Math 234
February 21
I. Find the intervals of increase and decrease for the given function. Also
determine the critical numbers of the given function and classify each critical
point as a relative maximum, a relative minimum, or neither
1. f (x) = x2
Answer:
f 0 (x) = 2x and f 0 (x) = 0 when x = 0
_
+
f’
0
f
Min
Increasing: (0, ∞)
Decreasing: (−∞, 0)
Critical numbers: x = 0 (Minimum)
2. f (x) = x3
Answer:
f 0 (x) = 3x2 and f 0 (x) = 0 when x = 0
+
+
f’
0
f
Increasing: (−∞, 0) ∪ (0, ∞)
Critical numbers: x = 0 (Neither)
3. f (x) = x4 + 1
Answer:
f 0 (x) = 4x3 and f 0 (x) = 0 when x = 0
_
+
f’
0
f
Min
1
Increasing: (0, ∞)
Decreasing: (−∞, 0)
Critical numbers: x = 0 (Minimum)
4. f (x) = x4 + −4x
Answer:
f 0 (x) = 4x3 − 4 = 4(x3 − 1) and f 0 (x) = 0 when x = 1
_
+
f’
1
f
Increasing: (1, ∞)
Decreasing: (−∞, 1)
Critical numbers: x = 1 (Minimum)
5. f (x) = x3 /3 − 9x + 2
Answer:
f 0 (x) = x2 − 9 = (x + 3)(x − 3) and f 0 (x) = 0 when x = 3, −3
f’
_
+
−3
+
3
f
Max
Min
Increasing: (−∞, −3) ∪ (3, ∞)
Decreasing: (−3, 3)
Critical numbers: x = 3 (Minimum) x = −3 (Maximum)
6. f (x) = x5 − 5x4 + 200
Answer:
f 0 (x) = 5x4 − 20x3 = 5x3 (x − 4) and f 0 (x) = 0 when x = 0, 4
2
f’
_
+
+
0
4
Max
Min
f
Increasing: (−∞, 0) ∪ (4, ∞)
Decreasing: (0, 4)
Critical numbers: x = 0 (Maximum) x = 4 (Minimum)
7. f (t) = 1/(t2 + 1) − 1/(t2 + 1)2
Answer:
f 0 (t) = −2t(t2 + 1)−2 + 6t(t2 + 1)−3 = (t2 + 1)−2 (6t(t2 + 1) − 2t) =
(t2 + 1)−2 (6t3 + 4t) = (t2 + 1)−2 2t(3t2 + 2) and f 0 (t) = 0 when t = 0
_
+
f’
0
f
Min
Increasing: (0, ∞)
Decreasing: (−∞, 0)
Critical numbers: t = 0 (Maximum)
√
8. f (x) = 9 − x2
Answer:
√
f 0 (x) = (−2x)/(2 9 − x2 ) and f 0 (x) = 0 when x = 0 and f 0 (x) is
undefined when x = −3, 3
(Note that f (x) is not defined for x > 3 or x < −3.)
_
+
f’
−3
0
3
f
Max
Increasing: (−3, 0)
Decreasing: (0, 3)
Critical numbers: x = 0 (Maximum) x = −3, 3 (Neither)
9. f (x) = x + 9/x
3
Answer:
f 0 (x) = 1 − 9/x2 and f 0 (x) = 0 when x = −3, 3 and f 0 (x) is undefined
when x = 0
_
+
f’
_
+
0
−3
3
f
Neither
Min
Max
Increasing: (−3, 0) ∪ (0, 3)
Decreasing: (−∞, −3) ∪ (3, ∞)
Critical numbers: x = 3 (Maximum) x = −3 (Minimum) x = 0 (Neither)
10. f (x) = x/(x + 3)2
2 −2x(x+3)
2
2 −6x
2 +9
= x +6x+9−2x
= −x
and f 0 (x) = 0 when
f 0 (x) = (x+3)(x+3)
4
(x+3)4
(x+3)4
x = −3, 3 and f 0 (x) is undefined when x = −3
(Note that f (x) is undefinded for x = −3. )
_
_
+
f’
−3
3
f
Max
Neither
Increasing: (−3, 3)
Decreasing: (−∞, −3) ∪ (3, ∞)
Critical numbers: x = 3 (Maximum) x = −3 (Neither)
11. f (x) = x2 − 1/x2
4
f 0 (x) = 2x − 2x−3 = 2 x x−1
and f 0 (x) = 0 when x = 1, −1 and f 0 (x) is
3
undefined when x = 0
(Note that f (x) is undefined at x = 0.)
_
_
+
f’
−1
0
Min
Neither
+
1
f
Min
4
Increasing: (−1, 0) ∪ (1, ∞)
Decreasing: (−∞, −1) ∪ (0, 1)
Critical numbers: x = 1 (Minimum) x = −1 (Minimum) x = 0 (Neither)
12. f (x) = 3x4 − 8x3 + 6x2 + 2
f 0 (x) = 12x3 − 24x2 + 12x = 12x(x2 − 2x + 1) = 12x(x − 1)2 and
f 0 (x) = 0 when x = 1, 0
_
+
f’
+
0
1
Min
Neither
f
Increasing: (0, 1) ∪ (1, ∞)
Decreasing: (−∞, 0)
Critical numbers: x = 0 (Minimum) x = 1 (Neither)
√
13. f (x) = x√ 9 − x
√
f 0 (x) = 9 − x + 2√−x
= 18−2x−x
=
9−x
2 9−x
0
and f (x) is undefined when x = 9
(Note f (x) is undefined for x > 9 )
f’
and f 0 (x) = 0 when x = 6
_
+
6
f
18−3x
√
2 9−x
Max
9
Neither
Increasing: (−∞, 6)
Decreasing: (6, 9)
Critical numbers: x = 6 (Maximum) x = 9 (Neither)
14. f (x) = x2 /(x2 + x − 2)
2
2 (2x+1)
3
2 −4x−2x3 −x2
x(x−4)
x2 −4x
f 0 (x) = 2x(x +x−2)−x
= 2x +2x
= ((x−1)(x+2))
2 = (x−1)2 (x+2)2
(x2 +x−2)2
(x2 +x−2)2
and f 0 (x) = 0 when x = 0, 4 and f 0 (x) is undefined when x = 1, −2
(Note f (x) is undefined for x = 1, −2 )
5
f’
+
_
+
−2
0
_
+
1
4
f
Neither Max Neither Min
Increasing: (−∞, −2) ∪ (−2, 0) ∪ (4, ∞)
Decreasing: (0, 1) ∪ (1, 4)
Critical numbers: x = 0 (Maximum) x = 4 (Minimum) x = −2, 1
(Neither)
15. f (x) = (x2 − 1)4
f 0 (x) = 4(x2 − 1)3 2x = 8x(x2 − 1)3 and f 0 (x) = 0 when x = −1, 0, 1
_
_
+
f’
−1
0
Min
Max
+
1
f
Min
Increasing: (−1, 0) ∪ (1, ∞)
Decreasing: (−∞, −1) ∪ (0, −1)
Critical numbers: x = −1, 1 (Minimum) x = 0 (Maximum)
II. To produce x units of a particular commodity, a monopolist has a total
cost of
C(x) = x2 + 3x + 17
and total revenue of R(x) = xp(x), where p(x) = 5 − 2x is the price at which
x units will be sold. Find the profit function and sketch the graph. For what
level of production is profit maximized ?
Answer:
First find the function for Profit, which is revenue minus cost.
6
P (x) = R(x) − C(x)
= xp(x) − C(x)
= x(5 − 2x) − (x2 + 3x + 17)
= 2x − 3x2 − 17
The graph looks like:
-17
-18
-19
-20
-21
-1.0
0
0.5
-0.5
1.0
0
P (x) = −6x + 2 and P (x) = 0 when x = 1/3. Also P 0 (x) > 0 when x < 1/3
and P 0 (x) < 0 when x > 1/3.
Therefore x = 1/3 is a critical point and a relative maximum. Therefore the
profit is maximized when x = 1/3.
III. The concentration of a drug t hours after being injected into the arm
of a patient is given by
0.15t
C(t) = 2
t + 0.81
Sketch the graph of the concentration function. When does the maximum
concentration occur?
Answer:
7
0.05
-2
1
-1
2
-0.05
C(t) =
t2
0.15t
+ 0.81
Therefore :
C 0 (t) =
=
0.15(t2 +0.81)−0.15t(2t)
(t2 +0.81)2
−0.15t2 +0.81(0.15)
(t2 +0.81)2
Thus C 0 (x) = 0 when
−0.15t2 +0.81(0.15)
(t2 +0.81)2
2
−0.15t + 0.81(0.15)
0.81(0.15)
0.81
t
=0
=0
= 0.15t2
= t2
= 0.9, −0.9
The problem is only defined for t > 0. For 0 < t < 0.9, C 0 (t) > 0 and for
t > 0.9, C 0 (t) < 0. Therefore there is a maximum at t = 0.9. The maximum
concentration will occur 54 mins after the injection.
IV. A company determines that if x thousands are spent on advertising
a certain product, then S(x) units of the product will be sold where
S(x) =
200x + 1500
0.02x2 + 5
a) Sketch the graph of S(x).
b) How many units will be sold if nothing is spent on advertising?
8
c) How much should spent on advertising to maximize sales? What is the
maximum sales level?
Answer:
a)
500
400
300
200
100
-10
-5
5
10
15
20
b)
There is no advertising when x = 0 and S(0) = 1500/5 = 300. There will be
300 units sold if nothing is spent on advertising.
c)
First we must find S 0 (x)
S 0 (x) =
=
=
=
=
200(0.02x2 +5)−(200x+1500)(0.04x)
(0.02x2 +5)2
4x2 +1000−4x2 −60x
(0.02x2 +5)2
−4x2 +−60x+1000
(0.02x2 +5)2
−4(x2 +15x−250)
(0.02x2 +5)2
−4(x+25)(x−10)
(0.02x2 +5)2
Thus S 0 (x) = 0 when x = −25, 10. When x < −25, then S 0 (x) < 0, and
when −25 < x < 10, then S 0 (x) > 0 and when x > 10 then S 0 (x) < 0
There there is a minimum at x = −25 and there is a maximum at x = 10.
S(10) = 200(10)+1500
= 3500
= 500.
0.02(10)2 +5
7
They should spend $10,000 on advertising which will result in 500 units being
sold.
9