| Solution |
|
|
|
| 76.
By the calculator we can find v'(4).
Answer C
|
|
|
|
|
|
| 77.
The first integral can be found from the picture of the function.
The signed area A is -2, signed area B is 2, and the signed area C is-2.
This totals to -2.
The second integral can be treated geometrically as a rectangle of width
6 and height 1. The area of this region is the definite integral. The
area is 6.
The total of the two areas is 4.
Answer C |
|
|
|
| 78.
Answer C |
|
|
|
| 79.
Graph I has a limit as x approaches 4 from either side, but it does not
equal the value of the function at 4.
Graph II also has a limit as x approaches 4 from either side, but there
is not value of the function at 4.
Graph III has a value of the function at 4 but there is not limit at 4
since there are two different limits at 4. As you approach 4 from the
left there is a higher limit than if you approach 4 from the right.
Answer D |
|
|
|
| 80. We are dealing with a
continuous function on the closed interval and a differentiable function on
the open interval. The picture shows the other two given facts.
We know that there is a c such that f(c)=0 by the intermediate value
theorem.
We know that there is not a c such that f'(c)=0
because the function might be monotonic and therefore only increasing on the
given interval.
We know that there is a c such that f(c) = 3 also be the intermediate
value theorem.
Since the slope of a segment between the two given points is 3 we know by
the Mean Value Theorem that a c exists where the slope of the function must
be equal to the slope of the line between the two given points.
The graph of f could be a linear function and this time c can be at the
endpoints of the interval. Therefore if c = 1 then f(c)=f(x) for all x
on the closed interval.
Answer B |
|
|
|
| 81. Let's look at graph of f
' (x) in the described window to understand the behavior of function f in
that window:
From the graph we can see that f ' changes sign four times in the
described interval.
Answer D |
|
|
|
| 82. A quick graph of this
function show:
|
|
| 'But a better view, by
restricting the domain, is
Finding the zero of the rate indicates the location where the balloon
stops decreasing in height and begins to rise.
Note that y=-1E-12 is essentially zero.
Therefore the integral would be
Answer A |
|
|
|
| 83. The graph of v(t) is 
Notice that v(t) is always positive. Therefore the net change and
total change in position are the same between t = 0 and t = 3.
The area of the region bounded by v and the x axis and t = 0 and t= 3 is
So the net change or total distance traveled over the time period or 3 is
60.256611. The average velocity would be the total distance traveled
divided by the time period. This would be 20.08553692.
Another way to think about this would be to think about finding the height of the rectangle
of the same width that has the same area of 60.256611/3. or 20.08553692
Answer A |
|
|
|
| 84. The temperature of the
pizza can be found using the accumulation function.
Let C(t) be the current temperature of the pizza.
so
Answer A |
|
|
|
| 85. For the trapezoidal
approximation to be an overestimate the function needs to be concave up.
For a right Reimann sum to be an underapproximation the function must be
decreasing. The function that has both of these qualities is graph A.
Answer A. |
|
|
|
86. The base of the solid will be
in region R. If every slice along the red line is the shape of a square the
length of each side of the square is
 The
volume can be found using
Answer B |
|
|
|
| 87.
Looking at the graph of f " tells us where the sign changes from
positive to negative.
 This occurs at
0.473. Answer B |
|
|
|
| 88. 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 C |
|
|
|
| 89. We are given that the
function f passes through (2,3) and has an instantaneous rate of change of
-5 at that point. If function g is defined as g(x)=xf(x) then
g(2)=2(3) or 6
g'(x)=xf '(x) + f(x).
g'(2)= 2f '(2)+f(2) or 2(-5)+3 or -7
The tangent line is y=-7(x-2)+6
Answer D |
|
|
|
| 90. If a function has a positive
first derivative the function must be increasing. If the function has
a negative second derivative the function must be increasing but at a
decreasing rate. Chart A: The y values are increasing at an increasing
rate.
Chart B: The y values are increasing at a decreasing rate.
Chart C: The y values are decreasing at an increasing rate.
Chart D: The y values are decreasing at a decreasing rate.
Chart E: The y values are decreasing at a constant rate.
Answer B |
|
|
|
| 91.
Checking the function a(t) we notice it is always increasing over its domain:
Using the Fundamental Theorem of Calculus
Using numerical integration on the calculator:
Using substitution:
Answer E |
|
|
|
| 92.
Graphically g'(x) is
We can see that g' is negative between 1.772 and 2.507
So between these two values since g' is negative, g is decreasing on this
interval.
Answer D |
|
