APPM 1340
Exam 1
Fall 2016
INSTRUCTIONS: Books, notes, and electronic devices are not permitted. This exam is worth 100 points.
Box your final answers. Write neatly, top to bottom, left to right, one problem per page. A correct answer
with incorrect or no supporting work may receive no credit. If you need to find a derivative then you must find
it via definition. SHOW ALL WORK
1. (20 points) Factor and reduce the following expressions. Write any trig expressions in terms of sine:


 2

✓ cos2 (✓) ✓
x(2x2 8)
x +x 2
8x + 10
(a)
(b)
(c)
(d)
2
3
2
2
✓
x
2x + 100x 200
x
4x + 3
4x2 + x 5
Solution:

✓ cos2 (✓)
(a)
✓2
(b)

(c)

(d)

x3
✓
=
✓(cos2 ✓
✓2
x(2x2 8)
2x2 + 100x
x2 + x 2
(x
=
x2 4x + 3
(x
200
=
sin2 ✓
✓
2x(x2 4)
x2 (x 2) + 100(x
=
1)
=
=
cos2 ✓
✓
1)(x + 2)
x+2
=
1)(x 3)
x 3
8x + 10
2(4x + 5)
2
=
=
4x2 + x 5
(4x + 5)(x 1)
x 1
1
2)
2x(x 2)(x + 2)
2x(x + 2)
= 2
2
(x 2)(x + 100)
x + 100
2. (24 points) Sketch a graph of the following relationships:
8
<
g(x)
x2 + 6x
x 2
(a) f (x) =
:
1
8
,x > 3
,x < 3
,x = 3
(b) g(x) =
g(x)
(c) y =
p
4
8
>
>
>
>
<
>
>
>
>
:
(d) y = x2
x2
(x + 2)
|x + 2|
,x < 0
(x 2)
|x 2|
,x > 0
4x + 7
Solution:
(a)
(b)
4
2
2
4
2
◦
2
4
4
2
1◦
2
◦
2
◦
1◦
4
◦
4
2
g(x)
(c) semi-circle
(d) parabola
2
12
10
1
8
2
1
1
1
2
6
4
2
2
1
2
3
4
3. (10 points) The following questions are not necessarily related:
p
(a)Consider the graph of the equation: y = x.
What is the equation of the secant line to this graph thru x = 1 and x = 4?
p
4x + 1 3
(b) Rationalize the numerator of the expression
x 2
Solution:
(a) x-values 1 and 4 produce points (1, 1) and (4, 2).
These lead to slope: m = 24 11 = 13
1
Point-slope form: y 1 = (x 1)
3
1
2
Slope-Intercept form: y = x +
3
3
Standard form: x 3y = 2
(b)
(x
p
4x + 1
x 2
3
p
4x + 1 + 3
p
=
4x + 1 + 3
(x
4x + 1 9
⇥p
⇤=
2) 4x + 1 + 3
(x
4(x 2)
4
⇥p
⇤= p
2) 4x + 1 + 3
4x + 1 + 3
2)
4x
⇥p
8
⇤=
4x + 1 + 3
4. (6 points; 2,4)
(a) What is the smallest value y = sin(2x) can become? Name an x-value that produces this minimum.
x3 + 8x + 10
(b) Consider the expression y =
.
x 1
Which x-values, greater than zero, produce positive y-values?
Solution: (a) sin(✓) has a maximum of 1 and a minimum of 1
3⇡
3⇡
3⇡
This minimum occurs when ✓ =
. Therefore, 2x =
=) x =
2
2
4
(b) For x-values greater than zero, y has a numerator that is always positive. Therefore, y will only
be negative when the denominator is negative, i.e. when x 1 < 0 =) x < 1
5. (7 points) Given f (x) =
Solution:
f (1 + h)
h
2
f (1 + h)
, find and simplify
x
h
f (1)
=
2
1+h
h
2
1
=
2
f (1)
2(1 + h)
2h
2
=
=
h(1 + h)
h(1 + h)
1+h
6. (8 points) Person A and person B start a race at the same time. Person A finishes the race with an
average speed of 3 m.p.h. while person B finishes with an average speed of 8 m.p.h. Person B arrives at
the finish line 5 hours before person A. What distance was this race?
Solution: Distance equals rate times time: D = rt. So we have, 3(t + 5) = 8t =) 3t + 15 = 8t =)
5t = 15 =) t = 3.
Now that we have t, we can say that D = 8(3) = 24 miles
MORE ON THE BACK
7. (21 points) State whether the following are true or false AND explain (i.e. show work) why.

x2
2 2 cos(x) x2
1 cos(x)
1 x2
(a) If
<
,
then
>
24
2x2
x2
2 24
(b) If sin(3x)
(c)
1, then
5x2 sin(3x)
5x2 + 1
 2
2
x + 10
x + 10
tan(6t)
cos(4t) + 2 cos2 (2t)
=
sin(2t)
cos(6t)
Solution: (a) TRUE
x2
2 2 cos(x) x2
<
=)
24
2x2
x2
2(1 cos(x)) x2
<
=)
24
2x2
x2
2(1 cos(x))
x2
<
=)
24
2x2
2x2
x2
1 cos(x) 1
<
=)
24
x2
2
1 x2
1 cos(x)
<
2 24
x2
(b) TRUE
sin(3x)
1 =)
sin(3x)  1 =)
5x2
sin(3x)  5x2 + 1 =)
5x2 sin(3x)
5x2 + 1

x2 + 10
x2 + 10
(c) TRUE
tan(6t)
sin(6t)
sin(2t) cos(4t) + sin(4t) cos(2t)
=
=
=
sin(2t)
cos(6t) sin(2t)
cos(6t) sin(2t)
cos(4t) cos(2t) [2 sin(2t) cos(2t)]
cos(4t) + 2 cos2 (2t)
+
=
cos(6t)
cos(6t) sin(2t)
cos(6t)
8. (4 points) Name the zero’s of the following functional relationships. No explanation required.
4
20
3
f (x)
2
10
1
4
2
g(x)
15
5
2
4
3
2
1
1
2
3
Solution: The ”zero’s of a function” refers to the x-values that result in 0 as an output for the functional
relationship. Both graphs equal zero at x = ±2