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University of Sydney
School of Mathematics
and Statistics
© Copyright 2004-2006

Quiz 6: Continuous functions

Question

Question 1

Which of the following functions are continuous functions on the whole of their domain? (Tick each one that is continuous.)

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.

Your answers are correct

False. This function is not continuous at x = 2.
True. This function is continuous at every point in its domain.
False. This function is not continuous at x = 2.
True. This function is continuous at every point in its domain.
False. This function is not continuous at x = 2.

Question 2

At what value or values of x is the function

( { x+ 4 if x ≤ - 1 f(x) = x2 if - 1 < x < 1 ( 2- x if x ≥ 1

discontinuous?

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.

Your answers are correct

True.
False.
False.
False.
False.

Question 3

At what value or values of x is the function

( { |x2+ 1|- 1 if x < 0 f (x) = ( x + x if 0 ≤ x < 1 3 - x if 1 ≤ x

discontinuous?

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Your answer is correct.

Question 4

For what values of a will the function

{ x3 if x ≤ a f(x ) = x2 if x > a

be continuous for all x?

Not correct. Choice (a) is false.

As 2³

2² the function is not continuous if a = 2.

Not correct. Choice (b) is false.

As (

)³

(

)² the function is not continuous if a = √-
2

Not correct. Choice (c) is false.

As (-1)³

(-1)² the function is not continuous if a = -1.

Your answer is correct.

For the function to be continuous at x = a, we must have or a³ = a². Hence a = 0 or 1.

Not correct. Choice (e) is false.

As (-1)³

(-1)² the function is not continuous if a = -1.

Question 5

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.)

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 $′ 2 f (x ) = 3x + 4x$ , the critical points of f(x) occur at x = 0 and x = -. As x = - is a maximum f(x) must have a root close to x = -.
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.

Your answers are correct

True.
True. The intermediate value theorem applies to continuous functions. As every differentiable function is continuous this argument is correct.
False. The fact that a function has a local maximum or minimum is independent of the question of the existence of roots.
True.

Question 6

In which of the following cases can you correctly use L’Hôpital’s rule to evaluate the limit?

There is at least one mistake.
For example, choice (a) should be false.

L’Hôpital’s rule cannot be used here because, although $xli→mπx- π = 0,$ it is not true that $xli→mπcosx = 0$ .

There is at least one mistake.
For example, choice (b) should be true.

L’Hôpital’s Rule can be used.

x 1 xl→im∞ xe-x = xli→m∞-x = xli→m∞-x = 0. e e

Note that $lx→im∞ x = ∞ = xli→m∞ ex$ .

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, $lxim→1 ln(2x) = ln(2) ⁄= 0$ .

There is at least one mistake.
For example, choice (d) should be true.

Since $lxim→1lnx = 0 = xli→m1 x- 1$ we can apply L’Hôpital’s Rule. This shows that

1 lim -ln-x-= lim x-= 1. x→1 x - 1 x→1 1

There is at least one mistake.
For example, choice (e) should be true.

As $lim tanx = 0 = lim x x→0 x→0$ we can apply L’Hôpital’s Rule, to find that

tanx- sec2-x 1+-tan2x- lix→m0 x = lxi→m0 1 = xli→m0 1 = 0.

Your answers are correct

False. L’Hôpital’s rule cannot be used here because, although $xli→mπx- π = 0,$ it is not true that $xli→mπcosx = 0$ .
True. L’Hôpital’s Rule can be used. $x 1 xl→im∞ xe-x = xli→m∞-x = xli→m∞-x = 0. e e$ Note that $lx→im∞ x = ∞ = xli→m∞ ex$ .
False. L’Hôpital’s Rule can not be used here because although the denominator has limit 0, the numerator does not. That is, $lxim→1 ln(2x) = ln(2) ⁄= 0$ .
True. Since $lxim→1lnx = 0 = xli→m1 x- 1$ we can apply L’Hôpital’s Rule. This shows that $1 lim -ln-x-= lim x-= 1. x→1 x - 1 x→1 1$
True. As $lim tanx = 0 = lim x x→0 x→0$ we can apply L’Hôpital’s Rule, to find that $tanx- sec2-x 1+-tan2x- lix→m0 x = lxi→m0 1 = xli→m0 1 = 0.$