Download solutions

MATH 226: Calculus I
MIDTERM II
Fall 2015
NAME : SOLUTIONS
NOTE : There are 5 problems on this midterm (total of 6 pages). Use of
calculators will NOT be permitted. In order to receive full credit for any
problem, you must show work leading to your answer. You have 50 minutes
to complete this test.
Problem Possible
points
1
20
2
20
3
20
4
20
5
20
Total
100
Score
BONUS (5pts): What is the anti-derivative of f (x) =
tan−1 x + C
1
?
1 + x2
MATH 226 – MIDTERM II
Page 2
Problem 1. (20pts) Find the absolute maximum and minimum values of
f (x) = x2 − 2 ln x on the interval 1 ≤ x ≤ 3.
Solving
2
=0
x
2(x2 − 1)
=0
x
2(x − 1)(x + 1)
=0
x
f 0 (x) = 2x −
gives x = 1 is the only critical point of f (x),
since −1 and 0 are not in the domain.
Since f 0 (2) = 3 > 0, f is increasing on [1, 3].
This means f (1) = 1 is the absolute minimum of f
while f (3) = 9 − 2 ln 3 is the absolute maximum.
MATH 226 – MIDTERM II
Page 3
Problem 2. (20pts) Find all local extrema, intervals of increase and decrease, and identify any points of inflection.
f (x) = x4 − 4x2 = x2 (x2 − 4)
Solving
f 0 (x) = 4x3 − 8x = 0
4x(x2 − 2) = 0
√
√
4x(x − 2)(x + 2) = 0
√
gives 3 critical points at x = 0, ± 2.
Since
f 0 (−2) = −16 < 0,
f (1) = −4 < 0,
f (−1) = 4 > 0,
f (2) = 16 > 0,
√
√
2]
and
[0,
2]
f is decreasing on (−∞,
−
√
√
and increasing on [− 2, 0] and [ 2, ∞).
p
Since f 00 (x) = 12x2 − 8 = 0 when x = ± 2/3,
and f 00 (−1) = 4 = f 00 (1) > 0 and f 00 (0) = −8 < 0,
both points where f 00 (x) = 0 are inflection
points.
p
They have a common value of f (± 2/3)
=
−20/9.
p
Hence, there are 2 inflection points (± 2/3, −20/9).
MATH 226 – MIDTERM II
Page 4
Problem 3. (20pts) A rectangle is to be inscribed in a semicircle of radius 2.
What is the largest area the rectangle can have, and what are its dimensions?
The equation of the semicircle maybe be represented as
x2 + y 2 = 4,
y ≥ 0.
Let (x, y) be the corner of the inscribed
√ rectangle, with 0 < x < 2.
Since its base is 2x and height is y = 4 − x2 , its area is
√
A = 2x 4 − x2 .
Since,
√
1
dA
= 2 4 − x2 + (2x) √
· (−2x)
dx
2 4 − x2
√
2x2
= 2 4 − x2 − √
4 − x2
2(4 − x2 ) − 2x2
√
=
4 − x2
8 − 4x2
=√
4 − x2
4(2 − x2 )
= √
=0
4 − x2
√
√
when x = ± 2, the only critical point with x > 0 is x = 2.
At this critical point, the area √
A = 4, which is the absolute
maximum
√
0
0
because A (x) > 0 for 0 < x < 2 and A (x) < 0 for 2 < x < 2.
Alternatively, one can deduce A = 4 is an absolute maximum by checking
√
−x
4 − x2 (−8x) − (8 − 4x2 ) √4−x
2
A00 (x) =
2
4−x
2
(4 − x )(−8x) − (8 − 4x2 )(−x)
=
(4 − x2 )3/2
4x3 − 24x
4x(x2 − 6)
=
=
<0
(4 − x2 )3/2
(4 − x2 )3/2
√
when x = 2.
MATH 226 – MIDTERM II
Page 5
Problem 4. (20pts) Use L’Hopital’s rule to compute the following limit.
t2 − et
t→∞ 1 + et
lim
t2 − et
2t − et
=
lim
t→∞ 1 + et
t→∞
et
2 − et
= lim
t→∞
et
−et
= lim t
t→∞ e
= −1
lim
MATH 226 – MIDTERM II
Page 6
√
Problem 5. (15pts) Use Newton’s method to estimate 10 by using the
function f (x) = x2 − 10 with an initial guess x0 = 3 and finding x2 .
Since f 0 (x) = 2x,
f (3)
f 0 (3)
−1
=3−
6
19
=
6
x1 = 3 −
and
x2 =
=
=
=
=
19
f (19/6)
− 0
6
f (19/6)
361
− 10
19
− 36 19
6
2· 6
19
1 3
−
·
6
36 19
2 · 361 − 1
12 · 19
721
228