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Final

01:640:135 Final: 01:640:135 Final Exam 2006 SolutionsExam


Department
Mathematics
Course Code
01:640:135
Professor
All
Study Guide
Final

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1A
(16) 1. Calculate the following limits. Give a brief justication of your answers without
reference to calculator computations or graphing.
(a) lim
x!0
4x
sin3x
lim
x!0
4x
sin3x
= lim
x!0
4
3
sin 3x
3x
=
4
3
1
=
4
3
(b) lim
x!1
3x
4
+x
3
2x
2
+10
7x
5
4x
3
+2x+1
lim
x!1
3x
4
+x
3
2x
2
+10
7x
5
4x
3
+2x+1
= lim
x!1
3
x
+
1
x
2
2
x
3
+
10
x
5
7
4
x
2
+
2
x
4
+
1
x
5
=
0
7
=0
(c) lim
x!0
cos(2x)1
x
2
Using l'H^opital's Rule twice,
lim
x!0
cos(2x)1
x
2
= lim
x!0
2sin(2x)
2x
= lim
x!0
4cos(2x)
2
=
4
2
=2
(d) lim
x!3
j2x+6j
x+3
If xis slightly less than 3, then 2x+3<0, so j2x+6j=2x6. Thus
lim
x!3
j2x+6j
x+3
= lim
x!3
2(x+3)
x+3
= lim
x!3
2=2
(10) 2. Compute the derivative of
2
x1
directly from the denition.
2
x1
0
= lim
h!0
2
x+h1
2
x1
h
= lim
h!0
2(x1) 2(x+h1)
h(x+h1)(x1)
=
lim
h!0
2h
h(x+h1)(x1)
= lim
h!0
2
(x+h1)(x1)
=
2
(x1)
2
bvnbvnbv

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2A
(16) 3. Compute the derivatives with respect to xof the following functions. Algebraic
simplication of the answers need not be performed.
(a) e
x
ln(2x)
e
x
1
2x
2+e
x
ln(2x)
(b)
sinx
x
3
+2x
(x
3
+2x)cos(x)sin(x)(3x
2
+2)
(x
3
+2x)
2
(c)
Z
0
x
sectdt
Z
0
x
sectdt
0
=
Z
x
0
sectdt
0
=secx
(d)
Z
x
2
+x
0
sin(2t)dt
sin(2(x
2
+x))(2x+1)
(10) 4. Suppose that fis a function with rst and second derivatives. Suppose in addition
that the following values are known: f(1) = 0, f
0
(1) = 3, and f
00
(1) = 5. If g(x)=e
f(x)
,
what are g
0
(1) and g
00
(1)?
g
0
(x)=e
f(x)
f
0
(x)
so g
0
(1) = e
0
(3) = 3.
g
00
(x)=e
f(x)
f
00
(x)+e
f(x)
f
0
(x)
2
Thus g
00
(1) = e
0
(5) + e
0
(3)
2
=14

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3A
(15) 5. Find the following indenite integrals:
(a)
Z
x
4
+ sec(x)tan(x)
2
x
dx
x
5
5
+ secx2lnjxj+C
(b)
Z
3x
2
sin(x
3
+1)dx
If u=x
3
+1,then du =3x
2
dx and
Z
3x
2
sin(x
3
+1)dx =
Z
sin(u)du =cos(u)+C=cos(x
3
+1)+C
(c)
Z
cos(x)e
sin(x)
dx
If u= sin(x), then du = cos(x)dx and
Z
cos(x)e
sin(x)
dx =
Z
e
u
du =e
u
+C=e
sin(x)
+C
(18) 6. Compute the following:
(a)
Z
2
1
x
3
+5
x
dx
Z
2
1
x
3
+5
x
dx =
Z
2
1
x
2
+
5
x
dx =
x
3
3
+5lnx
2
1
=
8
3
+ 5ln2
1
3
+ 5ln1
=
7
3
+ 5ln2
(b) The area under the graph of y=4x+e
x
on the interval [0;2].
The area is
Z
2
0
(4x+e
x
)dx =(2x
2
+e
x
)j
2
0
=8+e
2
1=7+e
2
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