Final Exam 2013/2014 Q1: Find the following limits 2 1. lim x 5x 4 x 2 x 1
solution: (2) 2 5(2) 4 2 1 2 3
2 2. lim x 8x 15 x 3 x 3
is 0 0 2 (x 3)(x 5) lim x 8x 15 lim x 3 x 3 x 3 (x 3) lim(x 5) 2 solution:
x 3
3. lim
x 0
3x 5x 1 1
solution : is 0 0 multiply by conjugate lim
x 0
lim
x 0
1
3x 5x 1 1 3x 5x 1 1 lim 5x 1 1 5x 1 1 5x 1 1 x 0
3
3x
5x 1 1 5x
6
0 1 1 5
5
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4. lim 3x 22 x 1 ( x 1) solution: 3(1) 2 5 * lim 3x 22 x 1 ( x 1) (1 1) 2 0 3(1) 2 5 * lim 3x 22 x 1 ( x 1) (1 1) 2 0 lim 3x 22 x 1 ( x 1)
5. lim
x 0
tan(10x ) 2x
solution:
is 0 0
tan(10x ) 10 5 x 0 2x 2
lim
Q 2: Find the horizontal asymptotes ( if any ) for the function 2 f (x ) 3x 2x2 1
2 5x
solution: 2 * lim 3x 2x2 1 3 x 5 2 5x
(n m )
2 * lim 3x 2x2 1 3 x 5 2 5x
y 3 is a horizontal asymptote 5
2
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Q 3: Discuss the continuity of the function 2x 2 1 f (x ) 7 x 3 1
, if x 2 if x 2
at x 2
if x 2
solution: 1. f (2) 7 defined 2.
* lim (2x 2 1) 7 x 2
* lim ( x 3 1) 7 x 2
lim f ( x ) 7 x 2
3. f (2) lim f ( x ) 7 x 2
f ( x ) is continuous at x 2
dy Q 4: Find the first derivative for the following functions dx 1. y x 3 24 3 x x solution: y x 3 2x 4 3x
1 2
خطوة تجهيز لالشتقاق
3 dy 3x 2 8x 5 3 21 x 2 dx
2. f (x ) 2x 3 6x 5 solution: *d dx f '(x )
3
g (x )
g '(x ) 2 g (x )
6x 2 6 2 2x 3 6x 5
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3. y cos 2 (2x 1) solution y cos(2x 1)
2
خطوة تجهيز لنسهل التكامل
y ' 2 cos(2x 1) sin(2x 1) 2 y ' 4 cos(2x 1) sin(2x 1) 4. y ln(x 3 2x 1) solution: d ln(f (x) f '(x ) f (x ) dx 2 y ' 33x 2 x 2x 1
5.
y x 3x solution: ln y ln x 3x ln y 3x ln x
differentiate
y' (3) ln x (3x ) 1 y x y ' y 3ln x 3 y ' x 3x 3ln x 3
6. y x e 2 x solution: y ' (1) e 2 x (x ) e 2 x 2 y ' e 2 x 2xe 2 x
4
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Q 5: Use Implict Differentiation to find
dy for the question. dx
x 2 y y 3 5x 2 8 solution: 2xy x 2 y ' 3 y 2 y ' 10x x 2 y ' 3 y 2 y ' 10 x 2 xy y '(x 2 3 y 2 ) 10x 2xy y '
10x 2xy x 2 3y 2
Q 6 : Find the slope of the line tangent to the graph of f (x )
x at x 4 x 2
solution: f (4)
4 2 42
point (4, 2)
(1)(x 2) (1)(x ) 2 2 (x 2) 2 (x 2) slope f '(4) 2 2 1 2 (4 2) f '(x )
y y 0 m (x x 0 ) y 2 1 (x 4) 2 y 1 x 4 2
5
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Q 7 : Verify that the function f (x ) x 2 6x 8 satisfies the three hypotheses of Roll's Theorem on the interval 2, 4 . Then find all possible values of c that satisfy the conclusion of the theorem solution: f ( x ) is polynomial 1. f (x ) is continuous on 2, 4 2. f ( x ) is differentiable on 2, 4 3. f (2) (2) 2 6(2) 8 0 , f (4) (4) 2 6(4) 8 0 Roll's Theorem is applicable
f (2) f ( 4)
** To find c f '( x ) 2 x 6 f '(c ) 0 2c 6 0 2c 6
c 3 2, 4
,
Q 8 : Evaluate lim 1 tan x cos x x 0 3x sin x
solution:
using L'Hopital's rule
is 0 0
Using L'hopital rule 2 lim sec x sin x x 0 3 cos x sec 2 0 sin 0 3 cos 0 1 0 1 3 1 4
6
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Q 9 : Given that f (x ) x 3 3x 2 9x , Find 1. The critical numbers of f 2. The intervals on which f is increasing / decreasing 3. The relative extrema of f 4. The intervals on which f is concave up / concave down 5. The x coordinates of inflection points of f solution: f is continuous and differentiable on , 1. The critical numbers f '( x ) 3x 2 6x 9 2 f '( x ) 0 3x 2 6 x 9 0 x 2 2x 3 0 the critical number x 1 , x 3
** Looking for sign of f '(x ) 2.
f is increasing on ( , 3) (1, ) f is decreasing on ( 3,1)
3 . f has relative maximum at x 3 , f (3) 27 f has relative minimum at x 1 , f (1) 5 ** f ''( x ) 6 x 6 f ''( x ) 0 6 x 6 0 , 6 x 6
, x 1
** Looking for sign of f ''( x ) 4 . f is concave up on ( 1, ) f is concave down on (, 1) 5. f has inflection point at 1, f (1) 1,11
7
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Q 10 : Find the absolute extrema of the function f (x ) x 3 3x 6 defined on 2, 2 solution: f is continuous and differentiable on 2, 2 * Critical number f '( x ) 3x 2 3 f '( x ) 0 3x 2 3 0
, 3x 2 3 , x 2 1
x 1 2, 2
, x 1 2, 2
* Evaluate f ( 2) 4 f ( 1) 8 f (1) 4 f (2) 8 f has absolute maximum at x 1 , x 2 and equal to 8 f has absolute minimum at x 2 , x 1 and equal to 4
Exercises حلول نموذجية ألسئلة150 مذكرة محمد ندا ريض بمركز ( كوبي كوم ) مقابل السنة التحضيرية بجوار برغرايزر
8
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