Calculus 1 Quiz: Continuity Over An Interval

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Continuity Over An IntervalQuestion 1 of 20

Let $f(x) = \begin{cases} \frac{x^2-4}{x-2} & x < 2 \\ ax^2-bx+3 & 2 \le x < 3 \\ 2x-a+b & x \ge 3 \end{cases}$ . For which values of $a$ and $b$ is $f(x)$ continuous everywhere?

a = \frac{1}{2}, b = \frac{1}{2}

a = 1, b = 1

a = \frac{1}{2}, b = -\frac{1}{2}

a = \frac{3}{2}, b = \frac{1}{2}

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Calculus 1 Quiz

Calculus 1 Quiz: Continuity Over An Interval

Practice Continuity Over An Interval in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Continuity Over An Interval, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let $f(x) = \begin{cases} \frac{x^2-4}{x-2} & x < 2 \\ ax^2-bx+3 & 2 \le x < 3 \\ 2x-a+b & x \ge 3 \end{cases}$ . For which values of $a$ and $b$ is $f(x)$ continuous everywhere?

$a = \frac{1}{2}, b = \frac{1}{2}$ (correct answer)
$a = 1, b = 1$
$a = \frac{1}{2}, b = -\frac{1}{2}$
$a = \frac{3}{2}, b = \frac{1}{2}$

Explanation: First, check continuity at

x=2

. The left-hand limit is

\lim_{x\to 2^-} \frac{x^2-4}{x-2} = \lim_{x\to 2^-} (x+2) = 4

. The right-hand limit and function value are

a(2^2) - b(2) + 3 = 4a - 2b + 3

. So we must have

4a - 2b + 3 = 4

, which simplifies to

4a - 2b = 1

. Next, check continuity at

x=3

. The left-hand limit is

a(3^2) - b(3) + 3 = 9a - 3b + 3

. The right-hand limit is

2(3) - a + b = 6 - a + b

. Setting them equal:

9a - 3b + 3 = 6 - a + b

, which simplifies to

10a - 4b = 3

. Now we solve the system: (1)

4a - 2b = 1

and (2)

10a - 4b = 3

. Multiply equation (1) by 2 to get

8a - 4b = 2

. Subtract this from equation (2):

(10a - 4b) - (8a - 4b) = 3 - 2

, which gives

2a = 1

, so

a = \frac{1}{2}

. Substitute this into

4a - 2b = 1

4(\frac{1}{2}) - 2b = 1 \implies 2 - 2b = 1 \implies -2b = -1 \implies b = \frac{1}{2}

Question 2

Find the largest value $b$ such that the function $f(x) = \ln(x^2-x-12)$ is continuous on the interval $(-\infty, b)$ .

$-4$
$-3$ (correct answer)
$3$
$4$

Explanation: The natural logarithm function

\ln(u)

is continuous for

u > 0

. Thus, we need

x^2-x-12 > 0

. Factoring the quadratic, we get

(x-4)(x+3) > 0

. The roots are at

x=4

and

x=-3

. The parabola opens upwards, so the expression is positive when

x > 4

x < -3

. The domain of

f(x)

(-\infty, -3) \cup (4, \infty)

. The function is continuous on these two disjoint intervals. We are looking for the largest value

b

such that the interval

(-\infty, b)

is a subset of the domain. This corresponds to the interval

(-\infty, -3)

. Thus, the largest possible value for

b

-3

Question 3

Let $f(x) = e^x$ and $g(x) = \cos(x)$ . The Intermediate Value Theorem guarantees that the equation $f(x)=2g(x)$ has a solution on which of the following intervals?

$[-\pi, -\frac{\pi}{2}]$
$[-\frac{\pi}{2}, 0]$
$[0, \frac{\pi}{2}]$ (correct answer)
$[\frac{\pi}{2}, \pi]$

Explanation: To apply the IVT, we want to find a root of the function

h(x) = f(x) - 2g(x) = e^x - 2\cos(x)

. This function is continuous everywhere. We need to find an interval

[a, b]

where

h(a)

and

h(b)

have opposite signs. Let's test the intervals. A)

h(-\pi) = e^{-\pi} - 2\cos(-\pi) = e^{-\pi} - 2(-1) = e^{-\pi}+2 > 0

h(-\frac{\pi}{2}) = e^{-\pi/2} - 2\cos(-\frac{\pi}{2}) = e^{-\pi/2} - 0 > 0

. No guarantee. B)

h(-\frac{\pi}{2}) > 0

h(0) = e^0 - 2\cos(0) = 1 - 2(1) = -1 < 0

. The sign changes from positive to negative. However, this is not an option. Let me re-check. No, this interval isn't an option. Let's check C.

h(0) = -1 < 0

h(\frac{\pi}{2}) = e^{\pi/2} - 2\cos(\frac{\pi}{2}) = e^{\pi/2} - 0 > 0

. The sign changes from negative to positive on

[0, \frac{\pi}{2}]

, so the IVT guarantees a solution in this interval. D)

h(\frac{\pi}{2}) > 0

h(\pi) = e^{\pi} - 2\cos(\pi) = e^{\pi} - 2(-1) = e^{\pi}+2 > 0

. No guarantee.

Question 4

For what values of the constants $a$ and $b$ is the function $f(x)$ continuous on $(-\infty, \infty)$ ?

$a = \frac{3}{4}, b = -\frac{1}{4}$ (correct answer)
$a = \frac{1}{2}, b = -\frac{3}{2}$
$a = 1, b = 1$
$a = \frac{3}{4}, b = \frac{1}{4}$

Explanation: For continuity at

x=-1

, the left-hand limit must equal the right-hand limit:

a(-1)-b = 2(-1)^2+3a(-1)+b

, which simplifies to

-a-b = 2-3a+b

, or

2a-2b=2

, so

a-b=1

. For continuity at

x=1

, the left-hand limit must equal the right-hand limit:

2(1)^2+3a(1)+b = 4

, which simplifies to

2+3a+b=4

, or

3a+b=2

. We now have a system of two linear equations:

a-b=1

and

3a+b=2

. Adding the two equations gives

4a=3

, so

a=\frac{3}{4}

. Substituting this into

a-b=1

gives

\frac{3}{4}-b=1

, so

b = \frac{3}{4}-1 = -\frac{1}{4}

Question 5

Let $g(x) = \begin{cases} 2x+5 & \text{if } x \le -2 \\ ax^2+b & \text{if } -2 < x < 1 \\ 3-x & \text{if } x \ge 1 \end{cases}$ . For what values of $a$ and $b$ is the function $g$ continuous on $(-\infty, \infty)$ ? What is the value of $a-b$ ?

$-\frac{8}{3}$ (correct answer)
$-\frac{4}{3}$
$2$
$\frac{5}{3}$

Explanation: For

g

to be continuous, the pieces must meet at the breakpoints

x=-2

and

x=1

. At

x=-2

\lim_{x\to -2^-} g(x) = 2(-2)+5 = 1

\lim_{x\to -2^+} g(x) = a(-2)^2+b = 4a+b

. So, we must have

4a+b=1

. At

x=1

\lim_{x\to 1^-} g(x) = a(1)^2+b = a+b

\lim_{x\to 1^+} g(x) = 3-1 = 2

. So, we must have

a+b=2

. We now solve the system of linear equations:

$4a+b=1$
$a+b=2$ Subtracting (2) from (1) gives $3a = -1$ , so $a = -1/3$ . Substituting $a=-1/3$ into (2) gives $(-1/3)+b=2$ , so $b = 7/3$ . The question asks for $a-b$ : $a-b = -1/3 - 7/3 = -8/3$ .

Question 6

Let the function $f$ be defined by $f(x) = \begin{cases} ax^2 - 1 & \text{if } x < 2 \\ b & \text{if } x = 2 \\ x+a & \text{if } x > 2 \end{cases}$ . If $f$ is continuous on the interval $[0, 4]$ , what is the value of $b$ ?

$1$
$3$ (correct answer)
$5$
$7$

Explanation: For

f

to be continuous at

x=2

, the limit as

x

approaches

2

must exist and be equal to

f(2)

. This means the left-hand limit must equal the right-hand limit:

\lim_{x\to 2^-} f(x) = \lim_{x\to 2^+} f(x)

. Calculating the one-sided limits: Left-hand limit:

\lim_{x\to 2^-} (ax^2 - 1) = a(2^2) - 1 = 4a - 1

. Right-hand limit:

\lim_{x\to 2^+} (x+a) = 2+a

. Setting them equal:

4a - 1 = 2+a

, which simplifies to

3a = 3

, so

a=1

. For continuity,

f(2)

must equal this limit. We have

f(2)=b

. The limit is

2+a = 2+1 = 3

. Therefore,

b=3

Question 7

Let $f(x)$ and $g(x)$ be functions that are both continuous on the interval $[-5, 5]$ . Suppose $f(3)=4$ and $g(x) = x^2 - 13$ . Which of the following functions is NOT necessarily continuous on the interval $[-5, 5]$ ?

$h(x) = f(x) + g(x)$
$h(x) = f(x)g(x)$
$h(x) = \frac{f(x)}{g(x)}$ (correct answer)
$h(x) = g(f(x))$

Explanation: We use the properties of continuous functions. A) The sum of two functions continuous on an interval is also continuous on that interval. So,

h(x)=f(x)+g(x)

is continuous on

[-5, 5]

. B) The product of two functions continuous on an interval is also continuous on that interval. So,

h(x)=f(x)g(x)

is continuous on

[-5, 5]

. C) The quotient of two continuous functions,

f/g

, is continuous wherever the denominator

g(x)

is not zero. We must check if

g(x)=0

for any

x

[-5, 5]

. Set

g(x) = x^2 - 13 = 0

. This gives

x^2 = 13

, so

x = \pm\sqrt{13}

. Since

3 < \sqrt{13} < 4

, both

\sqrt{13}

and

-\sqrt{13}

are in the interval

[-5, 5]

. Because the denominator is zero at these points,

h(x) = f(x)/g(x)

is not continuous on the entire interval

[-5, 5]

. D) The composition of continuous functions is continuous. Since

f(x)

is continuous on

[-5, 5]

and

g(x)

is continuous everywhere, the composite function

h(x)=g(f(x))

is continuous on

[-5, 5]

. The value

f(3)=4

is irrelevant information.

Question 8

The function $f(x) = \lfloor x \rfloor + \sqrt{x - \lfloor x \rfloor}$ is defined for all real $x$ , where $\lfloor x \rfloor$ is the greatest integer function. Which statement accurately describes the continuity of $f(x)$ ?

The function is continuous for all real numbers. (correct answer)
The function has jump discontinuities at every integer.
The function is continuous from the right, but not the left, at every integer.
The function has removable discontinuities at every integer.

Explanation: To check for continuity at an integer

n

, we must check the left-hand limit, the right-hand limit, and the function value.

f(n) = \lfloor n \rfloor + \sqrt{n - \lfloor n \rfloor} = n + \sqrt{0} = n

. For the right-hand limit, as

x \to n^+

\lfloor x \rfloor = n

. So,

\lim_{x\to n^+} f(x) = \lim_{x\to n^+} (n + \sqrt{x-n}) = n+0=n

. For the left-hand limit, as

x \to n^-

\lfloor x \rfloor = n-1

. So,

\lim_{x\to n^-} f(x) = \lim_{x\to n^-} ((n-1) + \sqrt{x-(n-1)}) = (n-1) + \sqrt{n-(n-1)} = (n-1)+1=n

. Since the left-hand limit, right-hand limit, and function value are all equal to

n

at every integer

n

, the function is continuous at all integers. Between integers, the function is clearly continuous. Thus, the function is continuous for all real numbers.

Question 9

A function $f(x)$ is continuous on $[a, b]$ , and $N$ is a number such that $f(a) < N < f(b)$ . The Intermediate Value Theorem guarantees the existence of a number $c \in (a, b)$ with $f(c) = N$ . Which of the following is a necessary condition for this guarantee?

$f(x)$ must be differentiable on $(a, b)$ .
$f(x)$ must be monotonic on $[a, b]$ .
$f(x)$ must be continuous on $[a, b]$ . (correct answer)
$f(x)$ must be a polynomial function.

Explanation: This is a conceptual question about the preconditions for the Intermediate Value Theorem. The theorem's conclusion is guaranteed only if the function is continuous over the closed interval. Differentiability (A) is a stronger condition and is not required (e.g.,

f(x)=|x|

). Monotonicity (B) is also not required; the function can fluctuate. Being a polynomial (D) is sufficient because all polynomials are continuous, but it is not necessary. The core requirement is continuity on the closed interval.

Question 10

A function $f$ is continuous on $[a, b]$ , and the Extreme Value Theorem guarantees that it attains a maximum value $M$ and a minimum value $m$ on $[a, b]$ . What does the Intermediate Value Theorem add to our knowledge about the function's range?

It guarantees the function has a root between $a$ and $b$ .
It guarantees that $m < M$ if $f(a) \ne f(b)$ .
It guarantees that the function is one-to-one on $[a, b]$ .
It guarantees that the function's range is exactly the interval $[m, M]$ . (correct answer)

Explanation: When you encounter questions about the Extreme Value Theorem and Intermediate Value Theorem working together, focus on how these two fundamental theorems combine to give you complete information about a continuous function's behavior on a closed interval. The Extreme Value Theorem tells you that

f

achieves both a maximum value

M

and minimum value

m

somewhere on

[a,b]

. But this alone doesn't tell you whether

f

takes on every value between

m

and

M

. The Intermediate Value Theorem fills this gap: since

f

is continuous and takes on values

m

and

M

, it must take on every value between them. Therefore, the range is exactly

[m,M]

. Choice A is wrong because the Intermediate Value Theorem only guarantees a root if

f(a)

and

f(b)

have opposite signs. This isn't given in the problem. Choice B incorrectly connects the condition

f(a) \neq f(b)

to whether

m < M

. Even if

f(a) = f(b)

, the function could still vary in between, making

m < M

. Choice C is incorrect because continuous functions on closed intervals can definitely fail to be one-to-one. For example,

f(x) = x^2

[-1,1]

satisfies all conditions but isn't one-to-one. Remember this key insight: the Extreme Value Theorem gives you the "endpoints" of the range (

m

and

M

), while the Intermediate Value Theorem guarantees you get everything in between. Together, they completely determine the range as

[m,M]

Question 11

The temperature $T$ (in degrees Celsius) of a chemical reaction is modeled by a continuous function $T(t)$ for $t \ge 0$ , where $t$ is time in minutes. If $T(0) = 20$ and $T(10) = 60$ , which of the following statements is a necessary conclusion?

The maximum temperature on the interval $[0, 10]$ was exactly $60^\circ$ C.
The temperature was exactly $40^\circ$ C at time $t=5$ .
The temperature never dropped below $20^\circ$ C during the interval $(0, 10)$ .
The temperature was exactly $40^\circ$ C at some time $t \in (0, 10)$ . (correct answer)

Explanation: When you see a continuous function problem with two known values, immediately think about the Intermediate Value Theorem (IVT). This fundamental theorem states that if a function is continuous on a closed interval

[a,b]

and

k

is any value between

f(a)

and

f(b)

, then there exists at least one point

c

in the interval

(a,b)

where

f(c) = k

. Since

T(t)

is continuous and we know

T(0) = 20

and

T(10) = 60

, the IVT guarantees that for any temperature between 20°C and 60°C, there must be at least one time when the reaction reaches exactly that temperature. Since

40

lies between

20

and

60

, there must be some time

t \in (0,10)

when

T(t) = 40

. This makes D correct. Option A is wrong because the maximum could occur anywhere in the interval, not necessarily at the endpoint. Even though

T(10) = 60

, the temperature could have peaked higher at some interior point. Option B incorrectly assumes the temperature changes linearly. The IVT only guarantees that 40°C occurs somewhere in the interval, not specifically at the midpoint

t = 5

. Option C makes an unwarranted assumption about monotonicity. The function could dip below 20°C during the interval and still satisfy the given conditions, as long as it returns to higher values by

t = 10

. Key strategy: When you see continuous functions with boundary values, immediately consider the Intermediate Value Theorem. It's one of the most testable theorems in Calculus 1 because it connects continuity to guaranteed outcomes.

Question 12

Let $f(x) = \frac{x^2 - 4}{x^3 - 2x^2 - 8x}$ . On which of the following intervals is $f$ continuous?

$[-3, 0)$
$(-1, 3)$ (correct answer)
$(-4, 4)$
$[1, 5]$

Explanation: To find points of discontinuity, we factor the numerator and the denominator. Numerator:

x^2 - 4 = (x-2)(x+2)

. Denominator:

x^3 - 2x^2 - 8x = x(x^2 - 2x - 8) = x(x-4)(x+2)

. So,

f(x) = \frac{(x-2)(x+2)}{x(x-4)(x+2)}

. The function is undefined, and therefore not continuous, at

x=0

x=4

, and

x=-2

. At

x=-2

, the discontinuity is removable since the factor

(x+2)

cancels. At

x=0

and

x=4

, the discontinuities are non-removable (vertical asymptotes). A function is continuous on an interval if it is continuous at every point in that interval. We must find which option does not contain any of the points

x=-2, 0, 4

. A)

[-3, 0)

contains

x=-2

. B)

(-1, 3)

does not contain

-2, 0

, or

4

. C)

(-4, 4)

contains

x=-2

and

x=0

. D)

[1, 5]

contains

x=4

Question 13

Let $f(x) = \frac{1}{x-2}$ and $g(x) = \sqrt{x}$ . The composite function $h(x) = f(g(x))$ is continuous on which of the following sets?

$\{x \in \mathbb{R} \mid x \ge 0 \text{ and } x \ne 2\}$
$\{x \in \mathbb{R} \mid x > 0 \text{ and } x \ne 4\}$
$\{x \in \mathbb{R} \mid x \ge 0 \text{ and } x \ne 4\}$ (correct answer)
$\{x \in \mathbb{R} \mid x \ne 4\}$

Explanation: The composite function is

h(x) = f(g(x)) = f(\sqrt{x}) = \frac{1}{\sqrt{x}-2}

. For

h(x)

to be continuous, it must be defined. Two conditions must be met:

The inner function, $g(x)=\sqrt{x}$ , must be defined. This requires $x \ge 0$ .
The denominator of the resulting function cannot be zero. This means $\sqrt{x}-2 \ne 0$ , which implies $\sqrt{x} \ne 2$ , so $x \ne 4$ . Combining both conditions, the set of all points where $h(x)$ is defined and continuous is all real numbers $x$ such that $x \ge 0$ and $x \ne 4$ .

Question 14

On which of the following closed intervals is the function $h(x) = \ln(4-x^2)$ continuous?

$[-2, 2]$
$[-1, \sqrt{3}]$ (correct answer)
$[0, 3]$
$[-\sqrt{5}, -1]$

Explanation: The function

h(x) = \ln(u)

is continuous wherever

u > 0

. For

h(x) = \ln(4-x^2)

, we need

4-x^2 > 0

, which means

x^2 < 4

. The solution to this inequality is

-2 < x < 2

. So,

h(x)

is continuous on the open interval

(-2, 2)

. A function is continuous on a closed interval

[a, b]

[a, b]

is a subset of the domain of continuity. We must check which of the given closed intervals is a subset of

(-2, 2)

. A)

[-2, 2]

is not a subset because the endpoints

-2

and

2

are not in

(-2, 2)

. B) Since

\sqrt{3} \approx 1.732

, the interval

[-1, \sqrt{3}]

is a subset of

(-2, 2)

. C)

[0, 3]

is not a subset because it includes values greater than

2

. D)

[-\sqrt{5}, -1]

is not a subset because

-\sqrt{5} \approx -2.236

, which is less than

-2

Question 15

Let $g(x) = x^2 - 2x$ and $h(x) = x - 2$ . Let $f(x) = \max(g(x), h(x))$ . Which statement accurately describes the continuity of $f(x)$ ?

$f(x)$ is continuous everywhere except at the points where $g(x)=h(x)$ .
$f(x)$ is continuous for all real numbers. (correct answer)
$f(x)$ is not continuous because its graph has sharp corners.
$f(x)$ is continuous only on intervals where either $g(x) > h(x)$ or $h(x) > g(x)$ , but not at the transition points.

Explanation: Both

g(x)=x^2-2x

and

h(x)=x-2

are polynomials, so they are continuous for all real numbers. The function

f(x) = \max(g(x), h(x))

takes the value of either

g(x)

h(x)

, whichever is larger. A key theorem in calculus states that the maximum (and minimum) of two continuous functions is also a continuous function. To verify, we can check the points where the function might switch definitions, which is where

g(x)=h(x)

x^2 - 2x = x - 2 \implies x^2 - 3x + 2 = 0 \implies (x-1)(x-2)=0

. The transition points are

x=1

and

x=2

. At

x=1

g(1)=-1

and

h(1)=-1

. At

x=2

g(2)=0

and

h(2)=0

. Since

g

and

h

are continuous and they meet at the transition points, the combined function

f(x)

is continuous everywhere. The presence of sharp corners affects differentiability, not continuity.

Question 16

On which of the following intervals is the function $f(x) = \frac{\csc(x)}{2\cos(x) - 1}$ continuous?

$(\frac{\pi}{4}, \frac{3\pi}{4})$
$(\frac{\pi}{3}, \frac{5\pi}{3})$
$(\pi, \frac{5\pi}{3})$ (correct answer)
$(0, \pi)$

Explanation: The function can be written as

f(x) = \frac{1}{\sin(x)(2\cos(x) - 1)}

. Discontinuities occur when the denominator is zero. This happens if

\sin(x)=0

2\cos(x)-1=0

$\sin(x)=0$ when $x=n\pi$ for any integer $n$ . In the range $[0, 2\pi]$ , this occurs at $x=0, \pi, 2\pi$ .
$2\cos(x)-1=0$ when $\cos(x)=1/2$ . In the range $[0, 2\pi]$ , this occurs at $x=\pi/3$ and $x=5\pi/3$ . The points of discontinuity are $0, \pi/3, \pi, 5\pi/3, 2\pi$ , etc. We need to find an interval from the options that avoids all these points. A) $(\pi/4, 3\pi/4)$ contains $\pi/3$ . B) $(\pi/3, 5\pi/3)$ contains $\pi$ . C) $(\pi, 5\pi/3)$ is the interval between two consecutive discontinuities, so $f(x)$ is continuous here. D) $(0, \pi)$ contains $\pi/3$ .

Question 17

On which of the following intervals is the function $f(x) = \frac{x-2}{\sqrt{x^2 - 2x - 3}}$ continuous?

$(-1, 3)$
$[-2, -1]$
$(3, \infty)$
$[4, 10)$ (correct answer)

Explanation: A rational function with a square root is continuous on its domain. The domain requires the expression inside the square root to be strictly positive (since it's also in the denominator). We must solve the inequality

x^2 - 2x - 3 > 0

. Factoring the quadratic gives

(x-3)(x+1) > 0

. The roots are

x=-1

and

x=3

. Since the parabola opens upward, the inequality holds for

x < -1

x > 3

. The domain of

f

(-\infty, -1) \cup (3, \infty)

. The function is continuous on any interval that is a subset of this domain. The interval

[4, 10)

is a subset of

(3, \infty)

, so

f

is continuous on

[4, 10)

Question 18

Let $f(x) = \begin{cases} \sqrt{1-x} & \text{if } x \le 1 \\ \ln(x) & \text{if } x > 1 \end{cases}$ . Which of the following is the largest interval on which $f(x)$ is continuous?

$(-\infty, 1]$
$(0, \infty)$
$(1, \infty)$
$(-\infty, \infty)$ (correct answer)

Explanation: First, check the continuity of each piece on its given interval. The function

\sqrt{1-x}

is continuous for

1-x \ge 0

, which is

x \le 1

. The function

\ln(x)

is continuous for

x > 0

, so it is certainly continuous on its given interval

x > 1

. Next, check for continuity at the breakpoint,

x=1

. We need the one-sided limits to equal the function value.

f(1) = \sqrt{1-1} = 0

\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-} \sqrt{1-x} = \sqrt{1-1} = 0

\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} \ln(x) = \ln(1) = 0

. Since the left-hand limit, right-hand limit, and the function value are all equal to

0

x=1

, the function is continuous at

x=1

. Because the pieces are continuous on their intervals and the function is continuous at the breakpoint,

f(x)

is continuous for all real numbers. Thus, the largest interval of continuity is

(-\infty, \infty)

Question 19

Consider the function $f(x) = \begin{cases} x+1 & \text{if } 0 \le x < 2 \\ 4 & \text{if } x = 2 \end{cases}$ . Which statement explains why $f(x)$ is not continuous on the closed interval $[0, 2]$ ?

$f(x)$ is not continuous on the open interval $(0, 2)$ .
$\lim_{x\to 0^+} f(x) \ne f(0)$ .
$\lim_{x\to 2^-} f(x)$ does not exist.
$\lim_{x\to 2^-} f(x) \ne f(2)$ . (correct answer)

Explanation: For a function to be continuous on a closed interval

[a, b]

, it must be continuous on the open interval

(a, b)

, continuous from the right at

a

, and continuous from the left at

b

On $(0, 2)$ , $f(x)=x+1$ is a polynomial, so it is continuous.
At $x=0$ , we check for continuity from the right: $f(0)=0+1=1$ and $\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} (x+1) = 1$ . Since they are equal, $f$ is continuous from the right at $x=0$ .
At $x=2$ , we check for continuity from the left: $f(2)=4$ . The left-hand limit is $\lim_{x\to 2^-} f(x) = \lim_{x\to 2^-} (x+1) = 3$ . Since $\lim_{x\to 2^-} f(x) = 3$ and $f(2)=4$ , the condition $\lim_{x\to 2^-} f(x) = f(2)$ is not met. This is why the function is not continuous on $[0, 2]$ .

Question 20

Let $f(x) = (1+3x)^{1/x}$ for $x \ne 0$ . If $f$ is defined to be continuous at $x=0$ , what must be the value of $f(0)$ ?

$1$
$e$
$e^3$ (correct answer)
The limit does not exist, so continuity is not possible.

Explanation: For

f

to be continuous at

x=0

f(0)

must equal

\lim_{x\to 0} (1+3x)^{1/x}

. This limit is of the indeterminate form

1^\infty

. We can solve this by letting

y = (1+3x)^{1/x}

and finding the limit of

\ln(y)

\lim_{x\to 0} \ln(y) = \lim_{x\to 0} \frac{\ln(1+3x)}{x}

. Using L'Hôpital's Rule, this is

\lim_{x\to 0} \frac{3/(1+3x)}{1} = 3

. Since

\lim_{x\to 0} \ln(y) = 3

, we have

\lim_{x\to 0} y = e^3

. Alternatively, this is a direct application of the known limit form

\lim_{x\to 0} (1+ax)^{1/x} = e^a

, with

a=3

. Thus,

f(0)

must be

e^3