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MATH1901 Quizzes

Quiz 6: Continuous functions
Question 1 Questions
Which of the following functions are continuous functions on the whole of their domain? (Tick each one that is continuous.) (Zero or more options can be correct)
a)
f : ;f(x) = 2x + 3 x 2 if x2 and f(x) = 1 if x = 2.
b)
f : (,0) ;f(x) = 2x + 3 x 2
c)
f : (0,) ;f(x) = 2x + 3 x 2 if x2 and f(x) = 1 if x = 2.
d)
f : (,2) ;f(x) = 2x + 3 x 2
e)
f : [2,) ;f(x) = 2x + 3 x 2 if x2 and f(x) = 1 if x = 2.

There is at least one mistake.
For example, choice (a) should be False.
This function is not continuous at x = 2.
There is at least one mistake.
For example, choice (b) should be True.
This function is continuous at every point in its domain.
There is at least one mistake.
For example, choice (c) should be False.
This function is not continuous at x = 2.
There is at least one mistake.
For example, choice (d) should be True.
This function is continuous at every point in its domain.
There is at least one mistake.
For example, choice (e) should be False.
This function is not continuous at x = 2.
Correct!
  1. False This function is not continuous at x = 2.
  2. True This function is continuous at every point in its domain.
  3. False This function is not continuous at x = 2.
  4. True This function is continuous at every point in its domain.
  5. False This function is not continuous at x = 2.
At what value or values of x is the function
f(x) = x + 4if x 1 x2 if  1 < x < 1 2 x if x 1
discontinuous? (Zero or more options can be correct)
a)
1
b)
0
c)
1
d)
2
e)
f(x) is continuous everywhere

There is at least one mistake.
For example, choice (a) should be True.
There is at least one mistake.
For example, choice (b) should be False.
There is at least one mistake.
For example, choice (c) should be False.
There is at least one mistake.
For example, choice (d) should be False.
There is at least one mistake.
For example, choice (e) should be False.
Correct!
  1. True
  2. False
  3. False
  4. False
  5. False
At what value or values of x is the function
f(x) = |x + 1| 1if x < 0 x2 + x if 0 x < 1 3 x if 1 x
discontinuous? Exactly one option must be correct)
a)
1, 0 and 1
b)
1
c)
0
d)
0 and 1
e)
f(x) is continuous everywhere

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is correct!
For what values of a will the function
f(x) = x3if x a x2 if x > a
be continuous for all x? Exactly one option must be correct)
a)
1 and 2
b)
2 and 1
c)
1 and 1
d)
0 and 1
e)
1 and 0

Choice (a) is incorrect
As 2322 the function is not continuous if a = 2.
Choice (b) is incorrect
As (2)3(2)2 the function is not continuous if a = 2.
Choice (c) is incorrect
As (1)3(1)2 the function is not continuous if a = 1.
Choice (d) is correct!
For the function to be continuous at x = a, we must have or a3 = a2. Hence a = 0 or 1.
Choice (e) is incorrect
As (1)3(1)2 the function is not continuous if a = 1.
Which of the following are correct proofs, using the Intermediate Value Theorem, that the polynomial f(x) = x3 + 2x2 1 has at least one root? (More than one answer may be correct.) (Zero or more options can be correct)
a)
The function f(x) is continuous for all real x and f(0) = 1 and f(1) = 2. Therefore, since f(0) < 0 < f(1), by the Intermediate Value Theorem there is a number c in (0,1) such that f(c) = 0. In other words, c is a root of x3 + 2x2 1 = 0.
b)
The function f(x) is differentiable for all real x and f(0) = 1 and f(2) = 15. Therefore, since f(0) < 0 < f(2), by the Intermediate Value Theorem there is a number c in (0,2) such that f(c) = 0. In other words, c is a root of x3 + 2x2 1 = 0.
c)
As f(x) = 3x2 + 4x, the critical points of f(x) occur at x = 0 and x = 4 3. As x = 4 3 is a maximum f(x) must have a root close to x = 4 3.
d)
The function f(x) is continuous for all real x and f(0) = 1 and f(2) = 15. Therefore, since f(0) < 0 < f(2), by the Intermediate Value Theorem there is a number c in (0,2) such that f(c) = 0. In other words, c is a root of x3 + 2x2 1 = 0.

There is at least one mistake.
For example, choice (a) should be True.
There is at least one mistake.
For example, choice (b) should be True.
The intermediate value theorem applies to continuous functions. As every differentiable function is continuous this argument is correct.
There is at least one mistake.
For example, choice (c) should be False.
The fact that a function has a local maximum or minimum is independent of the question of the existence of roots.
There is at least one mistake.
For example, choice (d) should be True.
Correct!
  1. True
  2. True The intermediate value theorem applies to continuous functions. As every differentiable function is continuous this argument is correct.
  3. False The fact that a function has a local maximum or minimum is independent of the question of the existence of roots.
  4. True
In which of the following cases can you correctly use L’Hôpital’s rule to evaluate the limit? (Zero or more options can be correct)
a)
limxπ cosx x π
b)
limxxex
c)
limx1ln(2x) lnx
d)
limx1 lnx x 1
e)
limx0tanx x

There is at least one mistake.
For example, choice (a) should be False.
L’Hôpital’s rule cannot be used here because, although limxπx π = 0, it is not true that limxπ cosx = 0.
There is at least one mistake.
For example, choice (b) should be True.
L’Hôpital’s Rule can be used.
limxxex = lim x x ex = limx 1 ex = 0.
Note that limxx = = limxex.
There is at least one mistake.
For example, choice (c) should be False.
L’Hôpital’s Rule can not be used here because although the denominator has limit 0, the numerator does not. That is, limx1 ln(2x) = ln(2)0.
There is at least one mistake.
For example, choice (d) should be True.
Since limx1 lnx = 0 = limx1x 1 we can apply L’Hôpital’s Rule. This shows that
limx1 lnx x 1 = limx1 1 x 1 = 1.
There is at least one mistake.
For example, choice (e) should be True.
As limx0 tanx = 0 = limx0x we can apply L’Hôpital’s Rule, to find that
limx0tanx x = limx0sec2x 1 = limx01 + tan2x 1 = 0.
Correct!
  1. False L’Hôpital’s rule cannot be used here because, although limxπx π = 0, it is not true that limxπ cosx = 0.
  2. True L’Hôpital’s Rule can be used.
    limxxex = lim x x ex = limx 1 ex = 0.
    Note that limxx = = limxex.
  3. False L’Hôpital’s Rule can not be used here because although the denominator has limit 0, the numerator does not. That is, limx1 ln(2x) = ln(2)0.
  4. True Since limx1 lnx = 0 = limx1x 1 we can apply L’Hôpital’s Rule. This shows that
    limx1 lnx x 1 = limx1 1 x 1 = 1.
  5. True As limx0 tanx = 0 = limx0x we can apply L’Hôpital’s Rule, to find that
    limx0tanx x = limx0sec2x 1 = limx01 + tan2x 1 = 0.
Using L’Hôpital’s Rule find the value of limx021 + 3x 2 x . Type your answer into the box.

Correct!
Since limx021 + 3x 2 = 0 = limx0x we can apply L’Hôpital’s Rule. This gives
limx021 + 3x 2 x = limx0 3 1+3x 1 = limx0 3 1 + 3x = 3.

Incorrect. Please try again.
By L’Hôpital’s rule,

limx01 + 3x 1 x = limx0 3 21+3x 1

Using L’Hôpital’s Rule find the value of limx06ex 6x 6 x2 . Type your answer into the box.

Correct!
Notice that we can apply L’Hôpital’s Rule because limx06ex 6x 6 = 0 = lim x0x2. Therefore,

limx06ex 6x 6 x2 = limx06ex 6 2x = limx03ex 3 x Applying L’Hôpital’s rule a second time we have limx06ex 6x 6 x2 = limx03ex 1 = 3

Incorrect. Please try again.
You need to use L’Hôpital’s Rule more than once.
Find the value of the limit limx0cosx 1 x . Type your answer into the box.

Correct!
By L’Hôpital’s Rule,
limx0cosx 1 x = limx0sinx 1 = 0
Incorrect. Please try again.
Try using L’Hôpital’s Rule.
What is the value of the limit limx1 + 1 xx ? Exactly one option must be correct)
a)
1
b)
e
c)
d)
The limit does not exist.

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect