10550 PRACTICE FINAL EXAM SOLUTIONS
1. First notice that
x2 4
(x 2)(x + 2)
=
.
x2 5x + 6
(x 2)(x 3)
This function is undened at x = 2. Since, in the limit as x 2 , we only care about what happens
near x = 2 (an for x less than 2), we can cancel
lim
x2
x2
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Lecture 10 : Chain Rule
Here we apply the derivative to composite functions. We get the following rule of dierentiation:
The Chain Rule : If g is a dierentiable function at x and f is dierentiable at g(x), then the
composite function F = f g dened by F (x
Lecture 4 : Calculating Limits using Limit Laws
Using the denition of the limit, limxa f (x), we can derive many general laws of limits, that help us to
calculate limits quickly and easily. The following rules apply to any functions f (x) and g(x) and als
Lecture 3 : Limit of a Function
Limit of a Function
Consider the behavior of the values of f (x) = x2 as x gets closer and closer . and closer . to 3.
Example
Let f (x) = x2 . The table below shows the behavior of the values of f (x) as x approaches
3 fro
Lecture 5 : Continuous Functions
Denition 1 We say the function f is continuous at a number a if
lim f (x) = f (a).
xa
(i.e. we can make the value of f (x) as close as we like to f (a) by taking x suciently close to a).
Example Last day we saw that if f (
Lecture 2 : Tangents
Functions
The word Tangent means touching in Latin. The idea of a tangent to a curve at a point P , is a
natural one, it is a line that touches the curve at the point P , with the same direction as the curve.
However this description
Lecture 9 : Derivatives of Trigonometric Functions
In this section we will look at the derivatives of the trigonometric functions
sin x,
cos x,
tan x
, sec x,
csc x,
cot x.
Here the units used are radians and sin x = sin(x radians). Recall that sin x and
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Name:
Instructor:
Math 10550, Exam I
November 10, 2013
The Honor Code is in eect for this examination. All work is to be your own.
No calculators.
The exam lasts for 1 hour and 15 min.
Be sure that your name is on every page in case pages become detached.
MATH 10550
SOLUTIONS TO PRACTICE FINAL EXAM
1.Compute lim
x2
Solution
When x = 2,
x2 4
.
x2 5x + 6
x2 4
(x 2)(x + 2)
x+2
=
=
.
2 5x + 6
x
(x 2)(x 3)
x3
So
x2 4
2+2
lim 2
=
= 4.
x 5x + 6
x2
23
2. Compute lim
+
x0
x2 9
.
sin x
Solution
If x is close to 0 b
Name:
Instructor:
Math 10550, Exam III
November 19, 2013
The Honor Code is in eect for this examination. All work is to be your own.
No calculators.
The exam lasts for 1 hour and 15 min.
Be sure that your name is on every page in case pages become detache
MATH 10550, EXAM 2 SOLUTIONS
(1) Find an equation for the tangent line to
x2 + 2xy y 2 + x = 2
at the point (1, 2).
Solution: The equation of a line requires a point and a slope. The problem gives us the point so
we only need to nd the slope. Use implicit
SOLUTIONS TO EXAM 1, MATH 10550
1. Compute
lim
x1
x2
x2 1
.
+ 2x + 1
Answer: .
Solution:
lim
x1
x2 1
x2 +2x+1
= lim
x1
(x1)(x+1)
(x+1)(x+1)
= lim
x1
2. All the vertical asymptotes of the function f (x) =
Answer: x = 0 and x = 3
x1
x+1
= .
x2 1
are at
x3 9
Name:
Math 10550, Final Exam:
December 15, 2007
Instructor:
Be sure that you have all 20 pages of the test.
No calculators are to be used.
The exam lasts for two hours.
When told to begin, remove this answer sheet and keep it under the
rest of your test.
MATH 10550
EXAM III SOLUTIONS
1. Solving the equation x2 2 + cos( x ) = 0 using Newtons method
2
with initial approximation x1 = 1, what is x2 ?
Solution. Let f (x) = x2 2 + cos( x ). Then
2
f (x1 )
f (x1 )
f (x1 ) = f (1) = 1
x
f (x) = 2x sin( )
2
2
4
f
MATH 10550, EXAM 3 SOLUTIONS
1. In nding an approximate solution to the equation x3 +2x4 = 0
using Newtons method with initial approximation x1 = 1, what is x2 ?
Solution. Recall that
f (xn )
xn+1 = xn
.
f (xn )
Hence,
(1)3 + 2(1) 4
1
6
=1
=
x2 = 1
2 +
Name:
Instructor:
Math 10550, Self Diagnostic Exam
August 28, 2014
The Honor Code is in eect for this examination. All work is to be your own.
No calculators.
The exam lasts for 1 hour .
Be sure that your name is on every page in case pages become detache
Name:
Instructor:
Math 10550, Exam 1, Solutions
September 20, 2011.
The Honor Code is in eect for this examination. All work is to be your own.
No calculators.
The exam lasts for 1 hour and 15 min.
Be sure that your name is on every page in case pages bec