Break-even Point

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 3, Section 3.1

[Eigenvalue]

Note for 43 and 44: I used combinatorics to find the total possible number of function operations. There are 120--requiring different functions for the two functions--and 150--given that functions can repeat--possible binary function operations. I was only able to compute 13, however, I will have the other functions finished by tomorrow; it takes a while for function operations such as ([tex]f^{-1}[/tex] o g')(x). As with last time, if I miss a function operation or count one twice, please let me know.

Here is question 43:

[Eigenvalue]

Question 43, continued:

[Eigenvalue]

Question 44:

1. R(x)=C(x)
12x=10x+15000
2x=15000
x=7500
(7500,90000)
The break even point is $90000

2. R(x)>C(x)
12x>10x+15000
2x>15000
x>7500
The company must sell at least 7500 items

3. R(x)<C(x)
12x<10x+15000
2x<15000
x<7500

4. Let R(x)=y and switch x and y values to find [tex]R^{-1 }[/tex](x)
12y=x
y=[tex]\frac{1}{12}[/tex]x
[tex]R^{-1}[/tex](x)=[tex]\frac{1}{12}[/tex]x

5. Let C(x)=y and switch x and y values to find [tex]C^{-1 }[/tex](x)
10y+15000=x
y=0.1x-1500
[tex]C^{-1 }[/tex](x)=0.1x-1500

6. R'(x)=(12x)'=12

7. C'(x)=(10x+15000)'=10

[nycmath]

Question 44:

1. R(x)=C(x)
12x=10x+15000
2x=15000
x=7500
(7500,90000)
The break even point is $90000

2. R(x)>C(x)
12x>10x+15000
2x>15000
x>7500
The company must sell at least 7500 items

3. R(x)<C(x)
12x<10x+15000
2x<15000
x<7500

4. Let R(x)=y and switch x and y values to find [tex]R^{-1 }[/tex](x)
12y=x
y=[tex]\frac{1}{12}[/tex]x
[tex]R^{-1}[/tex](x)=[tex]\frac{1}{12}[/tex]x

5. Let C(x)=y and switch x and y values to find [tex]C^{-1 }[/tex](x)
10y+15000=x
y=0.1x-1500
[tex]C^{-1 }[/tex](x)=0.1x-1500

6. R'(x)=(12x)'=12

7. C'(x)=(10x+15000)'=10

Much easier to read than your pencil work. Please, use LaTex or flair pens.