Let
f
(
x
) =
x
3
+
x
2
 2
x
on [1,1].
Observe that
f
(1) = 2,
f
(1) = 0, and
k
= 1 is a number between
f
(1) and
f
(1).
Since f is a
polynomial, f is continuous on the interval [1,1].
Consequently, the
function f satisfies the hypotheses of the Intermediate Value Theorem.
Thus, we are entitled to invoke the magical conclusion that asserts that
there is at least one number
x
0
in (1,1) where f(
x
0
) = 1.
______________________________________________________________________
3. (5 pts.)
Pretend
f
is a magical function that has the property that
at x = 3 the tangent line f is actually defined by the equation
y = 2(x  1) + 5.
Obtain
(a)
f
(3) = 2(3  1) + 5 = 1
(b)
f
′
(3) = 2
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______________________________________________________________________
4. (25 pts.)
Compute the first derivatives of the following functions.
You may use any of the rules of differentiation that are at your disposal.
Do not attempt to simplify the algebra in your answers.
(a)
f
(
x
)
4
x
6
7
x
12
8tan(
x
)
f
(
x
)
24
x
5
84
x
13
8sec
2
(
x
)
(b)
g
(
x
)
(4
x
2
2
x
1
)sec(
x
)
g
(
x
)
(8
x
2
x
2
)sec(
x
)
(4
x
2
2
x
1
)sec(
x
)tan(
x
)
(c)
h
(
t
)
5
t
10
1
sin(
t
)
2
h
(
t
)
(50
t
9
)(sin(
t
)
2)
(5
t
10
1)cos(
t
)
(sin(
t
)
2)
2
(d)
y
cot
5
(2
θ
1)
dy
d
θ
5(cot(2
θ
1))
4
(
csc
2
(2
θ
1))(2)
10cot
4
(2
θ
1)
csc
2
(2
θ
1)
(e)
L
(
z
)
sin(4
z
8
)
4
csc
(
π
6
)
4cos(
z
2
)
dL
dz
(
z
)
cos(4
z
8
)(32
z
7
)
0
4(
sin(
z
2
))(
1
2
)
32
z
7
cos(4
z
8
)
2sin(
z
2
)
______________________________________________________________________
Silly 10 point Bonus Problem:
(Solution)
Theorem 2.5.5 asserts that if g is a function with
and
f
is a function that is continuous at
L
, then
lim
x
→
c
g
(
x
)
L
lim
x
→
c
f
(
g
(
x
))
f
(
L
).
It follows from
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 Fall '08
 STAFF
 Calculus, Derivative, Continuous function, lim g

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