MATH 31A (Butler)
Practice for Midterm IIa (Solutions)
1.
(a) Use linearization to give an estimate for
3
√
1017.
We first note that
3
√
1017 is close to
3
√
1000 = 10 and so we can use lineariza
tion for the function
f
(
x
) =
3
√
x
=
x
1
/
3
at
x
= 1000. Before we start we note
that
f
0
(
x
) =
1
3
x

2
/
3
=
1
3(
3
√
x
)
2
. So the linearization is
L
(
x
)
=
f
(
a
) +
f
0
(
a
)(
x

a
)
=
f
(1000) +
f
0
(1000)(
x

1000)
=
3
√
1000 +
1
3(
3
√
1000)
2
(
x

1000)
=
10 +
1
300
(
x

1000)
.
So we have that
3
√
1017
≈
L
(1017) = 10 +
1
300
(1017

1000) = 10 +
17
300
.
(b) Is the estimate given in part (a) too large or too small? Explain.
We note that
f
00
(
x
) =

2
9
x

5
/
3
. In particular, at
x
= 1000 we see that
f
00
(
x
)
is negative and so the curve is concave down, i.e., it is bending down. So we
would expect the actual function to be under the tangent line so that the
answer in part (a) would be an overestimate.
[Note:
3
√
1017
≈
10
.
056348 while 10 +
17
300
≈
10
.
056666. So an overestimate
as predicted, but still a good guess.]
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2.
What is the area of the largest rectangle that you can make where the bottom
edge is on the
x
axis and the top two vertices lie on the parabola
y
= 12

x
2
?
For this problem it is good to draw a picture (I recommend you draw one
before proceeding with the argument). If one corner of the rectangle on the
parabola is at the point (
x, y
) = (
x,
12

x
2
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 Winter '10
 BUTLER
 Math, Calculus, Derivative, Mean Value Theorem, Continuous function, largest rectangle

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