Use This Study Guide to Prepare for the Final Exam

Calculus Final Exam Study Guide 1

Use this study guide to prepare for the final exam.

1. Over what interval will the Intermediate Value Theorem (IVT) apply?

Everywhere

2. Some mathematicians wanted to use the IVT as the definition of continuous. i.e. if the IVT does not hold then the function can not be continuous.
For the following graph what is the counter example that shows that the IVT does not hold over a non-continuous graph?

f(-4) = 1, f(2) = -1 so there must be a point where f(x) = .25.

f(-4) = 1, f(-2) = 1 so there must be a point where f(x) = .25.
f(4) = -1, f(2) = -1 so there must be a point where f(x) = .25.

the IVT holds for the above graph.

3. Use the IVT (Intermediate Value Theorem) to show that has a zero.

You can not use the IVT to show there is a zero in [-1,1]

You can use the IVT to show there is a zero in [-1,2]

You can use the IVT to show that there is a zero in [-1, 0]

The IVT cannot be applied to find a zero anywhere.

4. Use the IVT (Intermediate Value Theorem) to show that has a zero.

f(x) is continuous on [0,1], f(0) < 0 , f(1) = 0, so we can use the IVT to show there is a zero in [0,1]

f(x) is continuous on [-1,2], f(-1) < 0 , f(2) > 0, so we can use the IVT to show there is a zero in [-1,2]

f(x) is continuous on [-2,0], f(-2) < 0 , f(0) = 0, so we can use the IVT to show there is a zero, between [-2,1].

The IVT cannot be applied to find a zero.

5. Use the IVT (Intermediate Value Theorem) to show that has a zero.

You can use the IVT to show there is a zero in [-1,1]

You can use the IVT to show there is a zero in [-1,2]
You can use the IVT to show that there is a zero in [-1, 0]

The IVT cannot be applied to find a zero anywhere.

6. , if . Use the definition of the derivative to determine if the function is differentiable at x = 0.
f(x) is differentiable at x = 0.

f(x) is not differentiable at x = 0 because the does not exist at x = 0.

f(x) is not differentiable at x = 0 because the increases without bound at x = 0.

f(x) is not differentiable at x = 0 because the at x = 0.

7. Which statement is true?
f(x) = is differentiable at x = 0.
f(x) = tan(x) is differentiable everywhere.
f(x) = x2 + 3x + 6, is differentiable everywhere.
If f(x) is differentiable at x = c, the can not be 0.
8. Where is the graph below not differentiable and why?

f(x) is not differentiable at x = 1 because f(x) has a cusp at x = 1.
f(x) is not differentiable at x = 1 because f(x) has a corner at x = 1.
f(x) is not differentiable at x = 1 because f(x) is discontinuous at x = 1.
f(x) is differentiable everywhere shown.
9. Where is the graph below not differentiable and why?

f(x) is not differentiable at x = 1 because f(x) has a cusp at x = 1.
f(x) is not differentiable at x = 1 because f(x) has a corner at x = 1.
f(x) is not differentiable at x = 1 because f(x) is discontinuous at x = 1.
f(x) is not differentiable at x = 1 because f(x) has a vertical tangent at x = 1.
10. Where is the graph below not differentiable and why?

f(x) is not differentiable at x = 0 because f(x) has a cusp at x = 0.
f(x) is not differentiable at x = 0 because f(x) has a corner at x = 0.
f(x) is not differentiable at x = 0 because f(x) is discontinuous at x = 0.
f(x) is differentiable at x = 0 because f(x) is continuous at x = 0.
11. For the following graph of f(x), which statement is true?

f(x) is not differentiable everywhere shown because f(x) is not continuous everywhere shown.
f(x) is not differentiable at x = 1 because f(x) has a cusp at x = 1.
f(x) is differentiable everywhere.
f(x) is not differentiable everywhere because f(x) has a vertical tangent line at x = 1, and corners at both x = 2 and x = -2.
12. Which statement below is true?
It is possible for a function to be differentiable but not continuous.
If it is differentiable everywhere then it must be continuous.
If a graph is continuous then it must be differentiable.
A cusp is a point where the graph is not continuous.
13. Why is the following function not differentiable at x = 1?

f(x) has a cusp at x =1
f(x) is discontinuous at x =1.
f(x) 's left handed limit does not exist at x =1?
f(x) has a corner at x =1.
14.  A person is standing and throws a ball into the air at time t = 0 s and from a height of 1 m. (see graph below)

At time t = 5 s what is the position of the ball?
1 m
6 m
13 m
cannot calculate with information given.
15. Which of the following statements describes position?
The position of an object is the distance the object has traveled.
The position of an object at a specific time is measured by finding the location of the object at that time.
The position of an object always tells us how fast the object is moving.
The position of the object always tells us how far the object has moved
16.  The graph below shows the function. . If f(x) describes the position of an object over time, at what time is the object moving the slowest and what is the object’s velocity?

-2s, velocity -8 m/s
-2s, velocity 8 m/s
0s, velocity 0 m/s
0 s, velocity 1 m/s
17.  The graph below shows the function. . If f(x) describes the position of an object over time, at what time is the object moving the fastest in the interval [-2,2] and what is the object’s instantaneous velocity?

12 m/s
10 m/s
1 m/s
can not be computed
18.  A person is standing and throws a ball into the air at time t = 0 s from a height of one meter. The graph shows the height of the ball over time. The function shown describes the motion of the ball.


19.  What is the difference between the average velocity between [1, 5] and the instantaneous velocity at 1?
4
1
0
cannot be found.
The graph below shows the velocity of an object over time.

What is the average acceleration on the interval [-1,1]?
-2
0
1
-1
20.  The graph below shows the velocity of an object over time.

What do we know about the acceleration of the object?
The acceleration is positive between [1, 2].
The acceleration is negative between [1, 2].
The acceleration can not be measured from the graph.
The acceleration is negative between [-2, 1].
22. The velocity of an object v(t) is shown in the graph below. and

What is the instantaneous acceleration at t = 1?
0
-1
1
-2
23.  The function below shows the height an object has traveled over time. For what interval is the velocity equal to the speed?

[0, 3]
[2, 4]
[0, 6]
The velocity is never the same as the speed of the object.
24.  Bob is driving and his progress is shown on the graph below. The graph shows how far he has gone over time. The slope of the line (or the derivative) is the rate of change in distance over time. Which statement is true?

Speed is not the same as velocity for the above graph.
Speed and velocity happen to be the same for the above graph.
The slope shows how fast he is changing speed. (acceleration).
Bob was driving up a hill the whole time.
25.  Find the equation of the tangent line to the curve f(x) = -2x2 – 1 at x = 1.
T(x) = -3 + 4(x + 1)
T(x) = 3 – 4(x – 1)
T(x) = -3 – 4(x – 1)
T(x) = -3 + 4(x – 1)
26.  Find the equation of the tangent line to the curve f(x) = -x2 + 2x at x = 2.
T(x) = 2(x – 2)
T(x)= -2(x – 2)
T(x) = 2x – 1(x – 2)
T(x) = x2 – 1(x – 2)
27.  Find the equation of the tangent line to the curve ; assume at x = 2.




28. Find the equation of the tangent line to the curve f(x) = 2x2 – x at x = 1.
T(x) = 2 + 3(x – 1)
T(x) = 4 + 3(x – 1)
T(x) = 1 + 3(x – 1)
T(x) = 1 – 3(x – 1)
29.  Find the equation of the tangent line to the curve f(x) = sin(x); assume f ’(x) = cos(x) at

T(x) = 1
T(x) = 0
T(x) = -1
30.  Many things grow exponentially. The general formula for exponential growth is: , where A is the starting value, r is the rate of growth and t is time. Assume that A = 1, and r = 1.06 (a 6% annual growth rate). . What is the tangent line to the curve at t = 5?
T(t) = e5.3 – (1.05) e5.3(x – 5)
T(t) = e5.3 + (1.05) e5.3(x – 5)
T(t) = e5.3 + (1.06) e5.3(x – 5)
T(t) = e6 + (1.06) e6(x – 5)
31.  Find the equation of the tangent line to the curve f(x) = x3 at x = 0.
T(x) = 0
T(x) = 1
T(x) = -1
T(x) = -x
32.  Use a tangent line approximation to estimate the value of
3.975
4.1275
3.9875
3.9575
33.  Find the equation of the tangent line to the curve f(x) = -x4 at x = 1.
T(x) = 1 + (4)(x – 1)
T(x) = 1 + (-4)(x – 1)
T(x) = -1 + (-4)(x – 4)
T(x) = -1 + (-4)(x – 1)
34.  A differentiable function has the value y(6) = 4 and the derivative value y'(6) = -5. Approximate the value of y(5.8).
4.8
5.0
5.8
5.1