Exam
Review
v (ft/sec)
4
5
10
t
(sec)
-4
A turtle starts 2 feet to the left of my mailbox. Where is
the turtle at t=5? Where is the turtle at t=10?
Find the limit if it
exists.
1.
2.
3.
4.
f ( x)
5.
lim f ( x )
x  1
lim f ( x )
x  1
lim f ( x )
x  1
lim f ( x )
x2
lim f ( x )
x2
6. lim f ( x )
x2
A function is continuous at a point if the limit is the same
as the value of the function.
lim f ( x) must exist and be equal to f (c)
xc
2
1
1
2
3
4

Jump Discontinuity is typically caused by a
piece-wise function whose pieces don’t meet.
The limit does not exist because the left-hand
limit does not equal the right-hand limit.
Example:
 x  3,
f ( x)  
 x  1,
x0
x0
Find the real number, c that makes the function everywhere
continuous.
 x  2,
f ( x)   2
 x  c,
x3
x3
y  x2  8
(3,1)
y  x2
Is the function continuous at x = -3?
 x  19  4
, x  3

f ( x)   x  3
1
, x  3
 8
lim f ( x) must equal f ( 3)
x 3
To be differentiable, a function must be continuous and
smooth.
Derivatives will fail to exist at:
f  x  x
f  x  x
corner
2
3
cusp
1, x  0
f  x  
 1, x  0
f  x  3 x
vertical tangent
discontinuity

Intermediate Value Theorem
If a function is continuous between a and b, then it takes
on every value between f  a  and f  b  .
f b
Because the function is
continuous, it must take on
every y value between f  a 
and f  b  .
f a
a
b

Use the Intermediate Value Theorem to explain
2
why the function f ( x)  x  3x  6 must have a
root (x-intercept) on the closed interval [1,2].
f (1)  2
f (2)  4
If f (1)  2 and f (2)  4, then by the
Intermediate Value Thm, the function
must take on all y values between -2 and 4,
so a y value of 0 must exist and it will
happen when x is between 1 and 2.
Mean Value Theorem for Derivatives
If f (x) is continuous over [a,b] and differentiable
over (a,b), then at some point c between a and b:
f c 
f b   f  a 
ba
The Mean Value Theorem says that at some point
in the closed interval, the actual slope equals the
The Mean Value Theorem only applies over a closed interval.
average slope.

y
Slope of tangent:
f  c
B
Slope of secant:
f b   f  a 
ba
A
0
y  f  x
a
c
b
x

The Mean Value Theorem is actually common
sense. Think of it like this. If, on a 2-hour car trip,
you averaged 50 miles per hour, then according to
the Mean Value Theorem, at least once during the
trip, your speedometer actually read 50 mph.
At what x-value(s) on the interval [-2,3] does the graph of
f ( x)  x2  2 x 1 satisfy the Mean-Value Theorem?
The Extreme Value Theorem
If a function f is continuous on a closed interval [a,b],
then f has a global maximum and a global minimum value
on that interval.
To find extreme values, test all critical points and
endpoints by comparing their outputs (height).
Find the global maximum and minimum of the given
function on the indicated interval.
f ( x)  x3  3x2  2, 0.5,4
A 25 ft ladder is leaning against a wall. If the foot of the
ladder is being pulled away from the bottom of the wall at a
rate of 14 feet per second, at what rate is the top of the
ladder moving down the wall when it is 7 feet above the
ground?
Water is flowing into a cone-shaped tank at the rate of 5
cubic inches per second. If the cone has an altitude of 4
inches and a base radius of 3 inches, how fast is the water
level rising when the water is 2 inches deep?
4  x  h   4 x3
3
lim
h 0
h
3x  h  3x
lim
h 0
h
3  h

lim
h 0
lim
h 0
3
 27
h
25  h  5
h
Do page 39 of AP Prep Book #1-5,7,8
For x3  3xy  y 3  8
dy
A. Find
.
dx
B. Write an equation of the line tangent to the graph at x  0.