(b) The graph is concave up for
u1D465 <
2
and concave down for
u1D465 >
2
, and the derivative is undefined at
u1D465
= 2
. This is the
case if the graph is vertical at
u1D465
= 2
. One possible graph is shown in Figure 2.84.
2
u1D465
u1D453
(
u1D465
)
Figure 2.83
2
u1D465
u1D453
(
u1D465
)
Figure 2.84
12.
We want to look at
lim
ℎ
→
0
(
ℎ
2
+ 0
.
0001)
1∕2
− (0
.
0001)
1∕2
ℎ
.
As
ℎ
→
0
from positive or negative numbers, the difference quotient approaches 0. (Try evaluating it for
ℎ
= 0
.
001
,
0.0001, etc.) So it appears there is a derivative at
u1D465
= 0
and that this derivative is zero. How can this be if
u1D453
has a corner
at
u1D465
= 0
?
The answer lies in the fact that what appears to be a corner is in fact smooth—when you zoom in, the graph of
u1D453
looks like a straight line with slope 0! See Figure 2.85.
−2
−1
0
1
2
1
2
u1D465
u1D453
(
u1D465
)
−0
.
2
−0
.
1
0
0
.
1
0
.
2
0
.
1
0
.
2
u1D465
u1D453
(
u1D465
)
Figure 2.85
: Close-ups of
u1D453
(
u1D465
) = (
u1D465
2
+ 0
.
0001)
1∕2
showing differentiability at
u1D465
= 0
13.
(a)
u1D445
u1D43Au1D440
u1D445
2
u1D45F
u1D454
Figure 2.86
(b) The graph certainly looks continuous. The only point in question is
u1D45F
=
u1D445
. Using the second formula with
u1D45F
=
u1D445
gives
u1D454
=
u1D43Au1D440
u1D445
2
.
Then, using the first formula with
u1D45F
approaching
u1D445
from below, we see that as we get close to the surface of the earth
u1D454
≈
u1D43Au1D440u1D445
u1D445
3
=
u1D43Au1D440
u1D445
2
.
Since we get the same value for
u1D454
from both formulas,
u1D454
is continuous.
(c) For
u1D45F < u1D445
, the graph of
u1D454
is a line with a positive slope of
u1D43Au1D440
u1D445
3
. For
u1D45F > u1D445
, the graph of
u1D454
looks like
1∕
u1D465
2
, and so
has a negative slope. Therefore the graph has a “corner” at
u1D45F
=
u1D445
and so is not differentiable there.

202
Chapter Two /SOLUTIONS
14.
(a) The graph of
u1D444
against
u1D461
does not have a break at
u1D461
= 0
, so
u1D444
appears to be continuous at
u1D461
= 0
. See Figure 2.87.
−2
−1
1
2
1
u1D461
u1D444
Figure 2.87
(b) The slope
u1D451u1D444
∕
u1D451u1D461
is zero for
u1D461 <
0
, and negative for all
u1D461 >
0
. At
u1D461
= 0
, there appears to be a corner, which does not
disappear as you zoom in, suggesting that
u1D43C
is defined for all times
u1D461
except
u1D461
= 0
.
15.
(a) Notice that
u1D435
is a linear function of
u1D45F
for
u1D45F
≤
u1D45F
0
and a reciprocal for
u1D45F > u1D45F
0
. The constant
u1D435
0
is the value of
u1D435
at
u1D45F
=
u1D45F
0
and the maximum value of
u1D435
. See Figure 2.88.
u1D45F
0
u1D435
0
u1D45F
u1D435
Figure 2.88
(b)
u1D435
is continuous at
u1D45F
=
u1D45F
0
because there is no break in the graph there. Using the formula for
u1D435
, we have
lim
u1D45F
→
u1D45F
−
0
u1D435
=
u1D45F
0
u1D45F
0
u1D435
0
=
u1D435
0
and
lim
u1D45F
→
u1D45F
+
0
u1D435
=
u1D45F
0
u1D45F
0
u1D435
0
=
u1D435
0
.
(c) The function
u1D435
is not differentiable at
u1D45F
=
u1D45F
0
because the graph has a corner there. The slope is positive for
u1D45F < u1D45F
0
and
the slope is negative for
u1D45F > u1D45F
0
.
16.
(a) Since
lim
u1D45F
→
u1D45F
−
0
u1D438
=
u1D458u1D45F
0
and
lim
u1D45F
→
u1D45F
+
0
u1D438
=
u1D458u1D45F
2
0
u1D45F
0
=
u1D458u1D45F
0
and
u1D438
(
u1D45F
0
) =
u1D458u1D45F
0
,
we see that
u1D438
is continuous at
u1D45F
0
.