# Math Help - Mathematical economics help please

1. ## Mathematical economics help please

Three questions I didn't get, help please

1. Let f be a function of one variable defined for all reals. Which of the following conditions is not sufficient to guarantee that f has an inverse?
a.f is symmetric to the origin.
b.f is strictly increasing
c.f(x)=ax^3, a =/ 0
d.f is one-to-one

2.If f(x) is replaced by y=f(x-3), the graph is
a. moved 3 units right
b. moved upwards by 3 units
c. moved downward 3 units
d. moved 3 units to the left

3. A firm's cost function is given by the following equation:

C(q) = 10+ 10Q - 4Q^2 + Q^3

If average total cost (ATC) = C/Q and is given by the sum of (AVC) and (AFC). In other words, ATC(Q)= AVC(Q) + AFC(Q)

then

ATC(Q) = ?
AVC(Q) = ?
AFC(Q) = ?

I'm completely lost on these three questions, help please

Thank you!

2. Originally Posted by domain07
Three questions I didn't get, help please

1. Let f be a function of one variable defined for all reals. Which of the following conditions is not sufficient to guarantee that f has an inverse?
a.f is symmetric to the origin.
b.f is strictly increasing
c.f(x)=ax^3, a =/ 0
d.f is one-to-one

2.If f(x) is replaced by y=f(x-3), the graph is
a. moved 3 units right
b. moved upwards by 3 units
c. moved downward 3 units
d. moved 3 units to the left

3. A firm's cost function is given by the following equation:

C(q) = 10+ 10Q - 4Q^2 + Q^3

If average total cost (ATC) = C/Q and is given by the sum of (AVC) and (AFC). In other words, ATC(Q)= AVC(Q) + AFC(Q)

then

$ATC(Q) = \frac{C(Q)}{Q} =\frac{10+ 10Q - 4Q^2 + Q^3}{Q}$
$AVC(Q) = \frac{10Q - 4Q^2 + Q^3}{Q} = 10 -4Q + Q^2$
$AFC(Q) = \frac{10}{Q}$

I'm completely lost on these three questions, help please

Thank you!
Mine in red and math code in the last part. Be sure to understand the difference between variable and fixed cost. In the total cost function, fixed cost is simply the constant term (10 in your case); all terms with Q in them make up variable cost... as it varies with quantity. Good luck!