Download MA 137 - Calculus I for the Life Sciences Spring 2012 TEST 3

MA 137 - Calculus I for the Life Sciences
TEST 3 Solutions
Spring 2012
10-Apr-2012
1. Find where the following functions are increasing and decreasing.
(a)
f (x) =
x2
,
1 + 2x3
1
x 6= − √
.
3
2
2x(1 + 2x3 ) − x2 (6x2 )
f (x) =
(1 + 2x3 )2
−2x(x3 − 1)
=
(1 + 2x3 )2
′
Critical points: x = 0 and x = 1
√
√
f (x) decreasing for −∞ < x < − 3 2, − 3 2 < x < 0 and x > 1
f (x) increasing for 0 < x < 1.
(b) g(x) = x3 − 3x2 − 9x + 4.
f ′ (x) = 3x2 − 6x − 9
Critical points: x = −1 and x = 3
f (x) decreasing for −1 < x < 3
f (x) increasing for −∞ < x < −1 and x > 3.
2. [Mean Value Theorem]
(a) Let f (x) = x2 −5x+1 on [0, 2]. Find a value, c, from the Mean Value Theorem
so that
f (b) − f (a)
f ′ (c) =
.
b−a
f ′ (x) = 2x − 5
f (2) = −5 and f (0) = 1
f (2) − f (0)
= −3
2−0
f ′ (c) = −3 ⇒ c = 1
10-Apr-2012
Test 3
Solutions
(b) If f (x) = |x|, then f (−1) = 1 and f (3) = 3 but f ′ (x) is never equal to
1
f (3) − f (−1)
= .
3 − (−1)
2
Why does this function not violate the Mean Value Theorem?
The function f (x) = |x| is not differentiable on the interval [−1, 3] so the
Mean Value Theorem does not apply.
3. Let
f (x) = 2x5 − 5x4 , x ∈ [−1, 4].
(a) Locate the critical points of f and determine the intervals on which f is increasing and the intervals on which f is decreasing.
f ′ (x) = 10x4 − 20x3
= 10x3 (x − 2)
So the critical points are x = 0 and x = 2. Now, use the first derivative test.
f ′ (x)
f (x)
−1 < x < 0 x = 0 0 < x < 2 x = 2
+
−
ր
ց
2<x<4
+
ր
So f (x) is increasing for x < 0 and for x > 2 and f (x) is decreasing for
0 < x < 2.
(b) Locate the possible inflection points for f and determine the intervals on
which f is concave up and the intervals on which f is concave down.
f ′′ (x) = 40x3 − 60x2 .
3
Possible inflection points are x = 0 and x = . We will check the second
2
derivative to determine which, if either, are inflection points.
f (x)
f (x)
′′
−1 < x < 0 x = 0 0 < x <
−
−
⌢
⌢
3
2
x=
3
2
3
2
<x<4
+
⌣
Since f does not change concavity at x = 0, the point (0, 0) is not an inflection
3
point. The other point, however, x = is an inflection point because f (x) is
2
3
3
concave down for x < and f (x) is concave up for x > .
2
2
2
Spring 2012
MATH 137 001-004
10-Apr-2012
Test 3
Solutions
(c) Evaluate f at the critical points and endpoints.
f (−1) = −7
f (0) = 0
f (2) = −16
f (4) = 768
(d) What are the global maximum values and global minimum values of f (x)?
The global maximum value is 768 which occurs at x = 4 and the global
minimum value is −16 which occurs at x = 2.
4. Find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum.
Let x and y be the two numbers. We know that
x + y = 9,
x ≥ 0 and y ≥ 0.
We are asked to maximize P = x2 y.
From our constraint equation, we have that y = 9 − x, so
P = x2 (9 − x) = 9x2 − x3
P ′ = 18x − 3x2
Set P ′ = 0.
18x − 3x2 = 0 or
x = 0 and x = 6.
There are three points to check: x = 0, x = 6 and x = 9, the first and last because
they are endpoints of the domain.
x
0
P (x) 0
6 9
108 0
So the global maximum value of this product occurs when x = 6 and y = 3.
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Spring 2012
MATH 137 001-004
10-Apr-2012
Test 3
Solutions
5. A rectangle has its base on the x-axis, its lower left corner at (0, 0), and its upper
2
right corner on the curve y = . What is the smallest perimeter the rectangle can
x
have?
y=
2
x
We are to minimize the perimeter of the rectangle whose sides are x and y. The
2
perimeter is given by P = 2x + 2y subject to the condition that y = . Therefore,
x
P = 2x + 2y
= 2x + 2f (x) = 2x +
P ′(x) = 2 −
4
x2
0=2−
4
x2
4
x
Set P ′ = 0.
√
√
Thus the critical points are x = ± 2. We can eliminate the critical point x = − 2
because x > 0. We can use the Second Derivative Test to check the other critical
point:
8
P ′′(x) = 3
x
√
4
′′
P ( 2) = √ > 0
2
√
√
Therefore, the minimum occurs at x = 2 and the minimum perimeter is 4 2.
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Spring 2012
MATH 137 001-004
10-Apr-2012
Test 3
Solutions
6. Find the terms P0 , P1 , P2 , P3 , P4 , P5 for each of the following recursions
(a) Pn+1 = 2Pn + 1, P0 = 1.
P0
1
P1
3
P0
P1
P2
P3
P4
P5
P2
7
P3
15
P4
31
P5
63
=1
= 2P0 + 1 = 3
= 2P1 + 1 = 7
= 2P2 + 1 = 15
= 2P3 + 1 = 31
= 2P4 + 1 = 63
1
(b) Pn+1 = Pn with P0 = 31, 250.
5
To go from one term to the next, divide by 5.
P0
P1
P2
P3 P4
31250 6250 1250 250 50
P5
10
(c) Pn = 2Pn−1 − Pn−2 + 3 where P0 = 1 and P1 = −1.
P0
1
P0
P1
P2
P3
P4
P5
5
P1
−1
P2
0
P3
4
P4
11
P5
21
=1
= −1
= 2P1 − P0 + 3 = 0
= 2P2 − P1 + 3 = 4
= 2P3 − P2 + 3 = 11
= 2P4 − P3 + 3 = 21
Spring 2012
MATH 137 001-004
10-Apr-2012
Test 3
Solutions
7. Find the limits of the following sequences:
2n
.
(a) an =
n+2
2n
lim
= 2.
n→∞ n + 2
(b) an =
n2
1
+1
lim
n→∞ n2
(c) an =
1
= 0.
+1
(−1)n
n3 + 5
(−1)n
(−1)
3
lim 3
= lim n
= 0.
5
n→∞ n + 5
n→∞
1+ 3
n
n
(d) an =
3n + 2−n
5n
3
3n + 2−n
= .
n→∞
5n
5
lim
8. Assume that lim an exists. Find all fixed points for each of the following sequences
n→∞
{an }.
1
(a) an+1 = an + 4
3
2
1
Solve a = a + 4, or a = 4, a = 6.
3
3
√
(b) an+1 = 3an − 2
√
Solve a = 3a − 2, or a2 = 3a − 2, a2 − 3a + 2 = 0. So a = 1 or a = 2.
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Spring 2012
MATH 137 001-004
10-Apr-2012
Test 3
Solutions
9. Use the stability criterion to characterize the stability of the equilibria of
an+1 =
5an 2
.
6 + an 2
First, find the fixed points by solving
5a2
6 + a2
3
6a + a = 5a2
a(a2 − 5a + 6) = 0
a(a − 2)(a − 3) = 0
a = 0, a = 2,
a=
a=3
Now, to determine the stability of the equilibrium points, we need to find the derivative at each of these points.
5x2
6 + x2
60x
f ′ (x) =
(6 + x2 )2
f ′ (0) = 0 < 1 so x = 0 is a stable equilibrium point.
f ′ (2) = 1.2 > 1 so x = 2 is an unstable equilibrium point.
f ′ (3) = 0.8 < 1 so x = 3 is a stable equilibrium point.
f (x) =
10. Suppose that a fish population develops according to the discrete logistic equation
That is,
Nt
.
Nt+1 = RNt 1 −
K
Assume that R = 3 and K = 3, 000.
(a) US Fish and Wildlife decides that, in order to keep the population healthy,
they must harvest a fixed number of fish per year. Call this number H. If
H = 440 write a recurrence relation that describes the long term population.
Since you are harvesting a constant 440 fish each period, the population can
be described by
Nt
Nt+1 = 3Nt 1 −
− 440.
3000
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Spring 2012
MATH 137 001-004
10-Apr-2012
Test 3
Solutions
(b) If H = 440 and N0 = 2400, what is the population each of the first four years?
N0
N1
N2
N3
N4
2400 1000 1560 1806 1716
BONUS: Find the equilibrium points for this population.
Find equilibrium points by solving
N
− 440
N = 3N 1 −
3000
N
N + 440 = 3N 1 −
3000
2
N
N + 440 = 3N −
1000
1000N + 440000 = 3000N − N 2
N 2 − 2000N + 440000 = 0
By the quadratic formula
√ N = 200 5 ± 14
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Spring 2012
{251.669, 1748.33}
MATH 137 001-004