Due: 03/03/2016 in class
Instructor: Janina Letz
MATH 1210, Spring 2016
Lab #7
Name:
Question:
1
2
3
4
Total
Points:
10
16
24
15
65
Score:
Instructions: Please show all of your work as partial credit will be given where appropriate, and there may
be no credit given for problems where there is no work shown. All answers should be completely simplified,
unless otherwise stated. No calculators or electronics of any kind are allowed.
1. (10 points) (Critical points) Find all critical points of
(
(x − 4)(x + 1)3
F (x) =
0,
x not an integer
x an integer
in the interval [−5, 5] and state why these points are critical.
Solution:
(
0
F (x) =
(x + 1)3 + (x − 4)3(x + 1)2 = 4(x + 1)2 (x −
not defined
11
4 )
x not an integer and not − 1, 4
otherwise
point
-4,-3,-2,0,1,2,3
11
4 ,-1
-5,5
1
reason
the derivative is not defined
F 0 is 0
endpoints
MATH 1210, Spring 2016
Lab #7
2. (Concavity graph-based) Sketch a graph with the following properties.
(a) (8 points) i. concave up on (0, 4)
ii. concave down on (−3, 0)
iii. differentiable on (−5, 5)
iv. increasing on (0, 5)
Solution:
(b) (8 points) i. continuous on (−5, 5)
ii. maximum at 2
iii. conave up on (−1, 5)
iv. concave down on (−5, −1)
Solution: Such a graph does not exist, since for a maximum it needs to be f 00 (2) < 0, but it
also should be f 00 (x) > 0 on (−1, 5).
2
Due: 03/03/2016 in class
Instructor: Janina Letz
MATH 1210, Spring 2016
Lab #7
Name:
3. (Inflection points and Concavity) Find the inflection points and intervals of concavity for the following:
(a) (6 points) f (x) = x3 − 56x2 + 3x − 10
Solution:
f 0 (x) = 3x2 − 56x + 3
f 00 (x) = 6x − 112
56
56
The only root of f 00 is 112
6 = 3 , so this is an inflection point and f is on (−∞, 3 ) concave
down and on ( 56
,
∞)
concave
up.
3
(b) (6 points) f (x) = (x + 1)4
Solution:
f 0 (x) = 4(x + 1)3
f 00 (x) = 12(x + 1)2
f 00 has a double root at -1, so this is not an inflection point and f is on (−∞, −1) and (−1, ∞)
concave up.
(c) (6 points) f (x) =
√
3
4x2 − 12x
Solution:
f 0 (x) =
f 00 (x) =
8
3(4x2 − 12x)
2
3
−
8x − 12
2
3(4x2 − 12x) 3
2(8x − 12)2
9(4x2 − 12x)
5
3
=
24(4x2 − 12x) − 8(16x2 − 48x + 36)
5
9(4x2 − 12x) 3
So the critical values are only for
0 = 8(12x2 − 36x − 16x2 + 48x − 36) = 8(−4x2 + 12x − 36) = −32(x2 − 3x + 9)
and
0 = 4x2 − 12x4x(x − 3)
The first condition gives no points, while the second gives 0 and 3. These are inflection point
and f is on (−∞, 0) and (3, ∞) concave down and on (0, 3) concave up.
(d) (6 points) f (x) = x − sin(x)
Solution:
f 0 (x) = 1 − cos(x)
f 00 (x) = sin(x)
So f has inflection points at kπ for k ∈ Z and f is on (2kπ, (2k + 1)π) concave up and on
((2k + 1)π, 2kπ) concave down for k ∈ Z.
3
MATH 1210, Spring 2016
Lab #7
4. Given two increasing functions f, g and a decreasing function h. Prove or disprove the following
properties. (To disprove a property give a counterexample)
(a) (3 points) F (x) = f (x) − g(x) is an increasing function
Solution: The statement is false.
Choose f (x) = x, g(x) = 2x. These are increasing functions, but F (x) = f (x) − g(x) = −x is
decreasing.
(b) (3 points) F (x) = f (g(x)) is an increasing function
Solution: The statement is true. There are two ways to prove it.
1. Take x0 < x1 . Then x̂0 = g(x0 ) < g(x1 ) = x̂1 and f (x̂0 ) < f (x̂1 ).
2.
F 0 (x) = f 0 (g(x)) g 0 (x) > 0
| {z } | {z }
>0
>0
(c) (3 points) F (x) = f (x) − h(x) is an increasing function
Solution: The statement is true. There are two ways to prove it.
1. Take x0 < x1 . Then f (x0 ) < f (x1 ) and h(x0 ) > h(x1 ), so −h(x1 ) > −h(x0 ). This gives
f (x0 ) − h(x0 ) < f (x1 ) − h(x1 ).
2.
F 0 (x) = f 0 (x) − h0 (x) > 0
| {z } | {z }
>0
<0
(d) (3 points) F (x) = −f (h(x)) is an increasing function
Solution: The statement is true. There are two possible ways to prove it.
1. Take x0 < x1 . Then h(x0 ) > h(x1 ) and f (h(x0 )) > f (h(x1 )). This gives
−f (h(x0 )) < −f (h(x1 )).
2.
F 0 (x) = − f 0 (h(x)) h0 (x) > 0
| {z } | {z }
>0
<0
(e) (3 points) F (x) = f (x) · h(x) is an increasing function
Solution: The satement is false.
Choose f (x) = x and h(x) = −x. Then F (x) = −x2 . This is neither increasing nor decreasing.
4