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Game Math Chapter 2 Solutions
Game Math Chapter 2 Solutions
Frank Luna
www.gameinstitute.com
Monday, January 3, 2004
1. Plot the following graphs together:
( )
, | 0 ,
≤≤ = − + −
y x
0
21
x x
4
3
1
x
4
2
2
( )
, | 4 ,
≤≤ =−+ −
203
x x
2
}
( )
xy x y
≤≤
5.70711,
=
1.5
}
( )
xy x y
≤≤ =
3
( )
xy y x
≤≤ =
7
( )
xy y x
≤≤ =
8
( )
xy x y
≤≤ =
.5
}
( )
xy y x
≤≤ =
0
}
Frank Luna
Page 1
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x y
x
2
{
x y
x y
{
, | 4.29289
{
, | 6 ,
}
{
, | 0 ,
}
{
, | 0 ,
}
{
, | 8 ,
{
, | 0 ,
736239473.004.png
Game Math Chapter 2 Solutions
2. The following questions all refer to the graph of a certain function f , as shown.
a. Approximately what is ( )
f
5
?
f ?
c. Is the function f invertible? Why or why not?
d. If the function represented the temperature of an alien home world (the
horizontal axis representing months, and the vertical axis representing degrees
Celsius), about how many months would you say it takes for the alien planet
to orbit its sun?
0
2a. 40
2b. 10
f x f ≠ . This function is not one-to-one
since, from the graph, we can readily see that there exists a case where 1
x ≠ then ( ) ( )
2
1
2
x ≠ but
2
f x f = . Thus the function is not invertible.
2
2d. Observe the two troughs that dip past –20 C˚. These two troughs can be interpreted
as corresponding to winter, thus the distance between these two troughs indicates a year
(one complete orbit around the sun). It is approximately 6 months.
3. Create an exponential function that models the population of rabbits on an island.
Suppose the population begins at 10, and that after 12 months, there are 1000 rabbits.
Frank Luna
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b. Approximately what is ( )
2c. No. Recall that for a function to be invertible it must be one-to-one and onto . A
function is one-to-one if 1
() ()
1
736239473.005.png 736239473.006.png
Game Math Chapter 2 Solutions
f x = , where a and b are the unknown constants; you'll need the natural log
function to solve for b .)
bx
We want to fit a function of the form ( )
f x = through the points ( )
bx
0, 10 and
( )
Condition 1:
10
= ⇔=
0
b
a
10
Condition 2:
1000 10
= ⇔= ⇔ = ⇔=
e
12
b
100
e
12
b
ln 100 12
b b
ln 100 12
( )
Thus the following gives us the exponential function we seek:
fx x
=
10 exp
ln 100
( )
1
12
Now we want to know which x (month) gives us a population of ( ) 1000000
fx =
. To
find this, we solve the following equation for x :
1000000 10 exp
=
x
ln 100
( )
12
100000 exp
=
⇔
x
ln 100
( )
12
⇔
ln 100000
( )
=
x
ln 100
( )
12
⇔
x
=
12 ln 100000
( )
( )
=
30
ln 100
4. Suppose the rabbit population in the preceding problem can't grow exponentially
because of predators and limited resources. Try modeling the population using a log
1 Note that exp ( x ) is another way to express e . We use this alternate notation because it gives us more
room to write complex exponent expressions; using the exponent notation means we have to write very
small.
Frank Luna
Page 3
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How many months until the rabbit population exceeds one million? (Hint: Try using
()
12, 1000 .
ae
( )
()
736239473.007.png 736239473.001.png
Game Math Chapter 2 Solutions
function, making sure the new function satisfies the original constraints. (Hint: If you run
into difficulty solving for the constants, try graphing the log function to see what could be
causing the problem.)
= you will see that it approaches
infinity as x goes to zero. So with this behavior we will have trouble fitting the curve to
the data point ( )
f xax
ln
0, 10 . To get around this problem we will add an offset of 2 to the
natural logarithm argument: () ( )
fx a bx
=
ln
+ . Thus, as x approaches to 0, ()
2
f x
approaches ()
a
ln 2
, and we no longer have the problem of approaching infinity.
So we want to fit a function of the form ( ) ( )
fx a bx
=
ln
+ through the points
2
0, 10 and ( )
12, 1000 .
Condition 1: ( )
10
=
ab a
ln 0 2
+ ⇔= ≈
10
14.4269
ln 2
()
Condition 2:
1000
=
10
() ( )
ln 12 2
b
+
ln 2
⇔
100
=
ln 12 2
( )
()
b
+
ln 2
⇔
exp 100 ln 2
( ( )
=+
12 2
b
b
=
exp 100 ln 2
( ( )
⇔
−
2
≈
1.05637 10
×
29
12
Thus the following gives us the logarithmic function we seek:
fx
() (
=
14.4269 ln 1.05637 10
× +
29
x
2
)
We can test it to verify that it agrees at the points ( )
0, 10 and ( )
12, 1000 :
() (
f =
0 14.4269 ln 1.05637 10 0 2 10
× ⋅ + ≈
29
)
() (
)
f =
12 14.4269 ln 1.05637 10 12 2 1000
× ⋅ + ≈
29
Frank Luna
Page 4
www.gameinstitute.com
If you graph a log function such as ( ) ( )
( )
736239473.002.png
Game Math Chapter 2 Solutions
5. Say you're building a space-based resource management game and you want the cost
of research and development to increase exponentially. Suppose that having spent
nothing on research, you want it to cost 10 credits to advance to the next level of
development, but that after you have spent 10,000 credits, you want it to cost twice that
much to advance to the next level. Develop a function to model this behavior.
This problem is similar to problem 3. We want to fit a function of the form ()
f x =
bx
through the points ( )
0, 10 and (
10000, 20000 .
)
Condition 1: ()
10
=
a
exp 0
b a
⇔ =
10
Condition 2:
20000 10 exp 10000
=
( )
b
⇔
2000 exp 10000
=
( )
b
ln 2000 10000
( )
⇔
=
⇔
b
b
=
ln 2000
( )
≈
0.00076
10000
Thus the following gives us the exponential function we seek:
() ( )
=
10 exp 0.00076
x
We can test it to verify that it agrees at the points ( )
0, 10 and (
10000, 20000 :
)
() ( )
0 10 exp 0.00076 0 10
⋅ =
( ) (
10000 10 exp 0.00076 10000 20000
=
⋅ ≈
)
6. Prove that you cannot force an exponential function to pass through just any three
points. (Hint: Find three points that no exponential function can pass through.)
Intuitively, it is not hard to see this. Consider three points on an arbitrary line. If we
superimpose an arbitrary exponential function we see that at most the curves can intersect
at two points (see figure). Thus no exponential function can pass through three points on
Frank Luna
Page 5
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f x
f =
f
736239473.003.png

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