American seniors have been worried about our nations ability to continue to pay out Social Security. As
one retirement-bound Los Angeles resident puts it, The money that I put aside now, its not like that
money is going to be waiting for me. Unfortunately
A conveyor belt is dropping sand in a conical pile.
V =
1 2
x h
3
Let us suppose that the height is always twice as big as the
base radius
h = 2x
( )2
1 2
1
h
3
y=
h
V = x h =
3
3
2
12
Suppose the volume of sand is increasing at 8 cubic meters per
minut
Let V denote the volume in a piston
Let P denote the pressure of the gas inside the piston
The value of P depends on the value of V
P is a function of V
P = f (V )
Let V denote the volume in a piston
Let P denote the pressure of the gas inside the piston
lim f (x) = L
xa
lim f (x) = L
xa
The limit of f (x), as x approaches a, is L
f (x) gets closer and closer to L
as x gets closer and closer to a.
How close can we get f (x) to L?
lim f (x) = L
xa
f (x) can be made arbitrarily close to L by making
x suffic
MA 241 Calculus I
Practice Test 2
Dr. E. Jacobs
Most of the following practice problems appeared on an old MA 241 Exam 2.
1. (13 points) Suppose a resistor of resistance x ohms and another resistor of resistance
y ohms are connected in parallel. If R is t
Combine the fractions and simplify:
f (x + h)(g(x + h) g(x) g(x)(f (x + h) f (x)
+
h
h
Combine the fractions and simplify:
f (x + h)(g(x + h) g(x) g(x)(f (x + h) f (x)
+
h
h
Simplifies to:
f (x + h)g(x + h) f (x)g(x)
h
Combine the fractions and simplify:
Chromium-254 is a radioactive isotope with a half-life of approximately 1 month.
Chromium-254 is a radioactive isotope with a half-life of approximately 1 month.
Lets suppose we begin with an 8 gram sample of chromium.
After 1 month, we will be left with
MA241 Calculus I Fall 2016
Exam I Solutions
Dr. E. Jacobs
1. (12 points) Calculate the following limits. If a finite limit does not exist, then state this
clearly.
x1
x1
1
= lim 2
= lim 2 = (no finite limit)
3
2
x0 x x
x0 x (x 1)
x0 x
a)
lim
b)
lim
1+
x0
2x2 |x 1|
lim
x1
x1
Absolute Value
cfw_
x if x 0
|x| =
x if x < 0
Absolute value of x 1
cfw_
x 1 if x 1 0
|x 1| =
(x 1) if x 1 < 0
Absolute value of x 1
cfw_
x 1 if x 1
|x 1| =
1 x if x < 1
cfw_
x 1 if x 1
1 x if x < 1
cfw_ x1
if x > 1
|x 1|
= x1
1x
x1
if
Interval Notation
The notation [a, b] refers to the set of all numbers x
such that a x b. This is referred to as the closed
interval between a and b
A function f is said to have a maximum value on an
interval [a, b] if there is a number c in [a, b] with
t
If y is a function of u and u is a function of x, then y is also a
function of x.
The Chain Rule:
dy
dy du
=
dx
du dx
y=
Find
dy
dx
at x = 0
3x + 1
y=
Find
dy
dx
3x + 1
at x = 0
Let u = 3x + 1 so y = u1/2
du
=3
dx
dy
1
1
= u1/2 =
at x = 0
du
2
2
dy
dy du