Continuity:
f(x)= {-3+x, x<-4
{x^-2, -4<x<2
{x+1, x>2
Step 1: x--> -4- (-3+(-4)) = -7
x--> -4+ (-4^2 - 2 = 14
The function is not continuous because it does not come together at x=-4 from the left and right sides.
Intermediate Value Theorem: f(x)= x^2-2x - 3 on [2,6]
f(2) = -3
f(6) = 21
The IVT allows us to say that along the interval [2,6] there is a solution N.
f(x)= x^2 + 4x -22 on [1,3]
f(1)= -17
f(3)= -1
The IVT doesn't prove that there is a solution along the interval [2,6]
Derivatives: Two types of derivatives.
One way to solve: f(x) - f(a)
----------------
x-->a x-a
Second Way to Solve: Difference Quotient f(x+h) -f(x)
-----------------
h h--> 0
Ex: f(x) = 3x+7
3(x+h) +7 -(3x+7)
---------------------- h --> 0
h
3x+3h +7 -3x - 7 h(3x+3 + 7 -3x -7
-------------------- -->
h
x=3 The slope of the function as a whole (derivative) is 3.
Instantaneous vs Average Velocity: Instantaneous Velocity is slope at one certain point. Average Velocity is the average slope over an equation as a hole.
The first piecewise could be continuous at 2, be sure to check that
ReplyDelete"it does not come together" could read "the limit does not exist at x = 2, since the limit from the left is not the same as the limit from the right" so that you sound more scholarly. No "hole" at the end - "whole".
ReplyDeleteYou did not answer the question "What is the hardest part of finding a derivative for you?"