Math 105a Exam 1
Solutions
1. The graph of
f
0
is provided on the left, while the graph of
f
is provided on the right.
(a)
f
is greatest at D
(b)
f
is least at A
(c)
f
0
is greatest at A
(d)
f
0
is least at C
(e)
f
00
is greatest at D
(f)
f
00
is least A
(g)
f
is increasing everywhere except C since
f
0
(
x
) = 0 at
x
=C.
(h)
f
is increasing most rapidly at A since
f
0
(
x
) is greatest there.
2. Use the limit denition of the derivative to nd
f
0
(
x
) for
f
(
x
) = 5
x
.
f
0
(
x
) = lim
h
!
0
f
(
x
+
h
)
f
(
x
)
h
= lim
h
!
0
5
x
+
h
5
x
h
= lim
h
!
0
5
x
5(
x
+
h
)
xh
(
x
+
h
)
= lim
h
!
0
5
h
xh
(
x
+
h
)= lim
h
!
0
5
x
(
x
+
h
)=
5
x
2
3. (a)
f
(
x
) = 3
x
13
x
2
1
x
+
e
2
f
0
(
x
) = 3
x
ln 3
26
x
+1
x
2
(b)
f
(
x
) = 5
x
x
3
+ 8
x
7
f
0
(
x
) = 5
x
ln 5
x
3
+ 8
x
7
+ 5
x
3
x
2
+ 56
x
6
= 5
x
x
2
8
x
5
ln 5 + 56
x
4
+
x
ln 5 + 3
(c)
f
(
x
) =
e
x
+
x
1
x
2
f
0
(
x
) = (
e
x
+
x
ln
)(1
x
2
)
(
e
x
+
x
) (
2
x
)
(1
x
2
)
2
(d)
f
(
x
) =
q
1 +
p
x
=
1 +
x
1
=
2
1
=
2
f
0
(
x
) = 1
2
1 +
x
1
=
2
1
=
2
1
2
x
1
=
2
=1
4
p
x
p
1 +
p
x
4. If
f
(
x
)=3
x
3
4
x
2
at
x
= 1, then
f
0
(
x
)=9
x
2
8
x
. At
x
= 1 the slope
m
is
f
0
(1) = 1. The line is
tangent at (1
; f
(1)) = (1
1) and
y
-intercept of the tangent line is
b
=
y
mx
=
1
(1)(1) =
2.
Therefore the equation of the tangent line is
y
=
x
2.
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