7. [8 points] Find all solutions to
x
+

x
+ 2

+

x

3

= 7.
Solution:
We can break the problem into three cases.
Case 1:
If
x
≥
3, then

x

3

=
x

3 and

x
+ 2

=
x
+ 2, so the equation becomes
x
+
x
+ 2 +
x

3 = 7. Solving this gives us
x
= 8
/
3. But 8
/
3
6≥
3, so this solution
is extraneous.
Page 3
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Case 2:
If

2
≤
x
≤
3, then

x

3

=

x
+ 3 but

x
+ 2

=
x
+ 2, so the equation
becomes
x
+
x
+ 2

x
+ 3 = 7. Solving this gives us
x
= 2. Since

2
≤
2
≤
3, this
solution is valid.
Case 3:
If
x
≤ 
2, then

x

3

=

x
+ 3 and

x
+ 2

=

x

2, so the equation
becomes
x

x

2

x
+ 3 = 7. Solving this gives us
x
=

6. Since

6
≤ 
2, this
solution is also valid.
So the only solutions are
x
= 2
,

6.
8. A young alien on a distant planet throws a ball into the air. It reaches its maximum
height of 30ft after 4 seconds. Assume that the ball is at 2ft when it is released at time
t
= 0.
(a) [6 points] Find a formula for the height of the ball as a function of time. (
Hint
:
The graph of your function will be a parabola.)
Solution:
We know that the formula for the height
h
at time
t
will be a
quadratic, so let
h
(
t
) =
at
2
+
bt
+
c
. The problem gives us enough informa
tion to solve for
a
,
b
and
c
. First, we know that
h
(0) = 2, so
a
·
0
2
+
b
·
0+
c
= 2.
Therefore,
c
= 2. Second, we are told that
h
(4) = 30, so
a
·
4
2
+
b
·
4 + 2 = 30.
Simplifying, 4
a
+
b
= 7. Finally, we are told that
h
has its maximum value at
t
= 4, meaning that the parabola
y
=
h
(
t
) =
at
2
+
bt
+ 2 has its vertex there.
Completing the square,
h
(
t
) =
a
±
t
2
+
b
a
t
²
+ 2 =
a
"
±
t
+
b
2
a
²
2

±
b
2
a
²
2
#
+ 2
.
Therefore, the vertex is at
t
=

b
2
a
, so 4 =

b
2
a
. Simplifying,
b
=

8
a
. (Note
that we could have gotten this equation a little easier by either (1) using the
fact that
h
0
(4) = 0 because
h
has a local maximum at
t
= 4 or (2) that by
symmetry, after 8 seconds the ball will be at height 2ft again, so
h
(8) = 2.)
We have found that 4
a
+
b
= 7 and
b
=

8
a
. Solving for
a
and
b
, we have
a
=

7
/
4 and
b
= 14. Thus
h
(
t
) =

7
4
t
2
+ 14
t
+ 2
.
(b) [2 points] How fast was the ball thrown?
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 GOMEZ,JONES
 Math, Calculus, Algebra, Trigonometry, Sin, Quadratic equation, Mathematics in medieval Islam

Click to edit the document details