞ ﺣﻼ وﺣﻴﺪا αﺑﺤﻴﺚ ≺ α ≺ 1 2 ب -ﺣﺪد ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ) (ζواﻟﻤﺴﺘﻘﻴﻢ ) ( Δاﻟﺬي ﻣﻌﺎدﻟﺘﻪ . y =x
ج -أﻧﺸﺊ ) (ζﺛﻢ )' (ζﻣﻨﺤﻨﻰ اﻟﺪاﻟﺔ f −1اﻟﺘﻘﺎﺑﻞ اﻟﻌﻜﺴﻲ ﻟﻠﺪاﻟﺔ . f
) دون ﺣﺴﺎب .( α
ﺗﻤﺮﻳﻦ: 15 ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fﻟﻠﻤﺘﻐﻴﺮ اﻟﺤﻘﻴﻘﻲ xاﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ * IRﺑﻤﺎ ﻳﻠﻲ: 2 ⎧ )[( x ∈ ]−∞, 0[ ∪ ]0,1 ⎪⎪ f ( x) = − x + x ⎨ ⎪ f ( x) = 1 + x [∞( x ∈ [1, + 2 x ⎪⎩ -1ﺑﻴﻦ أن اﻟﺪاﻟﺔ fﻣﺘﺼﻠﺔ ﻓﻲ اﻟﻨﻘﻄﺔ x0 = 1 -2أ -ﺑﻴﻦ أن اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻨﺪ اﻟﻨﻘﻄﺔ x0 = 1ﻋﻠﻰ اﻟﻴﺴﺎر . ب -ﺑﻴﻦ أن اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻨﺪ اﻟﻨﻘﻄﺔ x0 = 1ﻋﻠﻰ اﻟﻴﻤﻴﻦ ) ﻻﺣﻆ أن (1 + x − 2 x = ( x − 1) 2
-3أ -ﺑﻴﻦ أن (∀x ∈ ]−∞, 0[ ∪ ]0,1[) f '( x ) ≺ 0
x −1 ب -ﺑﻴﻦ أن 4 x ج -اﻋﻂ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ .f -4ﻟﻴﻜﻦ ) (ζاﻟﻤﻨﺤﻨﻰ اﻟﻤﻤﺜﻞ ﻟﻠﺪاﻟﺔ fﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ) (o, i, j
= )(∀x ∈ ]1, +∞[) f '( x
أ -ادرس اﻟﻔﺮوع اﻟﻼﻧﻬﺎﺋﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ ) . (ζ ب -أﻧﺸﺊ اﻟﻤﻨﺤﻨﻰ ) ) (ζﻧﻘﺒﻞ أن ﻟﻠﻤﻨﺤﻨﻰ ) (ζﻧﻘﻄﺔ اﻧﻌﻄﺎف وﺣﻴﺪة أﻓﺼﻮﻟﻬﺎ (3
ﺗﻤﺮﻳﻦ: 16 ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fاﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ IRﺑﻤﺎ ﻳﻠﻲ:
⎧ f ( x) = x − 1 + 2 1 − x ; x ≤ 1 ⎪ ⎨ x3 − 1 ; x ≥1 ⎪ f ( x) = 3 x +3 ⎩ GG وﻟﻴﻜﻦ ) ( Cﻣﻨﺤﻨﺎهﺎ ﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ) (o, i, j -1اﺣﺴﺐ ) lim f ( xوﺑﻴﻦ أن lim f (x ) = −∞ :
∞x →−
∞x →+
-2ادرس ﻗﺎﺑﻠﻴﺔ اﺷﺘﻘﺎق fﻋﻠﻰ اﻟﻴﻤﻴﻦ وﻋﻠﻰ اﻟﻴﺴﺎر ﻓﻲ 1ﺛﻢ اﻋﻂ ﺗﺄوﻳﻼ هﻨﺪﺳﻴﺎ ﻟﻠﻨﺘﻴﺠﺘﻴﻦ اﻟﻤﺤﺼﻞ ﻋﻠﻴﻬﻤﺎ. -3أ -ﺑﻴﻦ أن اﻟﺪاﻟﺔ fﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل [∞[1, +
−x ﺑﻴﻦ أن )1 − x (1 + 1 − x ج -اﻋﻂ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ . f -4أ -ادرس اﻟﻔﺮﻋﻴﻦ اﻟﻼﻧﻬﺎﺋﻴﻴﻦ ﻟﻠﻤﻨﺤﻨﻰ ) (C ب -ارﺳﻢ اﻟﻤﻨﺤﻨﻰ ) ) ( Cﻻﺣﻆ أن f(-3)=0 -5ﻟﺘﻜﻦ gﻗﺼﻮر اﻟﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل [∞[1, + = )f '( x
ﻟﻜﻞ xﻣﻦ
[]−∞,1
أ -ﺑﻴﻦ أن gﺗﻘﺎﺑﻞ ﻣﻦ [∞ [1, +ﻧﺤﻮ ﻣﺠﺎل Iﻳﻨﺒﻐﻲ ﺗﺤﺪﻳﺪﻩ. ب -ﺣﺪد ) g −1 ( xﻟﻜﻞ xﻣﻦ اﻟﻤﺠﺎل . I ج -ﺑﻴﻦ أن g −1هﻲ داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ
2 2
)3 3 ( x − x − x + 1 2
3
→ xﻋﻠﻰ اﻟﻤﺠﺎل
ﺗﻤﺮﻳﻦ: 17 ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fاﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ اﻟﻤﺠﺎل [∞ ]0, +ﺑﻤﺎ ﻳﻠﻲ: 1 x GG وﻟﻴﻜﻦ ) ( Cﻣﻨﺤﻨﺎهﺎ ﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ) (o, i, j -1أ -اﺣﺴﺐ ) lim f ( xو )lim f ( x f ( x) = x − x +
∞x →+
x →0 x;0
ب -ﺣﺪد اﻟﻔﺮﻋﻴﻦ اﻟﻼﻧﻬﺎﺋﻴﻴﻦ ﻟﻠﻤﻨﺤﻨﻰ ) (C ⎞ ⎛ 2x + x + 1 ⎜⎜ = ) f '(xﻟﻜﻞ xﻣﻦ [∞]0, + -2ﺑﻴﻦ أن ⎟⎟ ( x − 1) : ⎠ ) ⎝ 2x x ب -اﻋﻂ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ . f -3أ -ادرس اﻟﻮﺿﻊ اﻟﻨﺴﺒﻲ ﻟﻠﻤﻨﺤﻨﻰ ) (Cواﻟﻤﺴﺘﻘﻴﻢ ) ( Δذي اﻟﻤﻌﺎدﻟﺔ . y=x 1 7 5 ب -ارﺳﻢ اﻟﻤﻨﺤﻨﻰ ) f (4) = ) ( Cو = ) ( ( f 4 4 2 -4ﻟﺘﻜﻦ gﻗﺼﻮر اﻟﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل [∞[1, +
أ -ﺑﻴﻦ أن gﺗﻘﺒﻞ داﻟﺔﻋﻜﺴﻴﺔ g −1وﺣﺪد ﻣﺠﻤﻮﻋﺔ اﻟﺪاﻟﺔ g −1 GG ب -ارﺳﻢ ﻓﻲ ﻧﻔﺲ اﻟﻤﻌﻠﻢ ) ، (o, i, jاﻟﻤﻨﺤﻨﻰ اﻟﻤﻤﺜﻞ ﻟﻠﺪاﻟﺔ . g −1 -5ﻧﻌﺘﺒﺮ اﻟﻤﺘﺘﺎﻟﻴﺔ اﻟﻌﺪدﻳﺔ ( an ) n∈INاﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ :
ﺗﻤﺮﻳﻦ: 18
أ -ﺑﻴﻦ ان ( ∀n ∈ IN ) an ; 1 :
⎧a0 = 2 ⎨ ) ⎩an +1 = f (an
ب -ﺑﻴﻦ أن اﻟﻤﺘﺘﺎﻟﻴﺔ ( an ) n∈INﺗﻨﺎﻗﺼﻴﺔ. ج -اﺳﺘﻨﺘﺞ أن اﻟﻤﺘﺘﺎﻟﻴﺔ ( an ) n∈INﻣﺘﻘﺎرﺑﺔ ﺛﻢ ﺣﺪد ﻧﻬﺎﻳﺘﻬﺎ. ﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ IRﺑﻤﺎ ﻳﻠﻲ: ⎧⎪ f ( x) = x − 2 x − 1 ; x ≥ 1 ⎨ ⎪⎩ f ( x) = x + 2 x − x ; x ≺ 1 وﻟﻴﻜﻦ ) (Cﻣﻨﺤﻨﺎهﺎ ﻓﻲ م.م.م ) (o, i, j -1اﺣﺴﺐ ) lim f ( xو )lim f ( x ∞x →+
∞x →−
-2أ -ادرس اﺗﺼﺎل fﻓﻲ .1 ب -ادرس ﻗﺎﺑﻠﻴﺔ اﺷﺘﻘﺎق fﻋﻠﻰ اﻟﻴﻤﻴﻦ وﻋﻠﻰ اﻟﻴﺴﺎر ﻓﻲ ،1ﺛﻢ اﻋﻂ ﺗﺄوﻳﻼ هﻨﺪﺳﻴﺎ ﻟﻠﻨﺘﻴﺠﺘﻴﻦ اﻟﻤﺤﺼﻞ ﻋﻠﻴﻬﻤﺎ. -3أ -اﺣﺴﺐ ) f’(xﻟﻜﻞ xﻣﻦ }IR − {1 ب -اﻋﻂ ﺟﺪول ﺗﻐﻴﺮات . f -4أ -ﺣﺪد اﻟﻔﺮﻋﻴﻦ اﻟﻼﻧﻬﺎﺋﻴﻴﻦ ﻟﻠﻤﻨﺤﻨﻰ ) . (C ب -ﺣﺪد ﺗﻘﺎﻃﻊ اﻟﻤﻨﺤﻨﻰ ) (Cﻣﻊ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ. ج -ارﺳﻢ اﻟﻤﻨﺤﻨﻰ ) (C 2 -5ﻟﺘﻜﻦ gاﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل [∞ [ 2, +واﻟﺘﻲ ﺗﺤﻘﻖ 3 أ -اآﺘﺐ ) g(xﺑﺪﻻﻟﺔ x ب -اﻋﻂ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ . g
= )g (2
:1 ﺗﻤﺮﻳﻦ x ∈ Df ⇔ ( x ≠ 1 ( وx + 1)( x − 1) ≥ 0) (-أ-1
⇔ [ x ≠ 1 ( وx ≥ 1 أوx ≤ −1)]
⇔ x ∈ ]−∞, −1] ∪ ]1, +∞[ D f = ]−∞, −1] ∪ ]1, +∞[
إذن
x +1 (-ب =1 x →+∞ x − 1 lim f ( x ) = −∞ وlim f ( x) = +∞ إذن lim
x →−∞
x →+∞
x +1 = +∞ ﺑﻤﺎ أن x →1 x − 1 lim+ f ( x) = +∞ ﻓﺈن lim+
x →1
f ( −1) = 0
lim−
x →−1
و
f ( x) − f (−1) x +1 = lim− = 0 (- أ-2 x →−1 x +1 x −1 .-1 ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻞ اﻟﻴﺴﺎر ﻓﻲf إذن f g' (−1) = 0 و
x ∈ D f − {−1} ( -ب −2 x +1 ( x − 1) 2 + ( x + 1) f '( x) = x −1 x +1 2 x −1 x +1 ( x − 1) 2 − ( x + 1) 1 x − = x +1 ( x − 1) 2 x −1 إذن ( x + 1)( x − 2) D f − {−1} ﻣﻦx ﻟﻜﻞ f '( x) = x +1 2 ( x − 1) x −1 ( x + 1)( x − 2) هﻲ إﺷﺎرةf '( x ) ( إﺷﺎرة-ج
f (2) = 3 3
( -أ-3
x +1 − ( x + 2) x −1 x +1 ( x + 1) 2 − ( x + 2) 2 x −1 = lim x →+∞ x +1 ( x + 1) + ( x + 2) x −1 3x + 5 = lim x →+∞ x +1 + ( x + 2)( x − 1) ( x 2 − 1) x −1 5 3+ x = lim =0 x →+∞ 1 x +1 2 (x − ) + (1 + )( x − 1) x x −1 x y = x + 2 ( اﻟﺬي ﻣﻌﺎدﻟﺘﻪD) إذن اﻟﻤﺴﺘﻘﻴﻢ −∞ و+∞ ( ﺑﺠﻮارC ) ﻣﻘﺎرب ل (C ) ب( إﻧﺸﺎء
lim [ f ( x) − ( x + 2)] = lim ( x + 1)
x →+∞
x →+∞
ﺗﻤﺮﻳﻦ:2 ⎤ 2 ⎡ lim f ( x) = lim ( x − 4) ⎢ −1 + ⎥ = −∞ -1 ∞x →− ∞x →− ⎦ 4− x ⎣ -2 )f ( x) − f (4 x−4+2 4− x lim = lim x→4 x → 4 x−4 x−4 x≺ 4 x≺ 4 ⎞ 2 ⎛ = lim ⎜1 − ∞⎟ = − x→4 4 x − ⎝ ⎠ x≺ 4 fﻏﻴﺮ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻴﺴﺎر ﻓﻲ 4 ﻳﻘﺒﻞ ) (Cﻧﺼﻒ ﻣﻤﺎس ﻓﻲ اﻟﻨﻘﻄﺔ ) A(4,0ﻳﻮازي ﻣﺤﻮر اﻷراﺗﻴﺐ. -3أ ( -ﻟﻜﻞ xﻣﻦ
[]−∞, 4
−1 2 4− x
f '( x) = 1 + 2
1 4 − x −1 = 4− x 4− x ب (-إﺷﺎرة ) f’(xهﻲ إﺷﺎرة 4 − x − 1 4 − x −1 3− x = = 4 − x −1 4 − x +1 4 − x +1 ﻟﺪﻳﻨﺎ ∀x ≺ 4 4 − x + 1 0 : إذن إﺷﺎرة ) f’(xهﻲ إﺷﺎرة 3 − x x ≺ 3 ⇔ f '( x ) 0 3 ≺ x ≺ 4 ⇔ f '( x ) ≺ 0 = 1−
f (3) = −1 + 2 = 1
-4
⎞ ⎛ 4 2 4− x )f ( x = lim ⎜⎜ 1 + + ⎟⎟ ∞x →− x x ⎝ x ⎠
lim
∞x →−
⎛ 4 ⎞4 1 = lim ⎜⎜1 + − 2 2 − ⎟⎟ = 1 ∞x →− x ⎠x ⎝ x
)
(
∞lim ( f ( x) − x ) = lim −4 + 2 4 − x = + ∞x →−
∞x →−
إذن ﻳﻘﺒﻞ ) (Cﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ اﺗﺠﺎهﻪ اﻟﻤﺴﺘﻘﻴﻢ ذي اﻟﻤﻌﺎدﻟﺘﻪ y = x
x≺4 -5 f ( x) = 0 ⇔ x − 4 + 2 4 − x = 0
⇔ 2 4− x = 4− x ⇔ 4 = 4− x ⇔ x=0 إذن ) (Cﻳﻘﻄﻊ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ ﻓﻲ أﺻﻞ اﻟﻤﻌﻠﻢ. 1 f (0) = 0 -6و 2 1 (T ) : y = x 2 f (−5) = − 3 -7
= )f '(0
:3 ﺗﻤﺮﻳﻦ x ∈ D f ⇔ x 2 + x 0 (-1 ⇔ x ( x + 1) 0
x ∈⇔ x ]−∞, −1[ ∪ ]0, +∞[
D f = ]−∞, −1[ ∪ ]0, +∞[ إذن ( ﻧﻌﻠ ﻢ أن-2 1 وlim x 2 + x = +∞ lim = 2 x →+∞ x →+∞ x +x lim f ( x ) = −∞ إذن x →+∞
1 = +∞ وlim+ x 2 + x = 0 وﺑﻤﺎ أن lim+ 2 x →0 x →0 x + x lim f ( x) = +∞ ﻓﺈن
x→0+
⎡ ⎛ 1⎞ ⎤ f ⎢ 2 ⎜ − ⎟ − x ⎥ = f (−1 − x ) (-3 ⎣ ⎝ 2⎠ ⎦ 1 = − (x + 1) 2 − 1 − x 2 (x + 1) − 1 − x 1 = 2 − x 2 + 2x + 1 − 1 − x = f ( x ) x + 2x − 1 − x ⎡ ⎛ 1⎞ ⎤ (∀x ∈ D f ) f ⎢ 2 ⎜ − ⎟ − x ⎥ = f (x ) إذن ⎣ ⎝ 2⎠ ⎦ . (C ) ( ﻣﺤﻮر ﺗﻤﺎﺛﻞΔ ) اﻟﻤﺴﺘﻘﻴﻢ D f ﻣﻦx ( ﻟﻴﻜﻦ-4 2x + 1
2x + 1 − (x 2 + x ) 2 2 x 2 + x ⎡ ⎤ 1 1 = −(2x + 1) ⎢ 2 + ⎥ 2 2 ⎣ (x + x ) 2 x + x ⎦
f '(x ) = −
1 1 + 0 (-5 2 (x + x ) 2 x 2 + x ]0, +∞[ ﻋﻠﻰ2x + 1 إﺷﺎرة ﻋﻜﺲ إﺷﺎرةf '(x ) إذن ﺗﻘﺒﻞ
(∀x ∈ D f ]0, +∞[)
2
(∀x〉 0) 2x + 1 0 : ﺑﻤﺎ أن ∀x 0 f '( x ) ≺ 0 : ﻓﺈن
x f’(x) f(x)
+∞
0
+∞
−∞
-6 ⎛ 1 ) f (x ⎞1 1 = lim ⎜⎜ 3 − − ⎟ = −1 ∞x →+ x →+∞ x + x 2 x ⎠⎟ x ⎝ ⎛ 1 ⎞ ⎟⎤ − ⎡ x 2 + x −x lim (f (x ) + x ) = lim ⎜ 2 ∞x →+ x →+∞ x + x ⎣ ⎠⎦ ⎝ x lim ( x 2 + x − x ) = lim ∞x →+ ∞x →+ x 2 +x +x 1 1 = lim ∞x →+ 2 1 1+ +1 x 1 إذن lim (f (x ) + x ) = − ∞x →+ 2 1 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﻤﺴﺘﻘﻴﻢ ) (Dاﻟﺬي ﻣﻌﺎدﻟﺘﻪ y = − x −ﻣﻘﺎرب ل ) (cﺑﺠﻮار ∞+ 2 lim
0 f (x ) = 0 (-7 1 f (x ) = 0 ⇔ 2 = x 2 +x x +x ⇔ (x 2 + x )2 = 1 ⇔ x 2 + x −1 = 0 Δ = 1+ 4 = 5 −1 + 5 إذن = x 2 −1 − 5 ﻏﻴﺮ ﻣﻘﺒﻮل ﻷﻧﻪ ﺳﺎﻟﺐ( ) اﻟﺤﻞ 2 ⎫⎪ ⎪⎧ −1 + 5 ⎨= S ⎬ ⎪⎭ ⎩⎪ 2 x
⎛ ⎞ −1 + 5 = ⎜⎜ Aهﻲ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ) (Cوﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ ﻋﻠﻰ *+ ⎟⎟ , 0 2 ⎝ ⎠
.
:4 ﺗﻤﺮﻳﻦ
lim+
x →−
1 2
x ∈ D ⇔ 2x + 1 0 -1 ⎤ 1 ⎡ ⇔ x ∈ ⎥ − , +∞ ⎢ ⎦ 2 ⎣ ⎤ 1 ⎡ D = ⎥ − , +∞ ⎢ ⎦ 2 ⎣ 1 2x + 1 = 0+ وlim+ (x + 1) = ﻟﺪﻳﻨﺎ 1 2 x →− 2
lim+ f (x ) = +∞ إذن
x →−
1 2
lim f ( x) = lim
x →+∞
1 x = +∞ 2 1 + x x2
1+
x →+∞
(C ) ( ﻣﻘﺎرب لD ) : y = −
1 -2 2
1 f ( x) x =0 lim = lim x →+∞ x →+∞ x 2x +1 ( ﻳﻘﺒﻞ ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ ﻓﻲ اﺗﺠﺎﻩ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞC )إذن 1+
. D ﻣﻦx ( ﻟﻴﻜﻦ-أ-3 1 2x + 1 − (x + 1) 2x + 1 f '(x ) = (2x + 1) 2x + 1 − x − 1 = (2x + 1) 2x + 1 =
x (2x + 1)
3 2
= x (2x + 1)
−3 2
(-ب
. D ﻣﻦx ( ﻟﻴﻜﻦ-أ-4 −5 3 f ''( x) = (2 x + 1) − × (2 x + 1) 2 × 2 2 3 2
−5 2
)= (1 − x )(2x + 1
−5 1 − ) f ''(x ) = (1 − x )(2x + 1) 2 2 1 بx − (- 2 f ''(x ) ≥ 0 ⇔ 1 − x 0 ⇔ x ≺1 إذن '' fﺗﻨﻌﺪم وﺗﻐﻴﺮ اﻹﺷﺎرة ﻓﻲ x 0 = 1
(∀x
2 2 3 = 3 3 ⎞⎛ 2 3 A ⎜⎜1,ﻧﻘﻄﺔ اﻧﻌﻄﺎف ل ) .(C وﻣﻨﻪ ﻓﺈن ⎟ ⎠⎟ 3 ⎝ = )f (1
-6أ g (-داﻟﺔ ﻣﺘﺼﻠﺔ وﺗﻨﺎﻗﺼﻴﺔ ﻗﻄﻌﺎ ﻋﻠﻰ . I و [∞J = g (I ) = [1, + وﻣﻨﻪ ﻓﺈن gﺗﻘﺎﺑﻞ ﻣﻦ Iﻧﺤﻮ .J ب (-ﻟﻴﻜﻦ xﻣﻦ Iو yﻣﻦ . J x +1 =y ⇔ ) y = g (x 2x + 1 .
x 2 + 2x + 1 = ⇔y2 2x + 1 2 2 ⇔ x + 2x (1 − y ) + 1 − y 2 = 0 Δ ' = y 2 ( y 2 − 1) ≥ 0 إذن x 1 = y 2 − 1 − y y 2 − 1و y 2 − 1
x 2 = y 2 −1+ y
اﻟﺤﻞ x1 = y 2 − 1 − y y 2 − 1ﻏﻴﺮ ﻣﻘﺒﻮل ﻷﻧﻪ ﺳﺎﻟﺐ . إذن x = x2 = y 2 − 1 + y y 2 − 1
وﻣﻨﻪ ﻓﺈن g −1 (x ) = x 2 − 1 + x x 2 − 1 :
ﺗﻤﺮﻳﻦ:5 0 h '(x ) = 3(1 − 2 x ) -1- I 0 ⇔ 1− 2 x
0 1 ≺ ⇔ x 2 1 ≺ ⇔0≺ x 4
∀x
) h '( x
⎞⎛1 h ⎜ ⎟ -2ﻗﻴﻤﺔ ﻗﺼﻮﻳﺔ ﻟﻠﺪاﻟﺔ . h ⎠⎝4 ⎞⎛1 إذن ⎟ ⎜ (∀x ∈ + ) h ( x ) ≤ h ⎠⎝4 + أي أن (∀x ∈ ) h ( x ) ≤ 0
-1 II )f (x ) − f (0 ⎛ 4x − 1 ⎞ ⎜ = lim+ ∞− 4x ⎟ = − x →0 x →0 x ⎝ x ⎠ إذن اﻟﺪاﻟﺔ fﻏﻴﺮ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻴﻤﻴﻦ ﻓﻲ اﻟﺼﻔﺮ. ﻳﻘﺒﻞ) (Cﻧﺼﻒ ﻣﻤﺎس ﻋﻤﻮدي ﻓﻲ اﻟﻨﻘﻄﺔ ذات اﻷﻓﺼﻮل 0 lim+
*+
-2أ (-ﻟﻜﻞ xﻣﻦ 1 × )f '(x ) = 4 x + (4x − 1 − 8x 2 x 8x − 4x − 1 − 16x x 2 x ⎞1 ⎛ ⎟ 4 ⎜ 3x − 4x x − 12x − 16x x − 1 ⎠4 = ⎝ = 2 x 2 x ) 2h ( x وﺑﺎﺗﺎﻟﻲ ﻓﺈن ﻟﻜﻞ xﻣﻦ : *+ = ) f '(x x =
ب(-
4 1 1 = lim = lim =0 2 x x →+∞ x x x →+∞ 2x إذن ∞lim f ( x ) = −
lim
∞x →+
∞x →+
∞+
0 1/2
∞+ ) f (x 1 ⎞ 1 ⎛ 4 ⎜ = lim − ج− 4 + 2 ⎟ = −∞ ( - ∞x →+ ∞x →+ x ⎠ 2x ⎝ x x x ﻳﻘﺒﻞ ) (Cﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ ﻓﻲ اﺗﺠﺎﻩ ﻣﺤﻮر اﻷراﺗﻴﺐ . lim
-3أ g (-داﻟﺔ ﻣﺘﺼﻠﺔ وﺗﻨﺎﻗﺼﻴﺔ ﻗﻄﻌﺎ ﻋﻠﻰ . I إذن gﺗﻘﺎﺑﻞ ﻣﻦ Iﻧﺤﻮ . J ⎤ ⎤⎞ ⎛ 1 ⎥ ⎟ ⎜ J = g (1) ⎥ lim g (x ), g ⎦⎠ ⎝ 4 ∞⎦ x →+ ⎤1 ⎤ وﻣﻨﻪ ⎥ J = ⎥ −∞, ⎦4 ⎦
ب (-ﻟﺪﻳﻨﺎ gﺗﻘﺎﺑﻞ ﻣﻦ Iﻧﺤﻮ Jو 0 ∈ J إذن 0ﻳﻘﺒﻞ ﺳﺎﺑﻖ وﺣﻴﺪ ﻓﻲ . I ﻳﻌﻨﻲ أن اﻟﻤﻌﺎدﻟﺔ x ∈ I g ( x ) = 0ﺗﻘﺒﻞ ﺣﻼ وﺣﻴﺪا α 3 7 1 2 2 +1 = ) ( gو g( ) = 3− ≺ 0 ﻟﺪﻳﻨﺎ 0 4 4 2 4 ⎡⎤1 3 إذن ⎢ α ∈ ⎥ , ⎣⎦2 4 -4
x )f’(x )fx
ﺗﻤﺮﻳﻦ:6 -1ﻟﺪﻳﻨﺎ ∞lim ( x 2 − 1 ) = lim x 2 = + ∞x →+
ﻳﻌﻨﻲ أن ∞x 2 − 1 = +
∞x →+
lim
∞x →+
إذن ∞lim f ( x ) = +
∞x →+
1 x − x 2 −1
= lim
∞x →−
وﺑﻤﺎ أن ∞x 2 − 1 = +
)x 2 − (x 2 − 1 x − x 2 −1
lim f (x ) = lim
∞x →−
∞x →−
lim
∞x →−
ﻓﺈن ∞lim(x − x 2 − 1 = − وﻣﻨﻪ lim f ( x ) = 0
∞x →−
-2أ(-
+ x −1 −1 x −1 ⎞ x +1 ∞⎟ = + ⎠ x 2 −1 2
f (x ) − f (1) lim x = x −1 ⎛ ⎛ ⎞ x 2 −1 lim ⎜1 + ⎟ = lim ⎜1 + ⎜ x →1 ⎝ x − 1 ⎟⎠ xx →11 ⎝ x 1 lim
ﻷن lim( x + 1) = 2و lim x 2 − 1 = 0+ x →1 x 1
x →1
)f (x ) − f (1 ∞= + x −1
lim x →1 x 1
وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﺪاﻟﺔ fﻏﻴﺮ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ ﻳﻤﻴﻦ 1و ) (Cﻳﻘﺒﻞ ﻧﺼﻒ ﻣﻤﺎس ﻋﻤﻮدي ﻣﻮﺟﻪ ﻧﺤﻮ اﻷراﺗﻴﺐ اﻟﻤﻮﺟﺒﺔ ﻋﻨﺪ اﻟﻨﻘﻄﺔ )A (1,1
)f (x ) − f ( −1 x + x 2 −1 +1 = lim x →−1 x →−1 x +1 x +1 x ≺ −1 x ≺ −1 lim
⎛ ⎞ x −1 = lim ⎜1 + ∞⎟ = − x →−1 ⎠ x 2 −1 ⎝ x ≺ −1 ﻷن lim (x − 1) = −2و lim x 2 − 1 = 0+
x →−1 x ≺ −1
x →−1
)f (x ) − f (−1 ∞= − x →−1 x +1 x ≺ −1 lim
اﻟﺪاﻟﺔ fﻏﻴﺮ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ ﻳﺴﺎر 1و ) (Cﻳﻘﺒﻞ ﻧﺼﻒ ﻣﻤﺎس ﻋﻤﻮدي ﻣﻮﺟﻪ ﻧﺤﻮ ﻣﺤﻮر اﻷراﺗﻴﺐ اﻟﻤﻮﺟﺒﺔ ﻋﻨﺪ اﻟﻨﻘﻄﺔ )B (−1, −1
بx ∈ ]−∞,1[ ∪ ]1, +∞[ (-
x
f '(x ) = 1 +
x 2 −1
x2 −1 + x
= )(∀x ∈ ]−∞,1[ ∪ ]1, +∞[ ) f '( x
ج ( -ﻟﺪﻳﻨﺎ 0 :
(∀x ∈ ]+1, +∞[ ) x 2 − 1 + x
وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن 0
) (∀x ∈ ]1, +∞[ )f '(x
x2 −1
إذن fداﻟﺔ ﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ [∞]1, + ) [∞(∀x ∈ ]−∞, −1[ ∪ ]1, + x 2 −1 − x 2
= ) f '(x
) x 2 − 1( x 2 − 1 − x −1
=
) x 2 − 1( x 2 − 1 − x 0
)
x 2 −1 − x
(
)[(∀x ∈ ]−∞, −1
إذن (∀x ∈ ]−∞, −1[) f '(x ) ≺ 0 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن fﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ
[]−∞, −1
-3أx ∈ ]1, +∞[ (-
)
x 2 −1 − x
=0
(
−1 x 2 −1 + x
)
lim [ f (x ) − 2x ] = lim
∞x →+
∞x →+
lim
∞x →+
ﻷن ∞x 2 − 1 + x = +
(
lim
∞x →+
إذن اﻟﻤﻨﺤﻨﻰ) (Cﻳﻘﺒﻞ ﻣﻘﺎرب ﺑﺠﻮار ∞ +ﻣﻌﺎدﻟﺘﻪ y = 2x
ب(-
g -1داﻟﺔ ﻣﺘﺼﻠﺔ وﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ . I و g (I ) = I إذن gﺗﻘﺎﺑﻞ ﻣﻦ Iﻧﺤﻮ .I وﻣﻨﻪ ﻓﺈن gﺗﻘﺒﻞ داﻟﺔ ﻋﻜﺴﻴﺔ g −1ﻣﻌﺮﻓﺔ ﻋﻠﻰ . I
ﺗﻤﺮﻳﻦ:7 -1أ= ]−∞, 0[ ∪ ]0, +∞[ (- ب ( -إذا آﺎن
x 2 +3 و −x
∈ xﻓﺈن
*
= Df
*
∈ −x
*
f (−x ) = −2x −
x2 + 3 = −2 x + )= − f ( x x إذن ) ∀ ∈ * f (− x ) = −f (x إذن fداﻟﺔ ﻓﺮدﻳﺔ. ⎛ ⎞⎞ 3 ⎛ ⎟ ⎟ x 2 ⎜1 + 2 ⎜ ⎟⎠ ⎝ x lim f (x ) = lim ⎜ 2x − -2 ∞x →+ ⎜ ∞x →+ ⎟ x ⎜⎜ ⎟⎟ ⎝ ⎠ ⎛ ⎞ 3 ∞= lim ⎜⎜ 2 x − 1 + 2 ⎟⎟ = + ∞x →+ ⎠ x ⎝ إذن ∞lim f ( x ) = + ∞x →+
x ∈ I (- 1-3
x +3 − 2x + 1 x 2
f (x ) − (2x − 1) = 2x −
x 2 +3 x − x 2 +3 = x x
= 1−
ب(-
−3
=
)x − (x + 3 2
2
)) x (x + x + 3 ﺑﻤﺎ أن ∞lim x ( x + x + 3 ) = + 2
(
x x + x 2 +3
= )f (x ) − (2x − 1
2
∞x →+
ﻓﺈن = 0
)
−3
lim
(
∞x →+
x x + x 2 +3
إذن lim [ f (x ) − (2x − 1)] = 0
∞x →+
وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ) ( Δﻣﻘﺎرب ﻟﻠﻤﻨﺤﻨﻰ ) (Cﺑﺠﻮار ∞+ ج (-ﻟﺪﻳﻨﺎ
)
−3
(
x x + x 2 +3
أي أن f ( x ) ≺ 2 x − 1
) (∀x ∈ I
) (∀x ∈ I
إذن اﻟﻤﻨﺤﻨﻰ ) (Cﻳﻮﺟﺪ ﺗﺤﺖ اﻟﻤﺴﺘﻘﻴﻢ ) ( Δﻋﻠﻰ اﻟﻤﺠﺎل
[∞]0, +
− x2 + 3 -4أ(-
x2 x +3
x2
)x 2 − (x 2 + 3
2
x ≠ 0; f '( x) = 2 −
= 2−
x 2 x 2 +3 3 = 2+ 2 x x 2 +3 3 x2 + 3
ب(-
2
(∀x ∈ I ) f '( x) = 2 + x
∞+
0 +
∞+
)f(x ∞−
-5أ (-ﺗﻘﺎﻃﻊ ) (Cﻣﻊ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ ﻋﻠﻰ . I x ∈I ⎪⎧ ⎧x ∈ I ⇔⎨ 2 ⎨ 2 ⎩ f (x ) = 0 ⎪⎩ x − x + 3 = 0 ⎧⎪ x 2 = x 2 + 3 ⎨⇔ ⎪⎩ x ∈ I ⎧ 4x 4 − x 2 − 3 = 0 ⎨⇔ ⎩ x ∈I اﻟﻤﻌﺎدﻟﺔ 4x 4 − x 2 − 3 = 0ﻳﺆول ﺣﻠﻬﺎ إﻟﻰ ﻣﻌﺎدﻟﺔ ﻣﻦ اﻟﺪرﺟﺔ اﻟﺜﺎﻧﻴﺔ . 3 ) x 2 = −أو 4 x 4 − x 2 − 3 = 0 ⇔ ( x 2 = 1 4 ) x = −1أو ⇔ ( x = 1 ⎫ f (x ) = 0 ⎬ ⇔ x =1 x ∈I ⎭ إذن اﻟﻤﻨﺤﻨﻰ ) (Cﻳﻘﻄﻊ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ ﻓﻲ اﻟﻨﻘﻄﺔ ) A (1, 0ﻋﻠﻰ اﻟﻤﺠﺎل I
ﻣﻌﺎدﻟﺔ ) (Tﻣﻤﺎس ) (Cﻋﻨﺪ اﻟﻨﻘﻄﺔ Aهﻲ: )y = f '(1)( x − 1) + f (1 T : y = 3x − 3 ب-
ﻟﺪﻳﻨﺎ ' fداﻟﺔ ﻓﺮدﻳﺔ إذن ﻣﻨﺤﻨﺎهﺎ ) (Cﻣﺘﻤﺎﺛﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ Oأﺻﻞ اﻟﻤﻌﻠﻢ. -6 gداﻟﺔ ﻣﺘﺼﻠﺔ وﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل . I و = )g (1 إذن gﺗﻘﺎﺑﻞ ﻣﻦ Iﻧﺤﻮ
x )f’(x
:8 ﺗﻤﺮﻳﻦ (-أ-1 lim (x + 2)(x − 3) = lim x = +∞ ﻟﺪﻳﻨﺎ 2
x →+∞
x →+∞
lim f ( x ) = +∞ إذن
x →+∞
2 (x + 2)(x + 3) f (x ) -ب = lim x →−2 x + 2 x →−2 x +2 x −2 x −2 lim
2( x + 2)(x + 3) 2(x + 3) = lim x →−2 ( x + 2) ( x + 2)( x + 3) x →−2 (x + 2)(x + 3) x −2 x −2
= lim
lim 2(3 − x ) = 10 ﺑﻤﺎ أن
x →−2
lim (x + 2)(x + 3) = 0+
x →−2 x −2
lim
x →−2 x −2
lim x →3 x 3
f (x ) = +∞ ﻓﺈن x +2
2 (x + 2)( x − 3) f (x ) = lim x → 3 x −3 x 3 x −3 = lim x →3 x 3
lim x →3 x 3
2(x + 2) = +∞ (x + 2)(x + 3) f (x ) = +∞ x −3
2 (x + 2)(3 − x ) f (x ) = lim x →3 x − 3 x →3 x −3 x ≺3 x ≺3
lim
lim x →3 x ≺3
−2(x + 2) = −∞ (x + 2)(3 − x )
f ( x) = −∞ lim x − 3 x→3 x≺3 (-أ-2
⎧⎪f (x ) = 2 (x + 2)(3 − x ); −2 ≺ x ≺ 3 ⎨ ⎪⎩ f (x ) = 2 (x + 2)(x − 3); x 3 x ∈ ]−2,3[ إذا آﺎن f '(x ) =
2 [ (x + 2)(3 − x ) ]
ﻓﺈن 2 (x + 2)(3 − x ) 1 − 2x (∀x ∈ ]−2,3[) f '(x ) = أي أن (x + 2)(3 − x )
( ( x − 2)(3 − x)
0 ) ﻷن1 − 2x ] هﻲ إﺷﺎرة−2,3[ ﻋﻠﻰf '( x ) وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن إﺷﺎرة x ∈ ]3, +∞[ إذا آﺎن
[(x + 2)(x − 3)] '(x ) = 2
'
ﻓﺈن 2 (x + 2)(x − 3) 2x − 1 (∀x ∈ ]3, +∞[) f '(x ) = (x + 2)(x − 3) f
x
3⇒ x
1 ⇒ 2x − 1 0 2 ⇒ f '( x ) 0
∀x ∈ ]3, +∞[ f '(x )
0 إذن -ب
(أ-3 2 (x + 2)(x − 3) f (x ) lim = lim x →+∞ x →+∞ x x
2 x2 − x − 6 x →+∞ x
= lim
⎛ 1 6 ⎞ 2 x 2 ⎜1 − − 2 ⎟ ⎝ x x ⎠ = lim x →+∞ x 1 6 2x x 2 1 − − 2 x x = lim x →+∞ x 1 6 = lim 2 1 − − 2 = 2 x →+∞ x x 1 6 lim = lim 2 = 0 ﻷن x →+∞ x x →+∞ x lim [ f ( x ) − 2x ] = lim ⎡⎣ 2 ( x + 2)( x − 3) − 2x ⎤⎦ x →+∞ x →+∞ (x + 2)(x − 3) − x 2 = 2 lim x →+∞ (x + 2)(x − 3) + x 6−x = 2 lim x →+∞ (x + 2)(x − 3) + x
6 −1 x = 2 lim = −1 ∞x →+ 1 6 1− − +1 x x إذن اﻟﻤﺴﺘﻘﻴﻢ ) (Dﻣﻘﺎرب ﻟﻠﻤﻨﺤﻨﻰ ) (Cﺑﺠﻮار ∞+
ب (-ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﻣﻦ [∞]3, + )f (x ) − (2x − 1) = 2 (x + 2)(x − 3) − (2x − 1 4( x + 2)(x − 3) − (2x − 1) 2 )2 (x + 2)( x − 3) + (2x − 1 −25 = ≺0 )2 (x + 2)(x − 3) + (2x − 1 =
ﻟﻜﻞ xﻣﻦ [∞f (x ) ≺ 2x − 1; ]3, + إذن اﻟﻤﺴﺘﻘﻴﻢ ) (Dﻳﻮﺟﺪ ﺗﺤﺖ اﻟﻤﻨﺤﻨﻰ ) (Cﻋﻠﻰ اﻟﻤﺠﺎل [∞. ]3, +
:9 ﺗﻤﺮﻳﻦ -1 x ∈ D f ⇔ x + 2x ≥ 0 ⇔ x ( x + 2) ≥ 0 2
x ∈ ]−∞, 2] ∪ [ 0, +∞[
D f = ]−∞, 2] ∪ [ 0, +∞[ lim (x 2 + 2x ) = lim x 2 = +∞ ﺑﻤﺎ أن
x →+∞
x →+∞
lim
x →+∞
x 2 + 2x = +∞ و lim x = +∞ و
x →+∞
lim f ( x ) = +∞ ﻓﺈن
x →+∞
lim (x + x 2 + 2x ) = lim
x →+∞
x 2 + 2x − x 2
-2 x 2 + 2x − x 2x 2x = lim x 2 + 2 − x x →−∞ x 1 + 2 − x x 2 = lim x →−∞ 2 − 1 + −1 x 2⎞ ⎛ lim ⎜1 + ⎟ = 1 ﻟﺪﻳﻨﺎ x →−∞ ⎝ x ⎠ ⎛ ⎞ 2 lim ⎜⎜ − 1 + − 1⎟⎟ = −2 إذن x →−∞ x ⎝ ⎠ x →+∞
= lim
x →−∞
lim f (x ) = −1 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن
x →−∞
y = −1 ﻣﻌﺎدﻟﺘﻪ−∞ ( ﻳﻘﺒﻞ ﻣﻘﺎرب ﺑﺠﻮارC) وﻣﻨﻪ ﻓﺈن اﻟﻤﻨﺤﻨﻰ -3
f (x ) − f (−2) x + x 2 + 2x + 2 lim = lim x →−2 x →−2 x +2 x +2 x ≺ −2 x ≺ −2 ⎛ x 2 + 2x = lim ⎜ 1 + x →−2 ⎜ x +2 x ≺ −2 ⎝
⎞ ⎛ ⎞ x 2 + 2x lim 1 = + ⎟ ⎜ ⎟ ⎟ x →−2 ⎜ (x + 2) x 2 + 2x ⎟ ⎠ ⎠ x ≺ −2 ⎝ ⎛ ⎞ x = lim ⎜1 + = −∞ ⎟ 2 x →−2 + 2 x x ⎠ x ≺ −2 ⎝ .-2 ﻏﻴﺮ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ ﻳﺴﺎرf إذن اﻟﺪاﻟﺔ ⎛ f (x ) − f (0) x 2 + 2x ⎞ lim = lim ⎜1 + ⎟ x →0 x →0 ⎜ ⎟ x x x 0 x 0 ⎝ ⎠
⎛ ⎛ ⎞ )x (x + 2 ⎞ x +2 = lim ⎜1 + ⎟ = lim ⎜1 + ⎟ 2 x →0 x →0 ⎝ x x + 2x ⎠ x 0 ⎠ x 2 + 2x ⎝ x 0 )f (x ) − f (0 lim إذن ∞= + x →0 x x 0
وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﺪاﻟﺔ fﻏﻴﺮ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ ﻳﻤﻴﻦ .0 -4أ (-ﻟﻴﻜﻦ xﻣﻦ x +1
[∞]−∞, −2[ ∪ ]0, + 2x + 2
= 1+
إذن x 2 + 2x )}(∀x ∈ D − {−2, 0
2 x 2 + 2x
f '(x ) = 1 +
0 ⇒ x + 1 0 ⇒ x + 1 + x 2 + 2x ) ∀x ∈ ]0, +∞[ f '(x
0 0
إذن fﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ﻟﻴﻜﻦ xﻣﻦ . D ) (x + 1) − (x 2 + 2x 2
(x + 1) − x 2 + 2x 1
x
[∞]0, + = x + 1 + x 2 + 2x
=
(x + 1) − x 2 + 2x ⎧⎪x + 1 ≺ −1 ≺ 0 ⎨ ⇒ x ≺ −2 2 ⎪⎩ − x + 2x ≺ 0 إذن ∀x ∈ ]−∞, −2[ f '(x ) ≺ 0 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﺪاﻟﺔ fﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ
∞+ ∞+
0
[]−∞, −2 ∞−
-2
+
x )f’(x
-1
)f(x
0
-2
-5أ(- lim [ f (x ) − (2x + 1) ] = lim ⎡ x + x + 2x − 2x − 1 ⎣ ∞x →+ )= lim ⎡ x 2 + 2 x − ( x + 1 ⎣ ∞x →+ x 2 + 2x − (x + 1)2 = lim ∞x →+ )x 2 + 2x + (x + 1 2
∞x →+
−1
= lim
x 2 + 2x + x + 1
∞x →+
ﺑﻤﺎ أن ∞ lim ( x + 1) = +و ∞x 2 + 2x = + ∞x →+
−1
lim
∞x →+
وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن = 0 x 2 + 2x + x + 1 ﻳﻌﻨﻲ أن lim [ f (x ) − (2x + 1)] = 0
lim
∞x →+
وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﻤﺴﺘﻘﻴﻢ اﻟﺬي ﻣﻌﺎدﻟﺘﻪ y = 2x + 1ﻣﻘﺎرب ﻟﻠﻤﻨﺤﻨﻰ ) (Cﺑﺠﻮار ∞+
-6أ g (-داﻟﺔ ﻣﺘﺼﻠﺔ وﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ [∞ [ 0, +و [∞g ([ 0, +∞[ ) = [ 0, + +
ﻧﺤﻮ
+
.
إذن gﺗﻘﺎﺑﻞ ﻣﻦ ﻟﻴﻜﻦ yﻣﻦ . + ﻧﻘﻮم ﺑﺤﻞ اﻟﻤﻌﺎدﻟﺔ ) x ≥ 0 y = g ( x ⎧⎪ y = x + x 2 + 2x ⎞ ) ⎛ y = g (x ⇔ ⎨ ⎜ ⎟ ⎝x ≥ 0 ⎠ ⎪⎩ x ≥ 0 ⇔ ⎡ y − x = x 2 + 2, x ≥ 0 ⎣ ⎦⎤ ⇔ ⎡⎣( y − x ) 2 = x 2 + 2x , x ≥ 0 2 2 2 )⎣⎡⇔ ( y − 2xy + x = x + 2x , x ≥ 0 ⎦⎤ ⇔ ⎡⎣(2 y + 2)x − y 2 = 0, x ≥ 0 ﺑﻤﺎ أن y ≥ 0ﻓﺈن 2 y + 2 ≠ 0
y2 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن 2y + 2 g −1 : + → + x2 2x + 2
x
= x
ﺗﻤﺮﻳﻦ:10 – a-1ﺗﺤﺪﻳﺪ . D ﺑﻤﺎ أن 27 + x 2 ≥ 0} :
و D = {x ∈ IR / 2x ≠ 0
وﺑﻤﺎ أن 27 + x 2 ≥ 0 :ﻟﻜﻞ xﻣﻦ IR ﻓﺈن D = {x ∈ IR / 2x ≠ 0} :
}= {x ∈ IR / x ≠ 0
*D = IR إذن : [∞= ]−∞, 0[ ∪ ]0, + -bﺣﺴﺎب ﻧﻬﺎﻳﺎت fﻋﻨﺪ ﻣﺤﺪات D ∞27 + x 2 = +
x +1 x 1 = lim و = x ∞→+ 2x 2x 2 إذن lim f (x ) = +∞ :
lim
∞x →+
lim
∞x →+
∞x →+
•
ﻟﺪﻳﻨﺎ lim 27 + x 2 = 27 : x →0
x +1 x +1 lim+و ∞= − وﻟﺪﻳﻨﺎ = +∞ : x →0 x →0 2x 2x إذن ∞ lim+ f ( x) = +و ∞lim− f ( x ) = − lim−
x→0
x→0
a-2اﻟﺘﺤﻘﻖ ﻣﻦ ﺻﺤﺔ اﻟﻤﺘﺴﺎوﻳﺔ ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﻣﻦ * IR x +1 ⎛ x +1 ⎞ x +1 27 + x 2 − ⎜ f (x ) − =⎟ 2 ⎝ 2 ⎠ 2x
x +1 ) x + 1 ( 27 + x 2 − x )( 27 + x 2 + x 2 ( 27 + x − x ) = = 2x 2x 27 + x 2 + x x + 1 27 + x 2 − x 2 = . 2x 27 + x 2 + x
⎞ 27 ⎛⎛ x +1 ⎞ x +1 ⎜ f (x ) −ﻟﻜﻞ xﻣﻦ *IR ⎜ إذن ⎟ : =⎟ 2 ⎠ ⎝ 2 ⎠ 2x ⎝ x + 27 + x -bاﻻﺳﺘﻨﺘﺎج ⎞ 27 ⎛⎛ x +1 ⎞ x +1 ⎜ f (x ) −ﻟﻜﻞ xﻣﻦ * IR ⎜ ﺑﻤﺎ أن ⎟ : =⎟ 2 ⎠ ⎝ 2 ⎠ 2x ⎝ x + 27 + x 27 x +1 1 وﺑﻤﺎ أن = lim limو = 0 2 x →+∞ 2 x ∞x →+ 2 x + 27 + x ⎞ ⎛x +1 27 ⎜ ﻓﺈن ⎟ = 0 : 2 x →+∞ 2x ⎠ ⎝ x + 27 + x ⎞⎛ x +1 ⎜ lim f (x ) − إذن ⎟ = 0 : ∞x →+ ⎠ ⎝ 2x x +1 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﻤﺴﺘﻘﻴﻢ ) ( Δ1ذا اﻟﻤﻌﺎدﻟﺔ = y 2 lim
هﻮ ﺑﺎﻟﻔﻌﻞ ﻣﻘﺎرب ﻣﺎﺋﻞ ﻟﻠﻤﻨﺤﻨﻰ ) (Cﺑﺠﻮار ∞+ -cﻟﻨﺒﻴﻦ أن ) ( Δ 2ﻣﻘﺎرب ﻣﺎﺋﻞ ﻟﻠﻤﻨﺤﻨﻰ ) (Cﺑﺠﻮار ∞− ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﻣﻦ *IR x +1 ⎛ x +1⎞ x +1 27 + x 2 + f (x ) − ⎜ − =⎟ 2 ⎠ 2x 2 ⎝ x +1 ) (x + 1)(27 + x 2 − x 2 = ) ( 27 + x 2 + x = 2x ) 2x ( 27 + x 2 − x x +1 27 . = 2x 27 + x 2 − x 27 x +1 1 وﺑﻤﺎ أن : lim = limو = 0 x →−∞ 2 x ∞x →− 2 27 + x 2 − x x +1 27 lim ﻓﺈن = 0 : x→−∞ 2 x 27 + x 2 − x ⎞⎛ x +1 lim f ( x) − ⎜ − إذن ⎟ = 0 : ∞x→− ⎠ 2 ⎝ x +1 y = −هﻮ ﺑﺎﻟﻔﻌﻞ ﻣﻘﺎرب ﻣﺎﺋﻞ ﻟﻠﻤﻨﺤﻨﻰ ) (Cﺑﺠﻮار ∞− وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﻤﺴﺘﻘﻴﻢ ) ( Δ 2ذا اﻟﻤﻌﺎدﻟﺔ 2 -a -3ﺣﺴﺎب ) f '( x اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ * IRوﻟﺪﻳﻨﺎ ﻟﻜﻞ xﻣﻦ * IR 11 2 0 2x x +1 = ) f '(x 27 + x 2 + . 2 4x 2x 2 27 + x 2 x +1 ) x 2 (x + 1) − (27 + x 2 − 27 + x 2 + = 2x 2 2 27 + x 2 2x 2 x 2 + 27 x 3 + x 2 − 27 − x 2 x 3 − 27 = = 2x 2 − x 2 + 27 2x 2 x 2 + 27 -bﺗﻐﻴﺮات f 3 إﺷﺎرة ) f '( xهﻲ إﺷﺎرة x − 27 3 وﻟﺪﻳﻨﺎ ﻟﻜﻞ xﻣﻦ * f '(x ) = 0 ⇔ x − 27 = 0 : IR ⇔ x 3 = 27 ⇔x =3 f '(x ) 0 ⇔ x 3 − 27 0 و 3 x 27 x 3 و f '( x ) ≺ 0 ⇔ x ≺ 3
=
إذن اﻟﺪاﻟﺔ fﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ اﻟﻤﺠﺎل [∞ [3, +وﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ آﻞ ﻣﻦ اﻟﻤﺠﺎﻟﻴﻦ -cﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ f
[ ]−∞, 0و ]]0,3
-a-4ﺗﻘﺎﻃﻊ ) (Cﻣﻊ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ ﻟﻴﻜﻦ xﻋﺪدا ﺣﻘﻴﻘﻴﺎ
(x + 1) 27 + x 2 ⇔ f (x ) = 0 ﺑﻤﺎ أن = 0 : 2x ⇔ x +1 = 0 ⇔ x = −1 ﻓﺈن ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ ﻳﻘﻄﻊ ) (Cﻓﻲ اﻟﻨﻘﻄﺔ )B ( −1, 0
ﺗﻤﺮﻳﻦ:11 * -1ﺗﺤﺪﻳﺪ D ﻟﻴﻜﻦ xﻋﺪدا ﺣﻘﻴﻘﻴﺎ 3 ﻟﺪﻳﻨﺎ 1 + x 2 ≥ 0و x ∈ D ⇔ x ≠ 0 1+ x 2 ≥ 0و x ≠ 0 وﺑﻤﺎ أن 1 + x 2 ≥ 0 :ﻟﻜﻞ xﻣﻦ IR *D = IR [∞= ]−∞, 0[ ∪ ]0, +
• اﻟﺘﺤﻘﻖ ﻣﻦ أن fﻓﺮدﻳﺔ ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﻣﻦ *IR
2(−x 2 ) − 1 ﻟﺪﻳﻨﺎ ( −x ) ∈ IR * :و . 1 + (−x )2 3 ) (−x
= ) f (−x
2x 2 − 1 ) 1 + x 2 = −f ( x −x 3 إذن اﻟﺪاﻟﺔ fهﻲ ﺑﺎﻟﻔﻌﻞ ﻓﺮدﻳﺔ.
=
-2ﺣﺴﺎب اﻟﻨﻬﺎﻳﺘﻴﻦ واﻟﺘﺄوﻳﻞ اﻟﻬﻨﺪﺳﻲ 2x 2 − 1 . 1+ x 2 ∞x →+ x3
lim f (x ) = lim
ﻟﺪﻳﻨﺎ:
∞x →+
2x 2 − 1 ⎞ 1 ⎛ ⎟ . x 2 ⎜1 + 2 3 ∞x →+ x ⎠ ⎝ x
= lim
1 2x 2 − 1 1 = lim .x 1 + 2 2 3 ∞x →+ x x x
1+
2x 2 − 1 .x ∞x →+ x3
= lim
2x 2 − 1 1 . 1+ 2 2 ∞x →+ x x 2x 2 − 1 2x 2 lim = lim وﺑﻤﺎ أن = 2 : ∞x →+ x →+∞ x 2 x2 1 lim 1 + 2 = 1 + 0 = 1 و ∞x →+ x ﻓﺈن : lim f ( x ) = 2.1 = 2
= lim
∞x →+
وهﺬا ﻳﻌﻨﻲ هﻨﺪﺳﻴﺎ أن اﻟﻤﺴﺘﻘﻴﻢ ذا اﻟﻤﻌﺎدﻟﺔ y = 2هﻮ ﻣﻘﺎرب أﻓﻘﻲ ﻟﻠﻤﻨﺤﻨﻰ ) (ζﺑﺠﻮار ∞+ 2x 2 − 1 وﻟﺪﻳﻨﺎ = −∞ : x3 إذن lim+ f ( x ) = −∞ :
lim+و lim+ 1 + x 2 = 1 x →0
x →0
x →0
وهﺬا ﻳﻌﻨﻲ هﻨﺪﺳﻴﺎ أن اﻟﻤﺴﺘﻘﻴﻢ ذا اﻟﻤﻌﺎدﻟﺔ x =0أي ﻣﺤﻮر اﻷراﺗﻴﺐ هﻮ ﻣﻘﺎرب رأﺳﻲ ﻟﻠﻤﻨﺤﻨﻰ ) . (ζ -3ﺗﻐﻴﺮات f اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ Dوﻟﺪﻳﻨﺎ ﻟﻜﻞ xﻣﻦ : D '
⎞ ⎛ 2x 2 − 1 2x 2 − 1 2 + + 1 x ') ( 1 + x 2 ⎜ = ) f '(x ⎟ 3 3 x ⎝ x ⎠
)4x .x 3 − 3x 2 (2x 2 − 1 2x 2 − 1 2x 2 1 + x + . 6 3 x x 2 1+ x 2 4x 2 − 6x 2 + 3 2x 2 − 1 2 x 2 = 1 + x + . x4 x2 1+ x 2 =
3 − 2x 2 2x 2 − 1 )(3 − 2x 2 )(1 + x 2 ) + x 2 (2x 2 − 1 2 + + = x 1 2 4 x4 x 1+ x 2 x 1+ x 2 3 + 3x 2 − 2x 2 − 2x 4 + 2x 4 − x 2 3 = = 4 4 x 1+ x 2 x 1+ x 2 إذن f '( x ) ; 0ﻟﻜﻞ xﻣﻦ * IR وهﺬا ﻳﻌﻨﻲ أن fﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ آﻞ ﻣﻦ اﻟﻤﺠﺎﻟﻴﻦ
[ ]−∞, 0و [∞]0, +
وﻣﻨﻪ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ f
ﻻﺣﻆ أن : ﻟﺪﻳﻨﺎ lim− f (x ) = +∞ :
x →0
و lim f (x ) = −2
∞x →−
ﻷن lim+ f ( x ) = −∞ :و lim f ( x ) = 2 x →0
و fﻓﺮدﻳﺔ.
∞x →+
=
-4أ -ﻟﻨﺒﻴﻦ أن gﺗﻘﺎﺑﻞ ﻣﻦ * IR +ﻧﺤﻮ ﻣﺠﺎل J ﺑﻤﺎ أن اﻟﺪاﻟﺔ gﻣﺘﺼﻠﺔ وﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ [∞]0, +
ﻓﺈﻧﻬﺎ ﺗﻘﺎﺑﻞ ﻣﻦ [∞ ]0, +ﻧﺤﻮ اﻟﻤﺠﺎل J = g (]0, +∞[) = ]−∞, 2[ : إذن ﻓﻬﻲ ﺗﻘﺒﻞ داﻟﺔ ﻋﻜﺴﻴﺔ g −1ﻣﻌﺮﻓﺔ ﻣﻦ [ ]−∞, 2ﻧﺤﻮ ب -ﻟﻨﺒﻴﻦ أن g −1ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻤﺠﺎل .J
[∞]0, +
ﻟﺪﻳﻨﺎ اﻟﺪاﻟﺔ gﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ [∞ ]0, +وﻟﺪﻳﻨﺎ ﻟﻜﻞ xﻣﻦ g '( x ) ≠ 0 إذن اﻟﺪاﻟﺔ g −1ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻤﺠﺎل . J ج -رﺳﻢ
[∞]0, +
)' (ζ
ﻣﻤﺎﺛﻞ ﻣﻨﺤﻨﻰ fﻋﻠﻰ [∞ ]0, +هﻮ اﻟﻤﻨﺤﻨﻰ )' (ζﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺴﺘﻘﻴﻢ ذي اﻟﻤﻌﺎدﻟﺔ ) y = xأﻧﻈﺮ اﻟﺸﻜﻞ(
ﺗﻤﺮﻳﻦ:12 ) f (x *-1ﺣﺴﺎب x
lim+
x →0
2 3
f ( x ) 6x = −4 x x −1
2
−1 6x 3 6 = − 4 = 6x 3 − 4 = 3 − 4 x x 6 وﺑﻤﺎ أن lim+ = 3 = +∞ : x →0 x ) f (x ﻓﺈن : ∞= + lim x → 0+ x ﻣﻠﺤﻮﻇﺔ :اﻟﺪاﻟﺔ fﻏﻴﺮ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻴﻤﻴﻦ ﻓﻲ 0 • اﻟﺘﺄوﻳﻞ اﻟﻬﻨﺪﺳﻲ اﻟﻤﻨﺤﻨﻰ ) (ζﻳﻘﺒﻞ ﻓﻲ اﻟﻨﻘﻄﺔ ) O(0,0ﻧﺼﻒ ﻣﻤﺎس رأﺳﻲ ﻣﻮﺟﻪ ﻧﺤﻮ اﻷﻋﻠﻰ.
-2ﺗﺤﺪﻳﺪ اﻟﻔﺮع اﻟﻼﻧﻬﺎﺋﻲ ) f (x 6 ﺑﻤﺎ أن = lim 3 − 4 : lim ∞x →+ x ∞→+ x x = 0-4 = -4 و
2 3
lim f ( x ) + 4x = lim 6x
∞x →+
∞x →+
∞= lim 6 3 x 2 = + ∞x →+
ﻓﺈن اﻟﻤﻨﺤﻨﻰ ) (ζﻳﻘﺒﻞ ﺑﺠﻮار ∞ +ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ اﺗﺠﺎهﻪ اﻟﻤﺴﺘﻘﻴﻢ ذو اﻟﻤﻌﺎدﻟﺔ y = −4x
-3ﺟﺪول ﺗﻐﻴﺮات f اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ
*+
IRوﻟﺪﻳﻨﺎ ﻟﻜﻞ xﻣﻦ
2
2 −1 f '( x) = 6. x 3 − 4 3 ⎞ ⎛ −1 ⎛ 1 ⎞ ⎟= 4 ⎜ x 3 − 1⎟ = 4 ⎜ 3 − 1 ⎜ ⎟ ⎠ ⎝ x ⎝ ⎠
) (1 − x =4 3
3
x إﺷﺎرة ) f '( xهﻲ إﺷﺎرة 1 − 3 x وﻣﻨﻪ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ f
*+
IR
⎛ 2 ⎞ ⎟ lim f (x ) = lim x ⎜ 6x 3 − 4 ﻣﻼﺣﻈﺔ :ﻟﺪﻳﻨﺎ: ∞x →+ ∞x →+ ⎝ ⎠ ⎛ 23 −1 ⎞ ⎛ −31 ⎞ ⎟ = lim x ⎜ 6x − 4 ⎟ = lim x ⎜ 6x − 4 ∞x →+ ⎝ ⎝ ∞⎠ x →+ ⎠ ⎛ 6 ⎞ ∞= lim x ⎜ 3 − 4 ⎟ = − ∞x →+ ⎝ x ⎠ 6 ) ﻷن lim x 3 − 4 = 0 − 4 = −4و ∞( lim x = + ∞x →+ ∞x →+ x -4أ -ﺗﺤﺪﻳﺪ ﻧﻘﻄﺘﻲ ﺗﻘﺎﻃﻊ ) (ζﻣﻊ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ. ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﻣﻦ IR + 2
ﻟﺪﻳﻨﺎ f (x ) = 0 ⇔ 6x 3 − 4x = 0 : ⎛ 2 ⎞ 2x ⇔ 2x 3 ⎜ 3 − 2 ⎟ = 0 ⎜ ⎟ ⎠x3 ⎝ 2 2 ⎞ 1− ⎛ ⇔ 2x 3 ⎜ 3 − 2x 3 ⎟ = 0 ⎝ ⎠ 2 1 ⎛ ⎞ ⇔ 2x 3 ⎜ 3 − 2x 3 ⎟ = 0 ⎝ ⎠ 1 3 = x 3أو ⇔ x = 0 2
2
3
⎞⎛ 1 ⎞ ⎛3 ⎟ ⎜ = ⎟ ⎜ x 3أو ⇔ x = 0 ⎠⎝ ⎠ ⎝2 27 = xأو ⇔ x = 0 8
27 إذن ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ ﻳﻘﻄﻊ ) (ζﻓﻲ اﻟﻨﻘﻄﺘﻴﻦ ) O(0,0و ), 0 8 ب -ﻣﻌﺎدﻟﺔ ) ( Δ 27 ﻣﻌﺎدﻟﺔ ) ( Δﻣﻤﺎس ) (ζﻓﻲ اﻟﻨﻘﻄﺔ ذات اﻷﻓﺼﻮل 8 ⎞ 27 ⎛ ⎞ ⎛ 27 ⎞ ⎛ 27 ⎟ ⎜ y = f ⎜ ⎟×⎜ x − ⎟ + f ⎠ 8 ⎝ ⎠ ⎝ 8 ⎠ ⎝ 8 ⎛4 ⎞ 27 ﻳﻌﻨﻲ أن : y = − ⎜x − ⎟+0 ⎝3 ⎠ 8 4 9 أي أن : y =− x + 3 2
هﻲ:
(A
*-5ﻟﻨﺒﻴﻦ أن gﺗﻘﺎﺑﻞ اﻟﺪاﻟﺔ gﻣﺘﺼﻠﺔ وﺗﻨﺎﻗﺼﻴﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل [∞I = [1, +
وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈﻧﻬﺎ ﺗﻘﺎﺑﻞ ﻣﻦ Iﻧﺤﻮ اﻟﻤﺠﺎل ]g (I ) = ]−∞, 2 أي أﻧﻬﺎ ﺗﻘﺒﻞ داﻟﺔ ﻋﻜﺴﻴﺔ g −1ﻣﻌﺮﻓﺔ ﻣﻦ ] ]−∞, 2ﻧﺤﻮ
•
ﺣﺴﺎب )( g −1 ) '(0
1 ﺑﻤﺎ أن : ))g ( g −1(0
= )( g −1) '(0
27 27 وﺑﻤﺎ أن g ( ) = 0 :ﻓﺈن 8 8
1 ﻓﺈن : ⎞ ⎛ 27 ⎟ ⎜' g ⎠ ⎝ 8
= )( g −1 )(0
= )( g −1) '(0
1 3 =− 4 4 − 3
=
[∞[1, +
ﺗﻤﺮﻳﻦ:13 -1أ -ﺣﺴﺎب ) lim f ( x
∞x→−
∞1 + x 2 = +
ﺑﻤﺎ أن lim − x = +∞ :
∞x →−
ﻓﺈن 1 + x 2 − x = +∞ : إذن f ( x ) = +∞ :
ﺑﻤﺎ أن :
lim
=0
1 + x2 − x2
1+ x 2 + x
lim
∞x→−
∞x→−
ب -ﻟﻨﺒﻴﻦ أن f ( x) = 0
1 + x2 + x 1
lim
∞x→−
lim
∞x→+
1 + x 2 − x = lim
∞x→+
lim
∞x→+
= lim
∞x →+
ﻓﺈن lim f (x ) = 02 = 0 :
∞x →+
) −2 f (x
= ) f '(x
-2أ -ﻟﻨﺒﻴﻦ أن : 1+ x 2 اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ IRوﻟﺪﻳﻨﺎ ﻟﻜﻞ xﻣﻦ : IR ' ) f '(x ) = 2( 1 + x 2 − x )( 1 + x 2 − x
⎛ 2x ⎞ ⎜ ) = 2( 1 + x 2 − x ⎟− 1 2 ⎝ 2 1+ x ⎠ ⎞ ⎟ ⎟ ⎠
⎛ x − 1+ x 2 ⎛ x ⎞ 2 ⎜ ) = 2( 1 + x − x ⎜ ) − 1⎟ = 2( 1 + x − x 2 ⎜ 1+ x 2 ⎝ 1+ x ⎠ ⎝ 2
−2( 1 + x 2 − x ) 2 1+ x 2
=
) −2( 1 + x 2 − x )( 1 + x 2 − x 1+ x 2
) −2f ( x 1+ x 2
ب *-ﻟﻨﺒﻴﻦ أن f '(x ) ≠ 0ﻟﻜﻞ xﻣﻦ IR ﻟﺬﻟﻚ ﺳﻨﺒﻴﻦ أن اﻟﻤﻌﺎدﻟﺔ f '( x ) = 0ﻻ ﺗﻘﺒﻞ أي ﺣﻞ ﻓﻲ . IR ) −2f (x ⇔ f '(x ) = 0 ﻟﺪﻳﻨﺎ = 0 : 1+ x 2 ⇔ f (x ) = 0
⇔ ( 1 + x 2 − x )2 = 0 ⇔ 1+ x 2 − x = 0 ⇔ 1+ x 2 = x ⎧1 + x 2 = x 2 ⎨⇔ ⎩x ≥ 0 ⎧1 = 0 ⎨⇔ ⎩x ≥ 0 وهﺬا ﻏﻴﺮ ﻣﻤﻜﻦ إذن اﻟﻤﻌﺎدﻟﺔ f '( x ) = 0ﻻ ﺗﻘﺒﻞ أي ﺣﻞ ﻓﻲ IR وهﺬا ﻳﻌﻨﻲ أن f '(x ) ≠ 0ﻟﻜﻞ xﻣﻦ . IR
= =
ﻃﺮﻳﻘﺔ ﺛﺎﻧﻴﺔ: ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﺣﻘﻴﻘﻴﺎ x 2 +1 x 2 ﺑﻤﺎ أن : x 2 +1
ﻓﺈن x 2 :
x 2 +1
ﻳﻌﻨﻲ أن :
x
وﺑﻤﺎ أن :
x ≥x
ﻓﺈن :
x
x 2 +1
ﻳﻌﻨﻲ أن x + 1 − x 0 : إذن f ( x ) 0 : وﻣﻨﻪ f (x ) ≠ 0 وﺑﺎﻟﺘﺎﻟﻲ f '(x ) ≠ 0 :ﻟﻜﻞ xﻣﻦ . IR • ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ f ) − 2f ( x = ) f '(xﻟﻜﻞ xﻣﻦ . IR ﺑﻤﺎ أن : 1+ x 2 ﻓﺈن إﺷﺎرة ) f '( xهﻲ إﺷﺎرة ) − f ( x وﺑﻤﺎ أن f ( x ) 0ﻟﻜﻞ xﻣﻦ . IR ﻓﺈن f '( x ) ≺ 0ﻟﻜﻞ xﻣﻦ . IR وﻣﻨﻪ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ . f 2
∞+
∞−
-
∞+
x ) f '( x ) f (x
0
) f (x *-3ﻟﻨﺒﻴﻦ أن = −∞ : ∞x →− x lim
f (x ) ( 1 + x 2 − x )2 = ﻟﻜﻞ xﻣﻦ * ، IR x x 1 + x 2 − 2x 1 + x 2 + x 2 1 + 2x 2 − 2x 1 + x 2 = x x 1 = + 2x − 2 1 + x 2 x 1 limو ∞lim 2 x − 2 1 + x 2 = − وﺑﻤﺎ أن = 0 : x→−∞ x ∞x→− )f ( x lim ﻓﺈن = −∞ : x→−∞ x • اﻟﺘﺄوﻳﻞ اﻟﻬﻨﺪﺳﻲ اﻟﻤﻨﺤﻨﻰ ) (ζﻳﻘﺒﻞ ﺑﺠﻮار ∞ −ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ اﺗﺠﺎهﻪ ﻣﺤﻮر اﻷراﺗﻴﺐ. =
-4أ -ﻣﻌﺎدﻟﺔ ﻟﻠﻤﺎس )(T ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻘﻴﻢ ) (Tﻣﻤﺎس )y = f '(0)( x − 0) + f (0 أي y = −2( x − 0) + 1 : أي : y = −2 x + 1 ب -رﺳﻢ ) (Tو
) (ζﻓﻲ اﻟﻨﻘﻄﺔ ذات اﻷﻓﺼﻮل 0هﻲ:
) (ζ
-5أ -ﻟﻨﺒﻴﻦ أن fﺗﻘﺎﺑﻞ ﻣﻦ IRﻧﺤﻮ ﻣﺠﺎل . J اﻟﺪاﻟﺔ fﻣﺘﺼﻠﺔ وﺗﻨﺎﻗﺼﻴﺔ ﻗﻄﻌﺎ ﻋﻠﻰ IR إذن ﻓﻬﻲ ﺗﻘﺎﺑﻞ ﻣﻦ IRﻧﺤﻮ اﻟﻤﺠﺎل [∞J = f (IR ) = ]0, + وﺑﺎﻟﺘﺎﻟﻲ ﻓﻬﻲ ﺗﻘﺒﻞ داﻟﺔ ﻋﻜﺴﻴﺔ ' fﻣﻌﺮﻓﺔ ﻋﻠﻰ ﻣﻦ [∞ ]0, +ﻧﺤﻮ IR
ب -ﺣﺴﺎب )) '(1
−1
(f
1 ﺑﻤﺎ أن : ))f '( f −1 (1 وﺑﻤﺎ أن ( f )(0) = 1 :ﻓﺈن f −1 (1) = 0 1 = )( f −1 )(1 ﻓﺈن : )f '(0 1 1 = =− 2 −2 ج -ﺣﺴﺎب ) f '( x
= )( f −1 ) '(1
ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﻣﻦ [∞ ]0, +و yﻋﻨﺼﺮا ﻣﻦ IR f −1 ( x) = y ⇔ f ( y ) = x ⇔ ( 1 + y 2 − y )2 = x
⇔ 1 + y 2 − y = − xأو ⇔ 1 + y 2 − y = x ﻟﺪﻳﻨﺎ 0 :
1+ y 2 − y
( 1+ y 2
) ﻷن y
أي 1 + y 2 − y = x : أي 1 + y 2 = x + y :
أي ( 1 + y 2 ) 2 = ( x + y ) 2 :
أي 1 + y 2 = 2 y x + y 2 :
1− x أي 2 y x = 1 − x :أي : 2 x 1− x = ) f −1 ( xﻟﻜﻞ xﻣﻦ IR إذن : 2 x
=y
ﺗﻤﺮﻳﻦ:14 -1أ -ﺣﺴﺎب ) lim f ( x
∞x →+
ﺑﻤﺎ أن lim = 3 x 2 + 1 = +∞ :
∞x →+
و ∞lim (x − 2) = +
∞x →+
ﻓﺈن lim f ( x) = +∞ :
∞x →+
ب -دراﺳﺔ اﻟﻔﺮع اﻟﻼﻧﻬﺎﺋﻲ
) f (x 2 3 x 2 +1 = lim 1 − + ∞x →+ x x x
ﻟﺪﻳﻨﺎ :
lim
∞x →+
2 3 x2 + 1 2 3 x2 + 1 = lim 1 − + = lim 1 − + 3 3 ∞x →+ ∞x →+ x x x3 x 2 1 1 = lim 1 − + 3 + 3 = 1 − 0 + 0 = 1 ∞x →+ x x x
و = lim f ( x) − x = lim − 2 + 3 x 2 + 1 ∞x →+
∞x →+
∞= + إذن ) (ζﻳﻘﺒﻞ ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ ﺑﺠﻮار ∞ +اﺗﺠﺎهﻪ اﻟﻤﺴﺘﻘﻴﻢ ذو اﻟﻤﻌﺎدﻟﺔ y =x -2أ -ﺣﺴﺎب ) f '( x +
اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ IR ﻟﺪﻳﻨﺎ ﻟﻜﻞ xﻣﻦ IR + '
1 ⎡ ⎤ ⎥ f '( x) = ⎢ x − 2 + ( x 2 + 1) 3 ⎣ ⎦ 1−1 −2 1 2x 2 2 3 )= 1 + 2 x( x + 1 = 1 + ( x + 1) 3 3 3
إذن :
3 3 ( x 2 + 1) 2 + 2 x 3 3 ( x 2 + 1) 2
=
ب -ﺟﺪول ﺗﻐﻴﺮات f ﺑﻤﺎ أن f '( x ) 0ﻟﻜﻞ xﻣﻦ IR ﻓﺈن ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ fﻳﻜﻮن آﺎﻟﺘﺎﻟﻲ : +
-1ﻟﻨﺒﻴﻦ أن fﺗﻘﺎﺑﻞ ﻣﻦ IR +ﻧﺤﻮ ﻣﺠﺎل .I ﺑﻤﺎ أن fداﻟﺔ ﻣﺘﺼﻠﺔ وﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ ، IR + ﻓﺈﻧﻬﺎ ﺗﻘﺎﺑﻞ ﻣﻦ IR +ﻧﺤﻮ اﻟﻤﺠﺎل I = f ( IR + ) = [ −1, +∞[ : إذن ﻓﻬﻲ ﺗﻘﺒﻞ داﻟﺔ ﻋﻜﺴﻴﺔ f −1ﻣﻌﺮﻓﺔ ﻣﻦ [∞ [ −1, +ﻧﺤﻮ IR +
-4أ -ﻟﻨﺒﻴﻦ أن اﻟﻤﻌﺎدﻟﺔ f(x) =0ﺗﻘﺒﻞ ﺣﻼ وﺣﻴﺪا. 3 5 ⎞⎛1 ﺑﻤﺎ أن f (1) = −1 + 3 2 0 :و f ⎜ ⎟ = − + 3 ≺ 0 2 4 ⎠⎝2 1 إذن f (1) . f ( ) ≺ 0 : 2 1 ⎡ ⎤ وﺑﻤﺎ أن fﻣﺘﺼﻠﺔ وﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ ⎦⎥⎢⎣ 2 ,1 1 ﻓﺈن اﻟﻤﻌﺎدﻟﺔ f(x)=0ﺗﻘﺒﻞ ﺑﺎﻟﻔﻌﻞ ﺣﻼ وﺣﻴﺪا αﺑﺤﻴﺚ ≺ α ≺ 1 2 ب -ﺗﺤﺪﻳﺪ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ) (ζو ) ( Δ ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﻣﻦ IR + ﻟﺪﻳﻨﺎ f ( x) = x ⇔ −2 + 3 x 2 + 1 = 0 :
⇔ 3 x2 + 1 = 2 ⇔ x2 = 7 ) ﻷن ⇔ x = 7 ( x ≥ 0
)
إذن ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ) (ζواﻟﻤﺴﺘﻘﻴﻢ ) ( Δاﻟﺬي ﻣﻌﺎدﻟﺘﻪ y =xهﻲ 7, 7 : ج -رﺳﻢ
(
A
) (ζو )' (ζ
ﻣﻠﺤﻮﻃﺔ :ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﺤﺎﻣﻞ ﻧﺼﻒ ﻣﻤﺎس ) (ζﻋﻨﺪ اﻟﻨﻘﻄﺔ ) B (0, −1هﻲ y = x − 1
ﺗﻤﺮﻳﻦ:15 -1ﻟﻨﺒﻴﻦ أن fﻣﺘﺼﻠﺔ ﻓﻲ .1 1+1 = )f (1 ﺑﻤﺎ أن : 2 1 2 = =1 2 1+ x و ﺑﻤﺎ أن : lim f ( x) = lim+ x →1+ x →1 2 x 1+1 2 = )= = 1 = f (1 2 1 2 ﻓﺈن اﻟﺪاﻟﺔ fﻣﺘﺼﻠﺔ ﻋﻠﻰ اﻟﻴﻤﻴﻦ ﻓﻲ . I ⎞2 ⎛ وﻟﺪﻳﻨﺎ lim f ( x) = lim− ⎜ − x + ⎟ : x →1− ⎝ x →1 ⎠x 2 )= −1 + = −1 + 2 = 1 = f (1 1 إذن اﻟﺪاﻟﺔ fﻣﺘﺼﻠﺔ ﻋﻠﻰ اﻟﻴﺴﺎر ﻓﻲ 1 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﺪاﻟﺔ fﻣﺘﺼﻠﺔ ﺑﺎﻟﻔﻌﻞ ﻓﻲ اﻟﻨﻘﻄﺔ x0 = 1ﻷﻧﻬﺎ ﻣﺘﺼﻠﺔ ﻋﻠﻰ اﻟﻴﻤﻴﻦ وﻋﻠﻰ اﻟﻴﺴﺎر ﻓﻲ هﺬﻩ اﻟﻨﻘﻄﺔ -2أ -ﻟﻨﺒﻴﻦ أن fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻴﺴﺎر ﻓﻲ 1 2 − x + −1 )f ( x) − f (1 x lim = lim− ﻟﺪﻳﻨﺎ : x →1− x → 1 x −1 x −1 − x2 + 2 − x )( x − 1)(− x − 2 = lim− = lim− x →1 x →1 )x( x − 1 )x( x − 1 − x − 2 −1 − 2 = lim− = = −3 x →1 1 x إذن اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﺑﺎﻟﻔﻌﻞ ﻋﻠﻰ اﻟﻴﺴﺎر ﻓﻲ 1 و f g' (1) = −3 : ب -ﻟﻨﺒﻴﻦ أن fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻴﻤﻴﻦ ﻓﻲ 1 1+ x −1 )f ( x) − f (1 x 2 lim = lim+ ﻟﺪﻳﻨﺎ: x →1+ x →1 x −1 x −1 1+ x − 2 x ( x − 1) 2 = lim+ = lim+ x →1 )2 x ( x − 1) x →1 2 x ( x − 1)( x + 1 x −1 1 −1 0 = = =0 2 x ( x + 1) 2(1 + 1) 4 إذن اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﺑﺎﻟﻔﻌﻞ ﻋﻠﻰ اﻟﻴﻤﻴﻦ ﻓﻲ 1 و f d ' (1) = 0 : -3أ -ﻟﻨﺒﻴﻦ أن f '( x ) ≺ 0
اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ ⎞2 ⎛ ⎟ f '( x) = ⎜ − x + ⎠x ⎝ ⎛ 2 ⎞ 2 ⎟ = −1 − 2 − ⎜ 1 + 2 ⎠ x ⎝ x
= lim+ x →1
ﻟﻜﻞ xﻣﻦ []−∞, 0[ ∪ ]0,1 [ ]−∞, 0[ ∪ ]0,1وﻟﺪﻳﻨﺎ xﻣﻦ []−∞, 0[ ∪ ]0,1
2 وﺑﻤﺎ أن 0 : x2 ﻓﺈن f '( x) ≺ 0 :ﻟﻜﻞ xﻣﻦ []−∞, 0[ ∪ ]0,1
1 +ﻟﻜﻞ xﻣﻦ
[]−∞, 0[ ∪ ]0,1
x −1 أ -ﻟﻨﺒﻴﻦ أن : 4x x fداﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ [∞ ]1, +وﻟﺪﻳﻨﺎ ﻟﻜﻞ xﻣﻦ [∞]1, + = )f '( x
ﻟﻜﻞ xﻣﻦ
[∞]1, +
'
⎞ ⎛ 1+ x ⎜ = )f '( x ⎟ ⎠⎝2 x 1 ⎤ x− ⎥ )(1 + x 1 2x −1 − x 2 x ⎥= . x ⎥ 2 2x x ⎥⎦
إذن 0
) f '( xﻟﻜﻞ xﻣﻦ
[∞]1, +
⎡ ⎢1 ⎢ = ⎢2 ⎢⎣ x −1 = 4x x
ج -ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ : f
ﻟﺪﻳﻨﺎ : ⎞2 ⎛ * ∞lim f ( x) = lim ⎜ − x + ⎟ = + ∞x →− ∞x →− ⎠x ⎝ ⎞2 ⎛ * ∞lim− f ( x) = lim− ⎜ − x + ⎟ = − x →0 ⎝ x →0 ⎠x ⎞2 ⎛ * ∞lim+ f ( x) = lim+ ⎜ − x + ⎟ = + x →0 ⎝ x →0 ⎠x 1+ x 1 x * ∞= + lim f ( x) = lim = lim + ∞x→+ x→+∞ 2 x x→+∞ 2 x 2 -4أ -دراﺳﺔ اﻟﻔﺮوع اﻟﻼﻧﻬﺎﺋﻴﺔ: *ﺑﻤﺎ أن lim+ f ( x ) = + ∞ :و ∞ lim− f ( x ) = − x →0
x →0
ﻓﺈن اﻟﻤﻨﺤﻨﻰ ) (ζﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ رأﺳﻴﺎ ﻣﻌﺎدﻟﺘﻪ x =0أي ﻣﺤﻮر اﻷراﺗﻴﺐ. 2 * ﻟﻜﻞ xﻣﻦ [، ]−∞, 0[ ∪ ]0,1 x
f ( x) = − x +
2 ﻟﺪﻳﻨﺎ = 0 : x→−∞ x إذن ) (ζﻣﻘﺎرﺑﺎ ﻣﻌﺎدﻟﺘﻪ y = -xﻳﻘﺒﻞ ﺑﺠﻮار ∞− lim
)f ( x 1+ x * ﻟﺪﻳﻨﺎ : = lim x ∞→+ x 2x x 1 1 + = 0+0 = 0 lim x →+∞ 2 x x 2 x إذن ) (ζﻳﻘﺒﻞ ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ اﺗﺠﺎهﻪ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ ﺑﺠﻮار ∞. + lim
∞x →+
ﻣﻠﺤﻮﻇﺔ :ﻣﻌﺎدﻟﺘﺎ ﺣﺎﻣﻠﻲ ﻧﺼﻔﻲ ﻣﻤﺎس ) (ζﻋﻨﺪ اﻟﻨﻘﻄﺔ ذات اﻷﻓﺼﻮل 1هﻤﺎ y = 1 :و y = −3 x + 4
ﺗﻤﺮﻳﻦ:16 *-1ﺣﺴﺎب ) f ( x
lim
∞x→+
x3 − 1 ﻟﺪﻳﻨﺎ : ∞x →+ x →+∞ x 3 + 1 x3 = lim 3 = 1 x→+∞ x * ﻟﻨﺒﻴﻦ أن lim f ( x) = −∞ : lim f ( x) = lim
∞x →−
ﻟﻜﻞ xﻣﻦ [f ( x) = x − 1 + 2 1 − x . ]−∞,1
) = 1 − x (2 − 1 − x
وﻟﺪﻳﻨﺎ 1 − x = +∞ : إذن f ( x ) = −∞ :
limو ∞lim 2 − 1 − x = −
∞x →−
∞x →−
lim
∞x→−
*-2دراﺳﺔ ﻗﺎﺑﻠﻴﺔ اﺷﺘﻘﺎق fﻋﻠﻰ اﻟﻴﻤﻴﻦ ﻓﻲ 1 x3 − 1 −0 3 )f ( x) − f (1 1 + x lim = lim+ x →1+ x →1 x −1 x −1 x3 − 1 ( x − 1)( x 2 + x + 1 = lim+ = lim ) 3 3 )x →1 ( x − 1)( x + 1 )x →1+ ( x − 1)( x + 1 x2 + x + 1 3 = x3 + 1 2
= lim+ x →1
3 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻴﻤﻴﻦ ﻓﻲ 1وﻟﺪﻳﻨﺎ : d 2 * دراﺳﺔ ﻗﺎﺑﻠﻴﺔ اﺷﺘﻘﺎق fﻋﻠﻰ اﻟﻴﺴﺎر ﻓﻲ 1 )f ( x) − f (1 x −1+ 2 1− x ﺑﻤﺎ أن : lim = lim− x →1− x → 1 x −1 x −1 x −1 2 1− x = lim− + x →1 x − 1 )−(1 − x = )f ' (1
2 1− x 1− x )2(1 − x = lim− 1 − x→1 x→1 (1 − x ) 1 − x (1 − x ) 1 − x 2 = lim− 1 − ∞= − x→1 1− x ﻓﺈن اﻟﺪاﻟﺔ fﻏﻴﺮ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻴﺴﺎر ﻓﻲ .1 * اﻟﺘﺄوﻳﻞ اﻟﻬﻨﺪﺳﻲ : ) (ζﻳﻘﺒﻞ ﻓﻲ اﻟﻨﻘﻄﺔ ) A(1, 0ﻧﺼﻔﻲ ﻣﻤﺎس أﺣﺪهﻤﺎ رأﺳﻲ ﻣﻮﺟﻪ ﻧﺤﻮ اﻷﻋﻠﻰ واﻵﺧﺮ ﻋﻠﻰ اﻟﻴﻤﻴﻦ ﻣﻌﺎدﻟﺔ ﺣﺎﻣﻠﻪ هﻲ = lim− 1 −
3 )( x − 1 2 -1-3ﻟﻨﺒﻴﻦ أن fﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ [∞[1, + =y
fداﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ
[∞[1, +
ﻟﻜﻞ xﻣﻦ [∞. ]1, + '
⎞ ⎛ x3 − 1 ⎟ f '( x) = ⎜ 3 ⎠⎝ x +1
)3x 2 ( x3 + 1) − 3x 2 ( x3 − 1) 3x 2 ( x3 + 1 − x3 + 1 = = ( x3 + 1) 2 ( x3 + 1)2 6 x2 = 3 ( x + 1) 2 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن f '( x) ; 0ﻟﻜﻞ xﻣﻦ [∞]1, + إذن fهﻲ ﺑﺎﻟﻔﻌﻞ داﻟﺔ ﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل
−x ب -ﻟﻨﺒﻴﻦ أن )1 − x (1 + 1 − x اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ وﻟﺪﻳﻨﺎ ﻟﻜﻞ xﻣﻦ
[∞[1, +
= ) f '( xﻟﻜﻞ xﻣﻦ
[]−∞,1
[]−∞,1
')f '( x) = ( x − 1 + 2 1 − x 1 −1 = 1 + 2. = 1− 2 1− x 1− x 1 − x − 1 ( 1 − x − 1)( 1 − x + 1 = = 1− x )1 − x ( 1 − x + 1 1− x −1 −x = = )1 − x ( 1 − x + 1 )1 − x ( 1 − x + 1
ج -ﺟﺪول ﺗﻐﻴﺮات f
-4أ -دراﺳﺔ اﻟﻔﺮﻋﻴﻦ اﻟﻼﻧﻬﺎﺋﻴﻴﻦ ﻟﺪﻳﻨﺎ lim f ( x ) = 1 : ∞x →+
إذن ) ( Cﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ أﻓﻘﻴﺎ ﻣﻌﺎدﻟﺘﻪ y = 1ﺑﺠﻮار ∞+ )f ( x x −1+ 2 1− x وﺑﻤﺎ أن : lim = lim ∞x →− ∞x →− x x 1 2 1−x = lim 1 − + ∞x →− x x )1 2(1 − x = lim 1 − + ∞x →− x x 1− x 1 2 1 ﻓﺈن : = lim 1 − + ( − 2). ∞x→− x x 1− x
[]−∞,1
وﺑﻤﺎ أن : lim f ( x) − x = lim x − 1 + 2 1 − x − x
∞x →−
∞x →−
∞= lim 2 1 − x − 1 = + ∞x→−
ﻓﺈن اﻟﻤﻨﺤﻨﻰ ) (Cﻳﻘﺒﻞ ﺑﺠﻮار ∞ −ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ اﺗﺠﺎهﻪ اﻟﻤﺴﺘﻘﻴﻢ ذو اﻟﻤﻌﺎدﻟﺔ y= x
-5أ -ﻟﻨﺒﻴﻦ أن gﺗﻘﺎﺑﻞ ﻣﻦ [∞ [1, +ﻧﺤﻮ ﻣﺠﺎل I ﺑﻤﺎ أن اﻟﺪاﻟﺔ gﻣﺘﺼﻠﺔ وﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل
[∞[1, +
ﻓﺈن ﻓﻬﻲ ﺗﻘﺎﺑﻞ ﻣﻦ [∞ [1, +ﻧﺤﻮ اﻟﻤﺠﺎل :
[J = f ([1, +∞[) = [ 0,1 إذن ﻓﻬﻲ ﺗﻘﺒﻞ داﻟﺔ ﻋﻜﺴﻴﺔ
−1
g
ﻣﻌﺮﻓﺔ ﻣﻦ [ [0,1ﻧﺤﻮ [∞[1, +
ب -ﺣﺴﺎب )g −1 ( x
ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﻣﻦ [ [ 0,1و yﻋﻨﺼﺮا ﻣﻦ ﻟﺪﻳﻨﺎ g −1 ( x) = y ⇔ g ( y ) = x :
y3 − 1 =x y3 + 1 )⇔ y 3 − 1 = x( y 3 + 1 ⇔ y 3 − 1 = xy 3 + x ⇔ y 3 − xy 3 = x + 1 ⇔ y 3 (1 − x) = x + 1 ⇔
[∞[1, +
x +1 1− x x +1 ⇔ y=3 1− x = ⇔ y3
1+ x إذن 1− x
3
= )g −1 ( x
ﻟﻜﻞ xﻣﻦ [[0,1
ج -ﻟﻨﺒﻴﻦ أن g −1داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ :
2 2
)( x − x − x + 1 2
3
3
→ x
ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﻣﻦ اﻟﻤﺠﺎل []0,1 ' ﻟﺪﻳﻨﺎ : '
⎞ ⎛ x +1 ⎡⎣ g ( x) ⎤⎦ = ⎜ 3 ⎟⎟ ⎜ 1 − x ⎝ ⎠ '
−1
1 ⎡ ⎤ 3 + x 1 ⎛ ⎞ ⎥ ⎜⎢ = ⎥ ⎠⎟ ⎢⎝ 1 − x ⎣ ⎦
1 1 2 − ⎞ 1 ⎛ x +1 ⎞ ⎛ x +1 1 −1 1 ⎛ x + 1 ⎞ 3 ⎜. ⎜ = ⎜⎟ = ⎟ ⎟ ⎠ 3 ⎝ −x + 1 ⎠ ⎝ −x + 1 ⎠ 3 (1 − x ) 2 ⎝ 1 − x 2 1 1 2 1 . = . = . 2 2 2 − )3 (1 − x 3 2 3 ⎛ x +1 ⎞3 ⎞ 3 3 ⎛ x +1 ⎜ ⎦⎤ )⎡⎣(1 − x ⎜ ⎟ ⎟ ⎠ ⎝ 1− x ⎠ ⎝ 1− x 2 1 2 1 = . = . 2 2 3 3 ⎛ (1 − x)3 ( x + 1) ⎞ 3 ⎡⎣(1 − x)2 ( x + 1) ⎦⎤ 3 ⎜ ⎟ 1− x ⎝ ⎠ 2 1 2 1 = . = . 2 2 3 3 ⎡⎣(1 − 2 x + x 2 )( x + 1) ⎦⎤ 3 ( x + 1 − 2 x 2 − 2 x + x3 + x 2 ) 3 2 1 2 = . =. 2 3 3 2 3 3 ( x3 − x 2 − x + 1) 2 ( x − x − x + 1) 3 2 إذن g −1هﻲ ﺑﺎﻟﻔﻌﻞ داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ x → 3 3 2ﻋﻠﻰ اﻟﻤﺠﺎل []0,1 2 )3 ( x − x − x + 1 1 −1 3
'
ﺗﻤﺮﻳﻦ:17 -1أ -ﺣﺴﺎب
) lim f (xو ∞x →+
ﺑﻤﺎ أن x ( x − 1) :
) lim f ( x
x → 0+
lim x − x = lim
∞x →+
∞x →+
∞= + 1 =0 و lim ∞x →+ x ﻓﺈن lim f ( x ) = +∞ : ∞x →+
1 وﺑﻤﺎ أن = +∞ : x ﻓﺈن lim+ f (x ) = +∞ :
lim+و lim x − x = 0
x → 0+
x →0
x →0
ب -ﺗﺤﺪﻳﺪ اﻟﻔﺮﻋﻴﻦ اﻟﻼﻧﻬﺎﺋﻴﻴﻦ * ﺑﻤﺎ أن lim+ f (x ) = +∞ :
x →0
ﻓﺈن ) (Cﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ رأﺳﻴﺎ ﻣﻌﺎدﻟﺘﻪ x = 0أي ﻣﺤﻮر اﻷراﺗﻴﺐ. ) f (x x 1 lim = lim 1 − + • ﺑﻤﺎ أن : ∞x →+ ∞x →+ x x x x 1 1 = lim 1 − + = 1− 0 + 0 = 1 ∞x →+ x x x 1 و− x : lim f (x ) − x = lim ∞x →+ ∞x →+ x ﻓﺈن ) (Cﻳﻘﺒﻞ ﺑﺠﻮار ∞ +ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ اﺗﺠﺎهﻪ اﻟﻤﺴﺘﻘﻴﻢ ذو اﻟﻤﻌﺎدﻟﺔ y = x -2أ -ﺣﺴﺎب ) f '(x *+ *+ اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ IRوﻟﺪﻳﻨﺎ ﻟﻜﻞ xﻣﻦ : IR 1 − 1 + 2 x f '(x ) = 1 − x 2 x 1 1 2x x − x − 1 − = 2 x 2x x 2x x وﻟﺪﻳﻨﺎ (2x + x + 1)( x − 1) = 2x x − 2x + x − x + x − 1 : = 1−
= 2x x − x − 1 ⎞ ⎛ 2x + x + 1 ⎜⎜ = ) f '(xﻟﻜﻞ xﻣﻦ [∞]0, + إذن ⎟⎟ ( x − 1) : x x 2 ⎝ ⎠ ب -ﺟﺪول ﺗﻐﻴﺮات : f إﺷﺎرة ) f '( xهﻲ إﺷﺎرة x − 1 وﻣﻨﻪ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ :f
-3أ -دراﺳﺔ اﻟﻮﺿﻊ اﻟﻨﺴﺒﻲ ﻟﻠﻤﻨﺤﻨﻰ ) (Cو
)(Δ
ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﻣﻦ *IR + 1 = f (x ) − x ﻟﺪﻳﻨﺎ − x : x 1− x = x إﺷﺎرة f ( x ) − xهﻲ إﺷﺎرة 1 − x وﻣﻨﻪ ﻧﺴﺘﻨﺘﺞ اﻟﺠﺪول اﻟﺘﺎﻟﻲ اﻟﺬي ﻳﻌﻄﻲ إﺷﺎرة f ( x ) − xواﻟﻮﺿﻊ اﻟﻨﺴﺒﻲ ﻟﻠﻤﻨﺤﻨﻰ ) (Cواﻟﻤﺴﺘﻘﻴﻢ
)(Δ
-4أ -ﻟﻨﺒﻴﻦ أن gﺗﻘﺒﻞ داﻟﺔ ﻋﻜﺴﻴﺔ g −1
ﺑﻤﺎ أن اﻟﺪاﻟﺔ gﻣﺘﺼﻠﺔ وﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل [∞[1, + ﻓﺈﻧﻬﺎ ﺗﻘﺎﺑﻞ ﻣﻦ [∞ [1, +ﻧﺤﻮ اﻟﻤﺠﺎل [∞[1, + [∞f ([1, +∞[) = [1, + إذن ﻓﻬﻲ ﺗﻘﺒﻞ داﻟﺔ ﻋﻜﺴﻴﺔ g −1ﻣﻌﺮﻓﺔ ﻣﻦ [∞ [1, +ﻧﺤﻮ [∞[1, + وﻣﻨﻪ ﻓﺈن D g −1 = [1, +∞[ :
ب -رﺳﻢ ) (C اﻟﻤﻨﺤﻨﻰ ) (Cهﻮ ﻣﻤﺎﺛﻞ ﻣﻨﺤﻨﻰ fﻋﻠﻰ [∞ [1, +ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺴﺘﻘﻴﻢ ) ( Δذي اﻟﻤﻌﺎدﻟﺔ y = x : g −1
g −1
-5أ -ﻟﻨﺒﻴﻦ ﺑﺎﻟﺘﺮﺟﻊ أن an 1ﻟﻜﻞ nﻣﻦ IN * ﻟﺪﻳﻨﺎ ، a0 = 2 :إذن a0 1 وهﺬا ﻳﻌﻨﻲ أن اﻟﺨﺎﺻﻴﺔ ﺻﺤﻴﺤﺔ ﻣﻦ أﺟﻞ n =0 • ﻟﻴﻜﻦ nﻋﻨﺼﺮا ﻣﻦ IN ﻟﻨﻔﺘﺮض أن an 1وﻟﻨﺒﻴﻦ أن an +1 1 ﺑﻤﺎ أن an 1 : و fﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل
[∞[1, +
ﻓﺈن f (an ) f (1) : أي أن an +1 1 إذن1 :
( ∀n ∈ IN ) an
ب -ﻟﻨﺒﻴﻦ أن ) (anﺗﻨﺎﻗﺼﻴﺔ ﻟﺬﻟﻚ ﺳﻨﺒﻴﻦ ﺑﺎﻟﺘﺮﺟﻊ أن an +1 ≺ an :ﻟﻜﻞ nﻣﻦ IN •
ﺑﻤﺎ أن a0 = 2 :
2 2 −1 و 2
= )a1 = f (a0 ) = f (2
)أﻧﻈﺮ اﻟﺸﻜﻞ(
ﻓﺈن a1 ≺ a0 : إذن اﻟﺨﺎﺻﻴﺔ ﺻﺤﻴﺤﺔ ﻣﻦ أﺟﻞ n = 0 • ﻟﻴﻜﻦ nﻋﻨﺼﺮا ﻣﻦ IN ﻟﻨﻔﺘﺮض أن an +1 ≺ an :وﻟﻨﺒﻴﻦ أن : ﻟﺪﻳﻨﺎ an +1 ≺ an : و fﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل
an + 2 ≺ an +1
[∞[1, +
إذن f (an +1 ) ≺ f (an ) : إذن an + 2 ≺ an +1 : ﻃﺮﻳﻘﺔ ﺛﺎﻧﻴﺔ:
ﻟﻴﻜﻦ nﻋﻨﺼﺮا ﻣﻦ IN ﺑﻤﺎ أن : 1 − −an 1 = an − an + = − an an an
an +1 − an = f (an ) − an
وﺑﻤﺎ أن 0 :
an
و ) 1 − an ≺ 0ﻷن 1
( an
1 − an ﻓﺈن ≺ 0 : an وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﻤﺘﺘﺎﻟﻴﺔ ) (anﺑﺎﻟﻔﻌﻞ ﺗﻨﺎﻗﺼﻴﺔ ج * -اﺳﺘﻨﺘﺎج أن ) (anﻣﺘﻘﺎرﺑﺔ ﺑﻤﺎ أن اﻟﻤﺘﺘﺎﻟﻴﺔ ) (anﺗﻨﺎﻗﺼﻴﺔ وﻣﺼﻐﻮرة ﺑﺎﻟﻌﺪد ) 1ﻷن 1 ﻓﺈﻧﻬﺎ ﻣﺘﻘﺎرﺑﺔ • ﺗﺤﺪﻳﺪ ﻧﻬﺎﻳﺔ اﻟﻤﺘﺘﺎﻟﻴﺔ ) (an ﺑﻤﺎ أن اﻟﺪاﻟﺔ fﻣﺘﺼﻠﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ﻓﺈن اﻟﻨﻬﺎﻳﺔ أي :
=
إذن= 1 :
ﺗﺤﻘﻖ 1
+
[∞[1, +
=) ( f
−أي = 0 :
1−
anﻟﻜﻞ nﻣﻦ ( IN
ﺗﻤﺮﻳﻦ:18
lim f (x ) = lim x − 2 1 − x
-1ﻟﺪﻳﻨﺎ :
∞x →+
⎞ ⎛ 2 x −1 ⎜⎜ 1 − ⎟ ⎠⎟ x 2 ⎝
و
∞x →+
⎞ ⎛ 2 1− x x = lim x ⎜⎜1 − ⎟⎟ = xlim ∞x →+ ∞→+ x ⎝ ⎠ ⎛ ⎞ 1 1 = lim x ⎜⎜1 − 2 − ∞⎟ = + ∞x →+ ⎠⎟ x x 2 ⎝ lim f (x ) = lim x + 2 1 − x ∞x →+−
∞x →−
⎞ ⎛ 2 1− x ⎞ ⎛ 2 1− x = lim x ⎜⎜1 − = lim x 1− ⎟ ⎜ ⎟ ∞x →− ⎝⎜ ∞− x ⎟⎠ x →− ⎠⎟ x 2 ⎝ ⎛ ⎞1 1 ∞= lim x ⎜⎜ 1 − 2 2 − ⎟⎟ = − إذن : ∞x→− x ⎠x ⎝ -2أ -اﻟﺪاﻟﺔ fﻣﺘﺼﻠﺔ ﻓﻲ 1ﻷن : ) lim1+ f ( x ) = f (1و )lim− f ( x ) = f (1 x →1
x →1
ب -اﻟﺪاﻟﺔ fﻏﻴﺮ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻴﻤﻴﻦ وﻋﻠﻰ اﻟﻴﺴﺎر ﻓﻲ .1 )f (x ) − f (1 )f ( x ) − f (1 lim+و ∞= − lim− ﻷن = −∞ : x →1 x →1 x −1 x −1 وهﺬا ﻳﻌﻨﻲ هﻨﺪﺳﻴﺎ أن اﻟﻤﻨﺤﻨﻰ ) (Cﻳﻘﺒﻞ ﻓﻲ اﻟﻨﻘﻄﺔ ) A (1,1ﻣﻤﺎﺳﺎ رأﺳﻴﺎ -3أ- ⎧ x −1 −1 ;x 1 = ) ⎪f '(x x −1 ⎪ ⎨ ⎪f '(x ) = 1 − x − 1 ; x ≺ 1 ⎩⎪ 1− x ب-
∞+ +
2 0
∞+
0 1 0 -2 1
∞−
+
x ) f '( x ) f (x
0 ∞−
-4أ (C ) -ﻳﻘﺒﻞ ﺑﺠﻮار ∞ +و ∞ −ﻓﺮﻋﻴﻦ ﺷﻠﺠﻤﻴﻴﻦ اﺗﺠﺎهﻬﻤﺎ اﻟﻤﺴﺘﻘﻴﻢ ذو اﻟﻤﻌﺎدﻟﺔ y =x ب -ﻳﻘﻄﻊ اﻟﻤﻨﺤﻨﻰ ) (Cﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ ﻓﻲ اﻟﻨﻘﻄﺘﻴﻦ ) I (2, 0و )J (−2 − 2 2,0
-5أ -ﻟﻴﻜﻦ xﻋﻨﺼﺮا ﻣﻦ اﻟﻤﺠﺎل
[∞[ 2, +
f ( x) = x − 2 x − 1
ﻟﺪﻳﻨﺎ :
1
1
= x − 2( x − 1) 2 = x − 2( x − 1)' ( x − 1) 2 1 +1 2
)( x − 1 x − 2. ﻳﻌﻨﻲ أن + c : 1 2 +1 2 x2 4 إذن = − . ( x − 1)3 + c 2 3 ﻣﻊ cﺛﺎﺑﺘﺔ ﺣﻘﻴﻘﻴﺔ 4 4 2 2 وﺑﻤﺎ أن g (2) = .ﻓﺈن : = − +c 2 3 3 3 إذن c=0 : x2 4 2, ∞+ = وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن − ( x − 1) x − 1 : ﻣﻦ x ﻟﻜﻞ g x ( ) [ [ 2 3 2
ب -ﻟﺪﻳﻨﺎ : ⎞ ⎟ ⎟ ⎠
⎞ ⎛ 1 4 ( x − 1)3 ⎟ = lim x 2 ⎜ − ⎟ x →+∞ ⎜ 2 3 x4 ⎠ ⎝
= )g ( x
x2 4 − ( x − 1)3 x→+∞ 2 3 ⎛ 1 4 ( x − 1)3 = lim x 2 ⎜ − ∞x →+ ⎜2 3x 2 ⎝
lim g ( x) = lim
∞x→+
⎞ ⎛ 1 4 ( x − 1)3 إذن : = lim x 2 ⎜ − ∞⎟ = + 4 ∞x →+ ⎜2 3 ⎟ x ⎝ ⎠ *ﻟﺪﻳﻨﺎ اﻟﺪاﻟﺔ gﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ [∞ ]2, +وﻟﺪﻳﻨﺎ ﻟﻜﻞ xﻣﻦ [∞ ) g '( x ) = f ( x) , ]2, +ﻷن gداﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ (f وﻣﻨﻪ ﻓﺈن إﺷﺎرة ) g '( xهﻲ إﺷﺎرة ) f ( xﻋﻠﻰ [∞]2, + وﺣﺴﺐ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ fأو اﻟﺘﻤﺜﻴﻞ اﻟﻤﺒﻴﺎﻧﻲ ﻟﻠﺪاﻟﺔ f ﻧﻼﺣﻆ أن f ( x) 0ﻟﻜﻞ xﻣﻦ [∞]2, +
وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﺪاﻟﺔ gﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ [∞[ 2, + وﺑﺎﻟﺘﺎﻟﻲ ﻧﺠﺪ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ : g