First translate the English phrase "Four times the sum of a number and 15 is at least 120" into a mathematical inequality.
"Four times..." means we're multiplying something by 4.
"... the sum of a number and 15..." means we're adding an unknown and 15 and then multiplying the result by 4.
"... is at least 120" means when we substitute the unknown for a value, in order for that value to be in the solution set, it can only be less than or equal to 120.
So, the resulting inequality is 4(x + 15) ≤ 120.
Simplify the inequality.
4(x + 15) ≤ 120
4x + 60 ≤ 120 <-- Using the distributive property
4x ≤ 60 <-- Subtract both sides by 60
x ≤ 15 <-- Divide both sides by 4
Now that we have the inequality in a simplified form, we can easily see that in order to be in the solution set, the variable x can be no bigger than 15.
In interval notation it would look something like this:
[15, ∞)
In set builder notation it would look something like this:
{x | x ∈ R, x ≤ 15}
It is read as "the set of all x, such that x is a member of the real numbers and x is less than or equal to 15".
Part A:
Let the length of one of the sides of the rectangle be L, then the length of the other side is obtained as follow.
Let the length of the other side be x, then

Thus, if the length of one of the side is x, the length of the other side is 8 - L.
Hence, the area of the rectangle in terms of L is given by

Part B:
To find the domain of A
Recall that the domain of a function is the set of values which can be assumed by the independent variable. In this case, the domain is the set of values that L can take.
Notice that the length of a side of a rectangle cannot be negative or 0, thus L cannot be 8 as 8 - 8 = 0 or any number greater than 8.
Hence the domain of the area are the set of values between 0 and 8 not inclusive.
Therefore,
Average rate of change of the function 
Solution:
Given function:
from x = 1 to x = 5
Substitute x = 1 and x = 5 in f(x).


Let us find the average rate of change of the function.
Average rate of change

Here a = 1 and b = 5.

Substitute f(5) and f(1).



Average rate of change of the function 