Proper way to solve function notations? I'm just starting to use function notation and I'm wondering if I'm solving correctly.
If $f(x) = 4x - 11$, determine 
a. $f (1/4)$
$f(x) = 4x - 11$
$f(1/4) = 4 (1/4) - 11$
$f(1/4) = 1 - 11 $
$f(1/4) = -10$
b. $f(-3)=4x-11$
$f(-3)=4(-3)-11$
$f(-3)= -12-11$
$f(-3) = -23$
The equation $C=10n+45$ represents the total cost, C dollars, for a sports banquet , when n people attend. 
a. Describe the function in words and then express the function using function notation.
b. Determine C(250) and explain what the result represents.
c(250)=10n+45
c(250)=10(250)+45
c(250)=2545
c. If the total cost of the banquet is C(n) = 4295, find n, the number of people attending the banquet.
4295 = 10n + 45
45+4295=10n+45+45
4340 / 10 90 / 10
434-9= 425
425 x 10 + 45 = 4295
425
 A: All your answers are correct.
However the redaction for the solution of $C(n)=4295$ is not very clear, particularly the line "4340 / 10 90 / 10". Here is how I would do it
$$\begin{array}{l|r}C(n)=4295 & \\ \iff 4295=10n+45 &\quad -45\\ \iff 4250 = 10n & \div 10\\ \iff 425 = n\end{array}$$
A: Just a piece of advice ... if you're going to do algebra, write down your work neatly on a piece of paper. If you want to write down informal calculations, do it on a piece of plywood. That kind of thing belongs on a construction site, not on a math assignment.

Just one further comment, what you did is correct. By adding $45$ to both sides you make the $45$ a $90$, which is divisible by $10$.
$$\begin{array}{lll}
4295&=&10n + 45\\
4295+45&=&10n + 45+45\\
4340&=&10n + 90\\
\frac{4340}{10}&=&\frac{10n + 90}{10}\\
434&=&n + 9\\
434-9&=&n + 9-9\\
425&=&n\\
n&=&425\\
\end{array}$$
But it is somewhat awkward. It is much cleaner subtracting $45$ from both sides to eliminate the addition on the right hand side.
$$\begin{array}{lll}
4295&=&10n + 45\\
4295-45&=&10n + 45-45\\
4250&=&10n + 0\\
4250&=&10n\\
\frac{4250}{10}&=&\frac{10n}{10}\\
425&=&n\\
n&=&425\\
\end{array}$$
