--f(a) exists because there is a point (a, f(a)) even though the point along the curve indicates a hole in the graph. True

--There is a function value for each point (x, f(x)) for 0<x<a.True

--The function f(x) is not continuous at x = a because of the hole in the graph.True

--The limit exists at x=a and has a value higher than f(a). It would be the value of the y-coordinate for the hole in the graph.True

--The limit of f ' (x) as x approaches a does not exist.  Two different slopes occur as your approach a, one from the right and a different one from the left. False

--Answer E

Matching the Taylor Series with P(x) yields

--Answer A

Answer C

One view indicates: 

So a better view is

Finding the zero of the rate indicates the location where the balloon stops decreasing in height and begins to rise.

Therefore the integral would be

A cannot be true since the table of values already indicates that a minimum value of 2 at 0 and 4.

B does not necessarily have to be true.  We are not sure of the behavior of the function at values in between those given in the chart.  The function could take on values larger than 4.

C  is not necessarily true because we only know the values of the function at the five given points, but not at points in between.

D is not necessarily true even though for the points given in the table this is true.  We are not sure what the function is doing in between the given points.

E is correct all the time because the function f is continuous and differentiable on the closed interval.  Therefore we are guaranteed there exists a c with 0<c<4 such that f ' (c)=0 since the slope between (0, f(0)) and (4, f(4)) is also zero.

shows us that f " changes sign 4 times (or the slope of f ') . 

Looking at the velocity graph we can see that the velocity equals zero the first time at t=.65513638

We can see that the particle traveled -.1835406 during time t=0 to t=.65513638.  Add this to the starting position of 3 and you end up at position 2.816459372.

This integral is referencing the average value of a function. 

means we are looking for a function where the average value of the function could possibly be equal to 1.  Another way to look at this integral would be think of it in terms of area. The area of the region bounded by x=2,x=4, the graph, and the x-axis would have to be 2. 

If a triangle were superimposed over the graph whose vertices were at (2,0), (3,2), and (4,0) the area of the triangle would be 2 so that tells us that the area of the given region in graph A is greater than 2.  This suggest that this function has an average greater than 1.

Similarly f a triangle were superimposed over the graph whose vertices were at (2,0), (3,2), and (4,0) the area of the triangle would be 2 so that tells us that the area of the given region in graph B is less than 2.  This would lead to an average less than 1.

It is impossible for the average value of the function to equal 1 in graphs D and E. 

Calculating the definite integral for graph C yields an area of exactly 2 so this graph would have an average value of 1. 

--Answer D

I.  This is not true because the accumulated area from x=0 to x=1 is positive so f(1) > f(0).

III.  This is not true since f(2) > f(1) and f(3) < f(2). This leads to a conclusion that f(3) > f(1).

II.  This is true since the area added between x=1 and x=2 is positive and we already know that f(1)>0 so f(2) >f(1)

91.

The maximum upward velocity is found at the time when h"(t) is changing from  positive to a negative.

t=.31640704

We need to find the point where the derivative of f equals 1.46209812.  Setting Y2 equal to the derivative of f and Y3 equal to 1.46209812 and looking for the intersection yields the point x = 2.164