5.
x 2 x2 2x 4
x 2 x2 2x 4
f(x)
2
x 2x 4
x3 x2 x2
x2 x 3 4 x 3
x3 x2
lim f(x) and lim f(x)
.x3x2
So, x 3 and x 2 are the only vertical asymptotes.Note that x 2
causes a division by zero, but there is no vertical asymptote there
because
lim f (x)
23 22
2
2
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Unit 5
Chapter 6 Applications of the Definite Integral
6.1
Area
y
y
y f ( x)
y f ( x)
0
a
b
x
y g ( x)
0
a
0
b
x
b
area f ( x) dx
0
an Rx region
a
Rx Regions We call a region an Rx region if it lies between the graphs of two functions
y f ( x)
Unit 2
Additional Problems
Calculator
1 + x2
f ( a + h) f ( a )
with h = 0.0001 to
1. Suppose f ( x) =
. Use the slope of the secant line
2
h
sin ( x )
3
approximate the slope of the line tangent to the graph of f at P ( 0.85, f (0.85) ) . Answer to the
n
Unit 2
Additional Problems
No Calculator
f ( x + h) f ( x )
to find f ( x).
h0
h
7
5
(b) f ( x) =
(c) f ( x)= 6
2x + 3
x +1
1. Use the definition f ( x) = lim
(a) f ( =
x)
1
x x3
2
2. The graph of y = f ( x) is given below. At what values of x does the g
Unit 1
Additional Problems
No Calculator
1
1. Simplify
( x + h)
2
1
+ 5 2 + 5
x
h
.
2. Solve: sin ( 2 x ) = sin x on [ 0, 2 ) .
3. A box with a square base and height h has a volume of 25. Find its surface area in terms of h.
4. What is the radius of a
P a g e |1
Geometry Examples Solutions
Example 1 Solution
Note that the length is 2w 5. So, the perimeter is
2 w 2 2 w 5 2 w 4 w 10 6 w 10.
Example 2 Solution
We have 2 r 25 . Dividing both sides by 2 gives r 12.5 in.
Example 3 Solution
Let each of the tw
Geometry Formulas
Perimeter: The perimeter of a simple closed figure is the distance once around the outside of
the figure.
Circumference of a Circle: C = 2 r
Area of Rectangle or Parallelogram: A = bh
1
Area of Triangle: A = bh
2
1
Area of Trapezoid:=
A
Page |1
Unit 1
Chapter 1 Precalculus Review
In this chapter, we review the essential topics of precalculus that will relate directly to many of
the calculus problems you will have to solve. This is not a comprehensive review and you will
need some additio
P a g e |1
Unit 4
Chapter 5 Integrals
5.1
Antiderivatives and Indefinite Integrals
Definition A function F is an antiderivative of f on an interval I if F ( x) f ( x)
for every x in I.
Example 1 Find three different antiderivatives of f ( x) 2 x .
Theorem
Calculus IExam 1 (Ch. 1, Ch. 2) June 19th, 2012
Solutions
1.
4
2 0x
4 2x
4 2x
0 xx
0
x
0
x
22x
x
,0
2, Note that x 0 is not a solution because of division
by 0.
2.
2 1 sin2 x sinx 1 0
2sin2 x sinx 1 0 2sin2 x sinx 1 0 2sin x 1 sin x 1
1
sinx , sinx 1 2
5
Derivative Rules
Definition
f ( x) = lim
h0
f ( x + h) f ( x )
h
Power Rule Dx ( x r ) = r x r 1
Constant Multiple Dx ( c f ( x) ) = c f ( x)
Sum/Difference Dx [ f ( x) g ( x) ] = f ( x) g ( x)
Product Rule Dx [ f= f ( x) g ( x) + f ( x) g ( x)
( x) g ( x
P a g e |1
Unit 2
Chapter 3 The Derivative
3.1
Tangent Lines and Rates of Change
Slope The slope of a non-vertical line measures how much y changes for a unit change in x
(called the rate of change of y with respect to x), and it is constant at all places
Unit 3
Additional Problems
Calculator
1. Suppose f ( x) x 2 2sin x. Use the graph of f to estimate the maximum and minimum
values of f on 0, 2 correct to four decimal places.
2. Find all numbers c that satisfy the conclusion of the Mean Value Theorem for
P a g e |1
Unit 3
Chapter 4 Applications of the Derivative
4.1
Extrema of Functions
Definition Let f be a function defined on an interval I , and let x1 , x2 denote numbers in I .
(i) f is increasing on I if f ( x1 ) f ( x2 ) whenever x1 x2 .
(ii) f is de
Trigonometric Identities
Reciprocal Identities
1
1
1
=
csc x
, sec x
, cot x
sin x
cos x
tan x
Tangent/Cotangent Identities
sin x
cos x
=
tan x
, cot x
cos x
sin x
Pythagorean Identities
sin 2 x + cos 2 x = 1 + tan 2 x = 2 x, 1 + cot 2 x = 2 x
1,
sec
csc