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MATH1111 Quizzes

Anti-differentiation (graphically and numerically) Quiz
Web resources available Questions

This quiz tests the work covered in the lecture on graphical interpretation of the anti-derivative and corresponds to Section 6.1 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are more web quizzes at Wiley, select Section 1.

There are not many relevant web resources but it may be useful to review the derivative function.

Visual Calculus, as usual, has links to an applet at http://archives.math.utk.edu/visual.calculus/4/ftc.9/ and some useful explanations.

Suppose F(x) = f(x) and F(2) = 4,F(3) = 8 and F(4) = 11.
Which of the following statements are correct? (Zero or more options can be correct)
a)
23f(x) = 7
b)
24f(x) = 7
c)
34f(x) = 3
d)
23f(x) = 4

There is at least one mistake.
For example, choice (a) should be False.
Note that 23f(x) = F(3) F(2) = 8 4 = 4
There is at least one mistake.
For example, choice (b) should be True.
24f(x) = F(4) F(2) = 11 4 = 7
There is at least one mistake.
For example, choice (c) should be True.
34f(x) = F(4) F(3) = 11 8 = 3
There is at least one mistake.
For example, choice (d) should be False.
Note that 23f(x) = F(3) F(2) = 8 4 = 4
Correct!
  1. False Note that 23f(x) = F(3) F(2) = 8 4 = 4
  2. True 24f(x) = F(4) F(2) = 11 4 = 7
  3. True 34f(x) = F(4) F(3) = 11 8 = 3
  4. False Note that 23f(x) = F(3) F(2) = 8 4 = 4
Consider the graph of f(x) below.
PIC
Let F(x) = f(x).
Which one of the following statements is true. Exactly one option must be correct)
a)
F has a local minimum at x = 1 and at x = 3 and a local maximum at x = 1.
b)
F has a local maximum at x = 1 and at x = 3 and a local minimum at x = 1.
c)
F has a local maximum at x = 0.4, a local minimum at x = 2.4 and a point of inflection at x = 1.
d)
F has a local minimum at x = 1, a point of inflection at x = 1 and a local maximum at x = 3.

Choice (a) is correct!
The derivative of F changes sign from negative to positive at both x = 1 and at x = 3 so there are minima at these points. The derivative changes sign from positive to negative at x = 1 so there is a maximum at this point.
Choice (b) is incorrect
Try again, look at the way the derivative changes sign at these points.
Choice (c) is incorrect
Try again, this is true of the function, f, the derivative of F.
Choice (d) is incorrect
Try again, look at the way the derivative changes sign at these points.
Consider the graph of y = f(x) below.
PIC

Suppose F(x) = f(x) and F(5) = 4. Determine the nature of the critical point of F at x = 9 and its coordinates.
Which one of the following statements is correct? Exactly one option must be correct)
a)
The point (9,10) is a local minimum of F.
b)
The point (9,10) is a local maximum of F.
c)
The point (9,2) is a local minimum of F.
d)
The point (9,2) is a local maximum of F.

Choice (a) is incorrect
Try again, you have not determined the nature of the critical point correctly.
Choice (b) is correct!
59f(x)dx = F(9) F(5) = F(9) 4. From the diagram we know that
59f(x)dx = 6. Hence F(9) = 6 + 4 = 10 and the critical point is (9,10).
The derivative changes from positive to negative at x = 9 so the critical point is a maximum.
Choice (c) is incorrect
Try again, you have not calculated the critical point or the nature of the point correctly.
Choice (d) is incorrect
Try again, you have not determined the coordinates of the critical point correctly.
Recall abf(x)dx = F(b) F(a).
Consider the graph of y = f(x) below.
PIC

Suppose F(x) = f(x) and F(2) = 1. Determine the nature of the critical point of F at x = 5 and its coordinates.
Which one of the following statements is correct? Exactly one option must be correct)
a)
The point (5,4) is a local minimum of F.
b)
The point (5,4) is a local maximum of F.
c)
The point (5,2) is a local minimum of F.
d)
The point (5,2) is a local maximum of F.

Choice (a) is incorrect
Try again, you have not calculated the critical point correctly.
Recall abf(x)dx = F(b) F(a).
Choice (b) is incorrect
Try again, you have not calculated the critical point or the nature of the point correctly.
Choice (c) is correct!
25f(x)dx = F(5) F(2) = F(5) 1. From the diagram we know that
25f(x)dx = 3. The area is beneath the x-axis so the integral is negative.
Hence F(5) = 3 + 1 = 2 and the critical point is (5,2).
The derivative changes from negative to positive at x = 5 so the critical point is a minimum.
Choice (d) is incorrect
Try again, you have the correct point but you have not determined the nature of the critical point.