8/29 | 8/31 | 9/1 | 10/2 | 10/3 | 10/5 | 10/6 | ||
9/5 | 9/7 | 9/8 | 10/9 | 10/10 | 10/12 | 10/13 | ||
9/11 | 9/12 | 9/14 | 9/15 | 10/16 | 10/17 | 10/19 | 10/20 | |
9/18 | 9/19 | 9/21 | 9/22 | 10/23 | 10/24 | 10/26 | 10/27 | |
9/25 | 9/26 | 9/28 | 9/29 | 10/30 | 10/31 | 11/1 | 11/2 |
x | f(x)= 5x^{2 } +3 |
3 | 48 |
2 | 23 |
1 | 8 |
0 | 3 |
-1 | 8 |
-2 | 23 |
-3 | 48 |
and a transformation (mapping) figure:
Based on this we concluded that the solution was {x | -2<x<2}
= (-2,2).
After the analysis of the transformation figure we considered when x+2<0.... x< -2. We graphed this on a number line.
----------------0 +++++++++++++++ (x+2)
-2 0
Then we considered
f(x)=
5x^{2 }+ 3 < 23; so 5x^{2 } - 20 < 0;
5(x+2)(x-2) < 0. We then scanned the line with information about the
factors (x+2) and (x-2) to investigate when the expression
5x^{2 } - 20 would be positive, negative, or 0.
-------------------------0+++++++++ (x-2)
----------------0 +++++++++++++++ (x+2)
-2 0 2
+++++++++++0--------0+++++++++ 5(x+2)(x-2)
Thus -5x^{2 } - 20 < 0 when 2<x<2... or x is in the interval (-2,2).
x | f(x)= 5x^{2 } +3 |
3 | 48 |
2 | 23 |
1 | 8 |
0 | 3 |
-1 | 8 |
-2 | 23 |
-3 | 48 |
With this definition we have
g(0)=5, g(1)=6, g(-1)=6,g(2)=9, g(1/2)=21/4, g(t)=t^{2}+5,
g(1+t)=(1+t)^{2}+5, etc.
x | f(x) |
2 | 5 |
1 | 3 |
0 | -2 |
-1 | 4 |
-2 | 0 |
x | L | A |
10 | 120 | 1200 |
20 | 100 | 2000 |
70 | 0 | 0 |
60 | 20 | 1200 |
50 | 40 | 2000 |
In fact, A = -2x^{2}+140x but A also can be expressed
by A= -2(x-35)^{2} +2450. It should be clear from thinking about
this expression for A that A is at most 2450, since for any x other than
35, A will be the result of subtracting a positive number from 2450.
Example: Average cost= C(x)/x