1. Continuity
The limit exists at x=a, f(a) exists and there are no holes or asymptotes, and the limit at x=a is equivalent to f(a).
Example where continuity does NOT work:
x^2+5, x<0
f(x)= 10, x=0
3x+5, x>0
Step 1: Limit as x approaches zero from theft and right is 5.
Step 2: f(0)=10. This means that there is no hole or asymptote.
Step 3: 5≠10 so the function is not contiunuous.
2. Intermediate Value Theorem
Example that gives a solution:
y=6x+9 on the interval [-3, 0]
y=6(-3)+9 → y=-18+9 → y=-9
y=6(0)+9 → y=0+9 → y=9
Since f is continuous on [-3, 0] and f(-3) = -9 < 0 < 9 = f(0), then there exists c in [-3, 0] such that f(c) = 0 (Because one solution is positive and one is negative and the function is continuous, it must cross the x-axis within the interval.)
Example that does not give a solution:
y=x^3+5 on interval [2, 5]
y=(2)^3+5 → y=8+5 → y=13
y=(5)^3+5 → y=125+5 → y=130
Since f is continuous on [2, 5] and f(2) = 13 > 0 < 130 = f(5), then it cannot be concluded that there exists c in [2, 5] such that f(c) = 0 (Because both solutions are positive, we can not determine if the function crosses the x-axis within the given interval.)
3. Derivatives
Definition: A function which gives the slope of a curve or the slope of the line tangent to a function.
Types of derivatives (example problem: f(x)=5x-5 at x=5):
The difference quotient. (H approaches 0.)
The limit of ((5(x+h) -5) - (5x-5))/h as h approaches 0
(5x+5h -5 -5x +5)/h = 5h/h = 5
The derivative that is found using the limit as x approaches a of the slope formula.
The Limit of ((5x-5)-(5(5)-5))/x-5 as x approaches 5
(5x-5)-(25-5)/x-5=(5x-5)-20/x-5=5x-25/x-5
=5(x-5)/x-5=5
The hardest part of finding the derivative is remembering to completely distribute when it is necessary.
4. Difference Between Instantaneous Velocity and Average Velocity
Instantaneous velocity is the slope of the tangent line at one point in a function. The average velocity on the other hand, is over an interval in a function.
