Blog Post 4

1. f(x) 3- {sqr(x) where x is greater than or equal to 4
               {x^2+3 where x is less than 4
Plug 4 into each equation, approaching the left and right
Equation #1 comes to 19
Equation #2 comes to 1
They are not equal, therefore f(x) is not continuous

2. f(x)=2x^2+5x-3 [0,4]
Plug 0 and 4 into the equation
f(0) comes out to equal -3
f(4) comes out to equal 49
Since one is negative and the other is positive, there is indeed a solution

f(x)= -3x^2+14x-8 [-3,0]
Plug -3 and 0 into the equation
f(-3) comes out to equal -77
f(0) comes out to equal -8
Since they are both negative, a solution cannot be proven.

3. Two types - Different quotient and slope of a tangent line
Linear derivative - f(x)=2x-3 as x approaches 1
2x-3- f(1) / x-1
2x-3 - 1 / x-1
2x-2 / x-1
(x-1)(x+1) / x-1
x+1 = 2
A derivative is the slope of a line at a certain point
The hardest part is the difference quotient and doing the algebraic part of distributing the xs and hs.

4. Instantaneous velocity is the slope of one certain point. Average velocity is the average slope of a point.

Revised Calc Post 3

If this doesn't show up, try here: https://docs.google.com/document/d/1iBl4ykvOscM1dmHnuCKQ1UwiRuIZBBqaywT-LJWiKKE/edit

Blog Post 3