MATH 150
Quiz Key #5
10/13-15/2015
(1) Find the domain of f tion.
( x ) = x 2
√
2 − x
− x − 12
. Give your answer in interval nota-
Solution: We need to determine which values of x make the denominator equal to 0 and which values make the radicand negative, to determine which x -values are not part of the domain. First, we find where the denominator is 0: x
2 − x − 12 = 0 = ⇒ ( x − 4 ) ( x + 3 ) = 0 = ⇒ x = 4 or x = − 3.
Therefore, the domain excludes 4 and -3. Now we need 2 x
− x ≥ 0 = ⇒ 2 ≥
= ⇒ x ≤ 2. Therefore, the domain of f ( x ) is ( − ∞
, − 3 ) ∪ ( − 3, 2 ] .
(2) Write the formula for the difference quotient and use it to find the difference quotient for f ( x ) = − 3 x 2 + 4 x − 2.
Solution:
Formula: f ( x + h ) − f ( x )
.
h
Therefore, to find the difference quotient, we need to find f
Simplified Difference Quotient: We know f ( x ) already.
( x + h ) and f ( x ) .
f ( x + h ) = − 3 ( x + h )
2
+ 4 ( x + h ) − 2.
Thus, f ( x + h ) = − 3 x
2
+ 2 xh + h
2
+ 4 x + 4 h − 2 = − 3 x
2 − 6 xh − 3 h
2
+ 4 x + 4 h − 2.
Therefore, the difference quotient for f ( x ) simplifies to
− 3 x
2 − 6 xh − 3 h
2
+ 4 x + 4 h − 2 − − 3 x
2
+ 4 x − 2 h
=
− 6 xh − 3 h
2
+ 4 h h
= h ( − 6 x − 3 h + 4 ) h
= − 6 x − 3 h + 4.
(3) f ( x
The function
) = x g ( x ) = −
1
3
( x − 4 )
3
− 5 is a transformation of the function
3
. Describe the transformations, in order, to get g ( x ) from f ( x ) .
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Solution: (1) Shift right 4 units.
(2) Shrink vertically by a factor of 3.
(3) Reflect around x -axis.
(4) Shift down 5.
[Any other ordering works as long as the vertical shrink and the reflection happen before the vertical shift.]
(4) Together Ryan and Natalie can sow 5 pounds of wildflower seeds in 7 hours. It would take Natalie by herself one hour longer than it would take
Ryan by himself to sow the seeds. How long would it take each of them individually to sow 5 pounds of wildflower seeds?
Solution: Think of this as a distance-rate-time problem. The “distance” is 5 pounds of wildflower seeds. If we let R be the amount of time it will take Ryan individually, then the amount of time it will take Natalie individually is R
Then we get the table below:
+ 1.
Dist rate time
Both 5 7
Natalie
Ryan
5
5
R + 1
R
Now, using the equation D
Ryan as follows:
Dist rate
= time r · t , we can fill in the rates for Natalie and
Both
Natalie
Ryan
5
5
5
5
R
+
5
R + 1
5
R + 1
5
R
R
7
+
R
1
Now, we can use the top row to write an equation for R , which we then solve:
5 5
5 = 7 + .
R R + 1
5 = 7 ·
5 ( R + 1 ) + 5 R
.
R ( R + 1 )
5 R ( R + 1 ) = 7 ( 5 R + 5 + 5 R ) .
5 R
2
+ 5 R = 70 R + 35.
5 R
2 − 65 R − 35 = 0.
R
2 − 13 R − 7 = 0.
Now using the quadratic formula:
R =
13 ±
√
169
2
But R has to be positive, so Ryan takes hours.
+ 28
=
13 ±
2
√
197
.
13 +
√
197
2 hours and Natalie takes
15 +
√
197
2
2

(5) A drone travels at a constant speed with a 10 mile per hour tailwind for 30 minutes and then returns against the wind on the same path in 36 minutes.
What is the constant speed, in miles per hour, of the drone still in the air?
Solution: We let x be the constant speed of the drone in the air. 30 minutes is half an hour and 36 minutes is as follows:
Dist
36
60
=
6
10
=
3
5 rate time of an hour. Then we set up a table
With wind
Against wind
3
5
1
2
(
( x x
+
−
10
10
)
) x x
+
−
10
10
3
5
1
2
Now, the distances are the same so we can set up the following equation:
1
2
( x + 10 ) =
3
5
( x − 10 ) .
To get rid of the denominators, we multiply by the least common denominator:
10.
10 ·
1
2
( x + 10 ) = 10 ·
3
5
( x − 10 ) .
5 ( x + 10 ) = 6 ( x − 10 ) .
5 x + 50 = 6 x − 60.
x = 110.
Therefore, the drone’s speed in the air is 110 miles per hour.
(6) Using the graph of f
2 f ( x + 3 ) +
( x ) [refer to quiz paper], graph the function
1. Describe the transformations used to get g ( x ) from f ( g x )
( x
.
) =
Solution: The transformations are (1) shift left 3, (2) stretch vertically by a factor of 2, and then (3) shift up 1.
The graph of f is a piecewise function consisting of three line segments: a line segment joining ( − 6, − 4 ) to ( − 2, 0 ) ; one joining ( − 2, 0 ) to ( 2, 0 ) ; and one joining ( 2, 0 ) to ( 6, 4 ) . Therefore, if we determine what happens to each of those points we will be able to sketch the graph of g
(1)
( x ) .
( − 6, − 4 ) 7→ ( − 9, − 4 ) .
(2) ( − 9, − 4 ) 7→ ( − 9, − 8 ) .
(3) ( − 9, − 8 ) 7→ ( − 9, − 7 ) .
(1)
(1)
(1)
( − 2, 0 ) 7→ ( − 5, 0 ) .
(2) ( − 5, 0 ) 7→ ( − 5, 0 ) .
(3) ( − 5, 0 ) 7→ ( − 5, 1 ) .
( 2, 0 ) 7→ ( − 1, 0 ) .
(2) ( − 1, 0 ) 7→ ( − 1, 0 ) .
(3) ( − 1, 0 ) 7→ ( − 1, 1 ) .
( 6, 4 ) 7→ ( 3, 4 ) .
(2) ( 3, 6 ) 7→ ( 3, 8 ) .
(3) ( 3, 8 ) 7→ ( 3, 9 ) .
Therefore, the graph of g is a piecewise function consisting of three line segments: a line segment joining ( − 9, − 7 ) to ( − 5, 1 ) ; one joining ( − 5, 1 ) to
( − 1, 1 ) ; and one joining ( − 1, 1 ) to ( 3, 9 ) .
(7)
(a) Is f ( x ) in problem (6) even, odd, or neither?
(b) On what intervals is the function f ( x ) in problem (6) increasing?
3

Solution:
(a) f ( x ) is symmetric about the origin, so it is odd.
(b) [ − 6, − 2 ] ∪ [ 2, 6 ] .
4