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M129: Applied Calculus
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Tutor Marked Assignment

Cut-Off Date:Dec 7th, 2013 Total Marks: 40

Contents
Feedback form …………………..…………………….…...….. 2
Question 1 …………………………………………………..……… 3
Question 2 …………………..………………..……………………… 3
Question 3 ……………………..………………..…………………… 4
Question 4 ………………..………………………..……………… 4
Question 5 …………………………………………………..……… 5
Question 6 …………………………………..……………………… 5
Question 7 …………………………………..…………………… 6
Question 8 ……………………………………..…………………… 6 Plagiarism Warning: As per AOU rules and regulations, all students are required to submit their own TMA work and avoid plagiarism. The AOU has implemented sophisticated techniques for plagiarism detection. You must provide all references in case you use and quote another person's work in your TMA. You will be penalized for any act of plagiarism as per the AOU's rules and regulations.

Declaration of No Plagiarism by Student (to be signed and submitted by student with TMA work):
I hereby declare that this submitted TMA work is a result of my own efforts and I have not plagiarized any other person's work. I have provided all references of information that I have used and quoted in my TMA work.

Name of Student:
Signature:
Date:

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M129 TMA Feedback Form

[A] Student Component

Student Name: ali bilal al-housseini

Student Number: 120081

Group Number:

[B] Tutor Component

Tutor Name:

QUESTION | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | MARK | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | SCORE | | | | | | | | | TOTAL | |

Tutor’s Comments:

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This TMA covers only chapters 0, 1, 2, 3, 4 and 5. The TMA consists of eight questions for a total of 40 marks. Please solve each question in the space provided. You should give the details of your solutions and not just the final results.
Q−1: [4+1 Marks] let fx= x2+x-6, gx=x3-2x2-3x, hx= -2x4-x3+ 3x2

a) Find the zeros of g(x) and h(x). b) Find the points of intersection, if any, of the graphs of y = f (x) and y = -4 .
a)- zeros of g(x) x3-2x2-3x=0 x(x2-2x-3)=0 xx-3(x+1)=0 x=0 x-3=0 x+1=0

x=3 x=-1

zeros of h(x)
-2x4-x3+3x2=0

2x4+x3-3x2=0

x2(2x2+x-3)=0

x2(2x+3)(x-1)=0

x2=0 2x+3=0 x-1=0

X=0 2x=-3 x=1

x=-32
b)- F(x) = x2+x-6

x2+x-6=-4 x2+x-2=0 (x+2)(x-1)=0

X+2=0 x-1=0

X=-2 x=1

Points of intersection :-

(-2,-4) . (1,-4)

Q−2: [2+2+1 Marks] Letfx= x , gx= 2-x. a) Find f(g(x)) and describe its domain. b) Find g (f (x)),and describe its domain c) Describe the domains of f(x)and g(x).

a)- f(g(x)) =

2-x ≥ 0

-x ≥ -2

X ≤ 2

Domain = (-∞,2]

b)- g(f(x)) =

2- ≥ 0

- ≥ -2

≤ 2

0 ≤ x ≤ 4

Domain = [0,4]

c)- domain of f(x)

x ≥ 0

domain = [0,∞]

domain of g(x)

2-x ≥ 0

-x ≥ -2

X ≤ 2

Domain = (-∞,2]

Q−3: [2+3 Marks] Find the slope of the tangent line to the following curves at the designated points: a) fx= x3-3x2+x-1 , (3,2) b) y2+x2=25, -3,4

a)- fx= x3-3x2+x-1 (3,2)

f'x= 3x2 -6+1

Slope = f' (3) = 3(3)2 – 6(3)+1=10 b)-

y2+x2=25 -3,4

2yy'+2x=0

2yy'=-2x

y'= =

y'=

slope =

Q−4: [2+3 Marks] Use the definition of the derivative to find f ′(x) if a) fx=4x2-3x+5. b) f(x) = a)- fx=4x2-3x+5 f'(x) =

f(x+h) = 4(x+h)2-3x+h+5

f(x+h) = 4(x2+2xh+h2)-3x+h+5

f(x+h) = 4x2+8xh+4h2-3x-3h+5

f'(x) = 4x2+ 8xh+4h2- 3x-3h+5-4x2+ 3x-5h

f'(x) = h(8x+4h-3)h

f'(x) = 8x-3

b)-

f(x) =

f(x + h) =

f'(x) =

f'(x) = = f'(x) = -1(x+ 2)2

Q−5: [5×1 Marks] Letfx= x4-4x3 a) Find f ′(x) and f ′′(x). b) Find the intervals on which f (x) is increasing or decreasing. c) Find the local maximum and minimum of f (x), if any. d) Find the intervals on which the graph of f (x) is concave up or concave down. e) Find the points of inflection, if any.

a)-

f'(x) = 4x3- 12x2

f ''(x) = 12x2 - 24x

b)- f'(x) = 0

4 x3 - 12 x2=0

x3 - 3 x2=0

x2 (x – 3 ) = 0

x2 = 0 x – 3 = 0

x = 0 x = 3

0 3 interval | ( - ∞, 0 ) | ( 0,3 ) | ( 3, ∞ ) | x | -1 | 1 | 4 | f '(x) | --- | --- | +++ | | ↓ | ↓ | ↑ |

f'(-1) = 4 ( -1 )3 - 12 ( -1 )2 = - 16

f'(1) = 4 (1)3 - 12 ( 1 )2 = - 8

f' (4) = 4 (4)3 - 12 ( 4)2 = 64

increasing & decreasing interval

( -∞,0 ) decreasing

( 0,3 ) decreasing

( 3,∞) increasing
c)-

f''(x) = 12 x2 - 24 x

f''(0) = 12 (0)2 - 24 (0) = 0

f ''(3) = 12 (3)2 – 24 (3) = 36

f (3) = 34 - 4 ( 3 )3 = - 27 ( 3 , -27 ) local min

d)-

f '' (x) = 0

12 x2 – 24 x = 0

x2 - 2 x = 0

x ( x - 2 ) = 0

x=0 x - 2=0 x = 2

0 2 Interval | ( -∞,0 ) | ( 0, 2) | ( 2, ∞) | x | -1 | 1 | 3 | f ''(x) | +++ | --- | +++ | | n | u | n |

f '' (-1) = 12 ( -1)2 – 24 ( -1) = 36

f '' (1) = 12 (1)2 – 24 (1) = - 12 f ''(3) = 12 ( 3)2 - 24 (3) = 36

concavity :

(-∞,0) concave up

( 0,2) concave down

( 2,∞) concave up

e)-

points of inflaction

f(0) = 0

f (2) = 24 - 4 (2)3 = 16 – 32 = - 16

( 0,0 ) , ( 2,-16 )

Q−6:[5 Marks]A window is being built and the bottom is a rectangle and the top is a semicircle. If there is 12 m of framing materials what must the dimensions of the window be to let in the most light?

Perimeter = 12

P = 2y + x + r

r =

p = 2y + x + *

p = 2y + 2.37 x

2y + 2.37 x = 12

y + 1.285 x = 6

y = 6 – 1.285 x

A = xy + r22

A = xy + ( x2) 22

A = xy + 0.393x2

A = x (6 – 1.285 x) + 0.393x2

A = 6 x – 1.285x2 + 0.393 x2 A = 6 x – 0.892x2 A' = 6 – 1.784 x A' = 0 6 – 1.784x = 0 1.784x = 6 x = = 3.363 y = 6 – 1.285 – 3.363 y = 1.679

Q−7:[2+3 Marks] a) Differentiate the function ft= etet+ e-t b) Use the logarithmic differentiation to differentiate the function

x+14(x-2)2 (x+4)2x-1(2x+7)

a)-

f(t)= etet+e-t

f'(t) = et+ e-t et- et(et- e-t)(et+ e-t)2

f'(t) = e2t+ e0- e2t+ e0(et+ e-t)2 = 1+1(et+ e-t)2 = 2(et+ e-t)2

b)-

y = ( x+1 )4 (x- 2 )2(x+ 4 )2x-1 (2x+7)

ln y = ln ( x+1 )4 (x- 2 )2(x+ 4 )2x-1 (2x+7)

ln y = ln ( x + 1 )4 + l ( x – 2 )2 + ln (x+4)2 – ln(x-1) – ln (2x +7) ln y = 4 ln ( x+1) + 2 ln ( x-2) + 2 ln ( x+4) – ln (x-1) – ln ( 2x+7)

y' = [][( x+1 )4 (x- 2 )2(x+ 4 )2x-1 (2x+7)]

Q−8:[5×1 Marks]LetP(t) be the population (in millions) of a certain city t years after 1950, and suppose that P(t) satisfies the differential equation P′(t) = 0.017 P(t), P(0)= 2560 a) Find the formula for P(t). b) What was the initial population (the population in 1950) ? c) What is the growth constant? d) What was the population in 1960? e) Estimate the population in the year 2020.

a)- p(t)= p.(t) ekt

p(t)= 2560 e0,017t

b)- p.= 2560

c)- k= 0,017

d)- p(10)= 2560 e0,017*10 = 3034

e)- p(70)= 2560 e0,017*70 = 8414…...

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