A function f(x) is continuous at x=a if:
1) The limit exists at x=a
2) f(a) exists (there is no hole/asymptote)
3) The limit at x=a is equivalent to f(a)
When a function isn't continuous:
{x2+3, 0 < x
f(x) = {5, 0 = x
{x+3, 0 > x
In order to check to see if a function is continuous, you must go through all three steps mentioned above.
Step 1
We will see if the limit exists at x=a. We must look at the limit as x approaches the left and right of 0. The limit of the function as x approaches 0 from the left is 3. The limit of the function as x approaches 0 from the right is 3. This means that the limit exists.
Step 2
Next, we must make sure f(0) exists. By looking at the equation, we can see that f(x) = 5 when x = 0. This means that f(0) = 5 and f(a) exists. There isn't a hole or an asymptote.
Step 3
Finally, we must check to see if the limit at x=a is equivalent to f(a). Our limit as x approaches 0 is 3. However f(0) = 5. 5 doesn't equal 3. Therefore, the function isn't continuous.
Intermediate Value Theorem
If a function is continuous on a closed interval [a,b], and f(a) < N < f(b) then there is a < c < b such that f(x) = N.
We can use this theorem to tell if there is a solution on the interval [-2, 3] for the function f(x) = 2x2 +3x - 5
For this equation, f(-2) = -3 and f(3) = 22. Due to the fact that we know this function is continuous on the interval and the values are on opposite sides of the x-axis, we know there must be a solution in between them. We write this as: since f is continuous on [-2, 3], and f(-2) = -3 < 0 < 22 = f(3), then there exists c ∈[-2, 3] such that f(c) = 0.
Let's use this theorem on another function.
f(x) = 3x3 + 2 on the interval [-5, -1]
For this equation, f(-5) = -373 and f(-1) = -1. Due to the fact that both numbers are negative, we can't prove there is a solution on this interval. However, this doesn't mean there isn't one.
Derivatives
The derivative is basically the slope of a certain point in a function. There are two ways to find the derivative. One is by using the difference quotient and the other is by finding the derivative in terms of slope. We will find a linear derivative to demonstrate these concepts.
Difference Quotient
When we find the derivative by using the difference quotient, we use:
lim f(x+h)-f(x)
h→0 h
Find the derivative of f(x) = 3x + 5
3(x + h) + 5 - (3x + 5)
h
3x + 3h + 5 -3x -5
h
h
3h
h
The derivative of this function is 3.
The Derivative in Terms of Slope
When we find the derivative in terms of slope, we use:
lim f(x) - f(a) at f(x) = a
x→a x - a
Find the derivative of f(x) = 3x + 5 at f(x) = 2
3x + 5 - (3(2) + 5)
x - 2
3x + 5 - 11
x - 2
3x - 6
x - 2
3(x - 2)
1(x - 2)
3
1
3
1
The derivative of this function is 3.
For me, the hardest thing about finding the derivative is foiling correctly while using the difference quotient.
Instantaneous Velocity and Average Velocity
Instantaneous velocity is the slope (or derivative) of an individual point. Average velocity is the derivative that can be used to find any point's slope for a particular function.
