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Date
Math 10550 Worksheet 7
1. One of the following functions has a slant asymptote. Identify the function and find the equation
of the corresponding slant asymptote.
h(x) =
3x2 + x + 2
,
x2 + 4
t(x) =
2x3 + x2 + 1
,
x2 + 2
r(x) =
x3 + x + 3
.
x4 + 2x3
Solution: For rational functions, slant asymptotes occur when the degree of the numerator is
one more than the degree of the denominator. This means t(x) has a slant asymptote.
Doing long division, we have
2x + 1
x +2
2x + x
+1
3
− 2x
− 4x
2
x − 4x + 1
− x2
−2
− 4x − 1
2
which yields the slant asymptote y = 2x + 1 .
3
2
2. An rectangular box with no top, with square base, and with a volume of 64 cubic feet is needed.
Material for the base costs $4 per square foot, and material for the sides costs $2 per square foot.
Determine the dimensions of the box that will minimize the cost of materials. Justify that your
answer is a minimum.
Solution: Let x be the length of the base, and y be the height of the box. We require the
volume, x2 y = 64. We have the cost of each square foot of base, is $4, and the cost of each
square foot of siding is $2. We want to minimize the total cost, i.e. minimize
C(x, y) = 4x2 + (4 × 2)xy.
Using the equation for volume, we substitute y =
64
x2
C(x) = 4x2 +
to get the new relation we wish to minimize
8 · 64
.
x
Now, to find the critical numbers we take the derivative
C 0 (x) = 8x −
8 · 64
.
x2
The critical numbers must satisfy
8x −
8 · 64
= 0 =⇒ 8x−2 (x3 − 64) = 0 =⇒ x3 − 64 = 0 =⇒ x = 4.
x2
We know this is a minimum, since C 0 (x) < 0 for all 0 < x < 4 and C 0 (x) > 0 for all 4 < x. So
= 64
= 4 feet.
each side should have length 4 feet and it should have a height of y = 64
42
16
3. Use Newton’s method with first approximation x1 = 2 to find the second approximation, x2 , to a
root of the equation x3 − 2x2 + 1 = 0.
Solution: Letting f (x) = x3 − 2x2 + 1, since f 0 (x) = 3x2 − 4x, we have
f (x1 )
f 0 (x1 )
x3 − 2x2 + 1
= x1 − 1 2 1
3x1 − 4x1
3
2 − 2(22 ) + 1
=2−
3(22 ) − 4(2)
8−8+1
=2−
12 − 8
1
7
=2− = .
4
4
x2 = x1 −
√
4. In an attempt to solve the equation 2 + tan x = 2 csc x on (0, π2 ] by Newton’s method, we begin
with x1 = π4 . Find the value of x2 in this process. Show all of your work.
√
Solution:
First, we rewrite the equation as 2 + tan x − 2 csc x = 0, and let f (x) = 2 + tan x −
√
2 csc x. Given x1 = π4 , using Newton’s method, the formula to find x2 is
x2 = x1 −
f ( π4 )
f (x1 )
π
=
−
.
f 0 (x1 )
4 f 0 ( π4 )
√
√ √
f ( π4 ) = 2 + tan( π4 ) − 2 csc( π4 ) = 2 + 1 − 2( 2) = 3 − 2 = 1. And since f 0 (x) = sec2 x +
√
√
√
√ √
2 csc x cot x. f 0 ( π4 ) = sec2 ( π4 ) + 2 csc( π4 ) cot( π4 ) = ( 2)2 + 2( 2) · 1 = 2 + 2 = 4. Hence,
x2 =
π−1
π 1
− =
.
4 4
4
5. Determine the dimensions of the rectangle of the largest area that can be inscribed in a right triangle
with base c units and height 2c units where c is an arbitrary positive constant. Justify that your
answer is a maximum. Use the variables’ names that are given in the picture below.
Solution: We want to maximize A = xy, the area of the inscribed rectangular. First of all,
we need to rewrite A in term of one, and only one, variable. To do this, we need to find a
(see picture
relationship between x and y. Using similar triangles, we obtain the ratio: yc = 2c−x
2c
c
1
1
below). And hence, y = 2c (2c − x) = 2 (2c − x) = c − 2 x.
So now we have A = x(c− 12 x) = cx− 21 x2 . To maximize this area, we need to compute A0 = c−x.
So, the critical point is x = c. Note that A00 = −1 < 0 for all x; so, A is always concave down.
And thus, we obtain the maximum value for A at the critical point x = c. In conclusion, the
dimensions of the inscribed rectangular that has the largest area are x = c and y = c − 12 c = 12 c.
x2 + 4
6. Sketch the graph of the function f (x) =
. Find the domain then indicate the x and y
x
intercepts, if relevant, and all horizontal, vertical and slant asymptotes. Also include local extrema
(i.e. local max and local min) and points of inflection along with the intervals where the function
is increasing, decreasing, concave up, and concave down.
Solution: We’ll determine some properties of the graph by algebraic methods so that we get a
sense of how to sketch it.
S
The domain of f is (−∞, 0) (0, ∞).
There are no x-intercepts. This is because the numerator of f (x) is always positive, and so the
quotient f (x) is never zero. There are no y-intercepts either since x = 0 is not in the domain of
f.
The vertical asymptote is x = 0. Next, we compute limx→0+ f (x) = +∞ since both numerator
and denominator are positive. Also, limx→0− = −∞ since x is negative but the numerator is
positive.
The power of x in the numerator is one more than that in the denominator, so this is the
situation in which we look for a slant asymptote. Performing long division,
x
x
x +4
− x2
2
we find that the slant asymptote is y = x.
To find critical points we consider the derivative.
f 0 (x) =
x2 − 4
2x2 − (x2 + 4)(1)
=
.
x2
x2
So, the critical numbers of f are x = ±2. Note that f 0 (x) is undefined at x = 0, but x = 0 is
not a critical number since 0 is not in the domain of f .
3
2
−4)
The second derivative is f 00 (x) = 2x −2x(x
= 8x
. And since x4 > 0 for all x, the graph is
x4
x4
concave down for x < 0 and concave up for x > 0.
The second derivative information then lets us classify x = 2 as a local minimum and x = −2
as a local maximum. Though f 00 (x) is undefined at x = 0, x = 0 is not an inflection point since
it is not in the domain of the function.
S
Finally, weS have f 0 (x) < 0 in the intervals (−2, 0) (0, 2) and f 0 (x) > 0 in the intervals
(−∞, −2) (2, ∞); and therefore, the function is decreasing and increasing on those intervals
respectively.
We can then sketch the graph (the dotted line is the slant asymptote).
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