Friday’s lesson was more about functions.

To review the problems we did in class:
1: If f(x) = x^2-2x find f(x-2)
F(x-2) = (x-2)^2 - 2(x-2)
= x^2- 4x + 4 - 2x + 2
= x^2 - 6x + 8

2: Domain of g(x) = (x^3-8)/√(x^2-9)
√(x^2-9) > 0
x = ± 3
D: (-∞, -3) U (3, ∞)

3: If a =2b^2 + 5 and b= c - 3, show a as a function of c
a = 2(c-3)^2 + 5
= 2(c^2 - 6c + 9) + 5
= 2c^2 - 12c + 18 +5
= 2c^2 - 12c +23

Later we learned the meaning of this statement:
A function F is increasing on an interval if, for any x and x_2 in the interval x < x_2 implies
f(x) < f (x_2).

This means than if a function is increasing in value on the graph of that function then the output of x must be lesser than the output of x_2 if x is in fact less than x_2.

For example: f(x) = x + 2, to show that this function is increasing we plug in two values.
f(3) = 3+2 f(4) = 4 + 2
f(3) = 5 f(4) = 6

Since 3 < 4 and f(3) < f(4), f(x) = x + 2 is an increasing function.