Download Midterm 2 - Loyola University Chicago

Midterm 2
Loyola University Chicago
Math 131, Section 009, Fall 2008
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Please do not start working until instructed to do so.
You have 50 minutes.
You must show your work to receive full credit.
Calculators are OK.
You may use one double-sided 8.5 by 11 sheet of handwritten (by you) notes.
Problem 1.
Problem 2.
Problem 3.
Problem 4.
Problem 5.
Total.
Problem 1.(10 points total) Find derivatives of the following functions. Give your answer in exact
form (do not use decimals).
³
a.(5 points) f (x) = arctan 3x2 + x
´
Solution: use chain rule
³
³
f 0 (x) = 2 arctan 3x2 + x
b.(5 points) f (x) =
µ
4x − 1
sin(x)
´´0
=
1
(6x + 1)
1 + (3x2 + x)2
¶3
Solution: use chain rule and quotient rule
0
f (x) =
õ
4x − 1
sin(x)
¶3 ! 0
=3
µ
4x − 1
sin(x)
¶2
4 sin(x) − (4x − 1) cos(x)
sin2 (x)
Problem 2.(10 points) The product of two positive numbers x and y is 270. What is the smallest
possible value of x + y?
We need to minimize x + y knowing that x, y are positive and xy = 270. So y = 270/x and we
need to minimize f (x) = x + 270/x with 0 < x < 270. So
f 0 (x) = 1 −
and f 0 (x) = 0 when x =
270
x2
√
270. This is the only critical point. To make sure it is a minimum, try
540
x3
√
√
√
√
and note that√f 00 ( 270) > 0. So x = 270, y = 270/ 270 = 270 and the smallest possible value
of x + y is 2 270.
f 00 (x) =
Problem 3.(10 points total) Find two points (x, y) on the circle
x2 + y 2 = 1
at which the tangent line to the circle has slope −2.
Solution: Implicit differentiation gives
2x + 2y
We need
dy
dx
= −2, so
x
y
dy
= 0,
dx
dy
x
=− .
dx
y
= 2, so x = 2y. Plug this in to the equation of the circle:
(2y)2 + y 2 = 1,
4y 2 + y 2 = 1,
5y 2 = 1,
√
√
So, y = 1/ 5 or −1/ 5.
The two points on the circle are then
µ
2 1
√ ,√ ,
5 5
¶
2
1
−√ , −√
5
5
µ
¶
1
y2 = .
5
Problem 4.(10 points total) The derivative of a certain function f (x) is given by the formula
f 0 (x) = x2 (x − 2)(x + 3).
a.(6 points) Find all critical points of f (x) and clearly identify them as a local minimum, local
maximum, or neither. Justify your answers.
Solution: f 0 (x) = 0 if x = 0 or x = 2 or x = −3. These are the critical points. To determine
if they are local minima, maxima, or neither, check the sign of f 0 (x) in between the critical points.
f 0 (−4) = 16(−6)(−1) > 0, f 0 (−1) = 1(−3)(2) < 0, f 0 (1) = 1(−1)(4) < 0, f 0 (3) = 9(1)(6) > 0.
Thus, x = −3 is a local maximum, x = 0 is not a local maximum nor a local minimum, x = 2 is a
local minimum.
b.(4 points) Find the second derivative f 00 (x) of the function f (x), and at least one inflection point
of f (x).
Solution: First, simplify f 0 (x) = x2 (x2 + x − 6) = x4 + x3 − 6x2 . Now, find f 00 (x):
f 00 (x) = 4x3 + 3x2 − 12x = x(4x2 + 3x − 12).
So, f 00 (x) = 0 when x = 0. This is a good candidate for an inflection point. To see that it really is
an inflection point, note that 4x2 + 3x − 12 is negative when x is close to 0, but x changes sign from
negative to positive at x = 0. So, f 00 (x) changes sign from positive to negative at x = 0.
Problem 5.(10 points) Waterpark management decided to install a rectangular billboard with a
picture of a dolphin under a popular waterslide. The end of the slide is three meters away from the
wall where it starts (see picture). According to the manual, the slide has the shape of a parabola
with equation y = 2x2 . Compute the area of the largest possible billboard that can fit into the space
between the slide and the wall.
16
12
8
4
1
2
3
Solution: Let x be the x-coordinate of the lower left-hand corner of the billboard. Then the
horizontal side of the billboard is 3 − x, while the vertical side of the billboard is 2x 2 . So, the area of
the billboard is
A(x) = (3 − x)2x2 = 6x2 − 2x3 .
To maximize this function on the interval [0, 3], find
A0 (x) = 12x − 6x2 = 6x(2 − x).
So A0 (x) = 0 if x = 0 or x = 2. x = 0 gives zero area, so it is obviously not the maximum. The
other endpoint, x = 3, also gives zero area. A(2) = 8, so this is the global maximum. (One could
also check A00 (x) = 12 − 12x at x = 2. One gets A00 (2) = −12 < 0 so x = 2 is a local maximum.)
So, the maximum area of the billboard is 8.