Name
MATH 105 MIDTERM II
KEY
Please work out each of the given
problems. Credit will be based on
the steps that you show towards the final answer.
Show your work.
PROBLEM 1 (7 Points
each) Please answer the following
true or false. If false, explain
why or provide a counter example. If
true, explain why.

Let f(x)
be differentiable on [0,1], then there is a positive c
less than 1 such that
f(1) = f(0) + f '(c)
Solution True,
by the mean value theorem, we have
f(1)
 f(0)
= f
'(c) for
some c in [0,1].
1 
0
and the result follows.

Let _{
}
f(x) be a differentiable function such that f
'(c) = f ''(c) = 0 _{
}
, then f has neither a relative maximum nor a
relative minimum at x = c
.
Solution False. The second derivative
test only tells you that we do not know about max and min there. For
example, y = x^{4} has
the above properties for c = 0 and
has a minimum there.

If
f ''(x) < 0 _{
}
for all x between 0 and 1 and f(0)
= 0
then f(1)
< 0.
Solution: False, just because a function is
concave down does not mean that it is decreasing. For example, y
= (x  1)^{2} + 1 is a counter
example.

If f(x)
_{
}
is a periodic differentiable function, that is f(x + p) = f(x)
_{
}
for some p
and all x, then f '(x) _{
}
has an infinite number of roots.
Solution: True, by Rolle's theorem,
since f(kp) = f((k + 1)p),
there is a c
between kp and (k
+ 1)p, such that f
'(c) = 0. This gives a root for any
integer k.
PROBLEM 2 Consider the
function (22 Points)
_{
}
Without the use of the graphing
capabilities of your graphing calculator, for the following two functions

Determine the relative extrema if any.
Solution
f '(x) = 2x 
1/x^{2
}Which has a root at (1/2)^{1/3}
hence the possible relative extrema is at
((1/2)^{1/3}
, (1/2)^{2/3} +(1/2)^{1/3})
f ''(x) = 2 + 2/x^{3
}and
f
''((1/2)^{1/3}) = 2 + 2 > 0
Hence there is a relative minimum there

Determine where the function is increasing and
decreasing.
Solution
We want to
determine when the derivative is positive and when it is negative. The
critical points of f are
(1/2)^{1/3}
and 0
For x <
0,
f '(x) = 2x 
1/x^{2
} < 0 (plug
in a test value such as x = 1)
For 0 < x
< (1/2)^{1/3}
f '(x) = 2x 
1/x^{2
}< 0 (plug
in a test value such as x = 0.1)
For (1/2)^{1/3} < x
f '(x) = 2x 
1/x^{2} > 0 (plug
in a test value such as x = 5)
Hence the function is increasing for (1/2)^{1/3}
< x and decreasing
for x < 0 and
0 < x
< (1/2)^{1/3}.

Determine the inflection points if any.
Solution
We set the second derivative equal to zero:
f ''(x) = 2 + 2/x^{3}
= 0
2x^{3} + 2 =
0
x^{3} = 1
x = 1^{
}Now plug back into the original function
to get the point
(1,
0^{ })
^{
}

Determine where the function is concave up and
concave down.
Solution
We want to
determine when the second derivative is positive and when it is
negative. The changes in sign can occur at
x = 1
or x = 0
For x <
1,
f ''(x) = 2 + 2/x^{3}
> 0 (plug
in a test value such as x = 2)
For 1 <
x < 0
f ''(x) = 2 + 2/x^{3}
< 0 (plug
in a test value such as x = 0.5)
For 0 < x
f ''(x) = 2 + 2/x^{3}
> 0 (plug
in a test value such as x = 5)
Hence the function is concave up for x <
1 and for 0
< x and concave down for
1 < x < 0.

Find any horizontal asymptotes.
Solution
We find the horizontal asymptotes by taking the limit
of f(x) as x approaches infinity. But this limit goes to infinity, so
there are no horizontal asymptotes.

Find any vertical asymptotes.
Solution
We find the vertical asymptotes by setting the
denominator equal to zero. Hence the vertical asymptote occurs at x
= 0. Notice that by looking at the left
and the right of the equation, we have a downup behavior near the
asymptote.

Use the above to graph the function, labeling
all important points and asymptotes.
Solution
PROBLEM 3 (20 Points)
As you are standing outside on a
beautiful sunny late afternoon, you notice that your shadow is growing in length
at a rate of 2 feet per hour. If
you are 6 feet tall, how fast is the angle of elevation of the sun decreasing
when your shadow is 8 feet long?
Solution
We have
tan q = 6/s
s = 6cot q
ds/dt = 6csc^{2} q
dq/dt
2 = 6(10/6)^{2} dq/dt
dq/dt = 0.12 rad/sec
PROBLEM 4 (20
Points Each)

Suppose that
x^{2}
+ y + sin(xy) = 2
Find the equation of
the tangent line to this curve at the point (0,2).
Solution
Taking the derivative of both sides implicitly with
respect to x, we get:
2x + y' + (y + xy')cos(xy) = 0
y' + 2cos(0) = 0
y' = 2
Now use the fact that the slope is 2 and the
yint is 2 to get
y =
2x + 2

Let _{
}
Find _{
}
Solution
We use the
quotient rule and the chain rule:
(1
 x)(2x)(sin(x^{2} + 1))  cos(x^{2} + 1)(1)
(1  x)^{2
}
(2x^{2}  2x)sin(x^{2} + 1) +
cos(x^{2} + 1)
=
(1  x)^{2
}
PROBLEM 5 (20
Points)
Sketch the graph of a function
with the following properties:
 _{
}
 _{
}
 _{
}
 _{
}
 _{
}
f ''(x) > 0 only for
x < 3, 1 < x
< 2, and
3 < x < 5
 _{
}
f '(x) > 0 only for x
< 2 and
x > 4
 _{
}
f '(2) = f '(2) = f '(4) = 0
 f ''(3) = f ''(2) = f ''(3) = 0
Solution
PROBLEM 6 (20
Points)
You plan to build a rectangular
office building of minimal cost with a with a wall down the middle to subdivide the building
into two offices as shown in the picture. The
outer walls cost $500 per foot to construct and the divider wall costs $100 per
foot to construct. If the combined
square footage of the two offices is to be 1,000 square feet, what dimensions
should you make the building?
Solution
Let w
be the width of the building and h be the length. Then the square footage
is
1000 = wh
or
w = 1000/h
and
the cost is
C = 500(2w + 2h) + 100h
= 1000w + 1100h
= 1,000,000/h + 1100h
Taking
a derivative and setting it equal to zero gives>
C'(h) = 1,000,000/h^{2}
+ 1100 = 0
1,000,000 + 1100h^{2} = 0
h^{2} = 10000/11
h = 30.15 feet
So
that
w = 1000/h = 1000/30.15 = 33.17
The
building should have dimensions 30.15 by 33.17
