Calculus 1 Quiz: Intermediate Value Theorem

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Intermediate Value TheoremQuestion 1 of 20

Let $f(x) = x + \ln(x)$ . For which of the following intervals does the Intermediate Value Theorem guarantee a solution to $f(x) = 2$ ?

[0, 1]

[1, 2]

[2, 3]

[e, 3]

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Calculus 1 Quiz

Calculus 1 Quiz: Intermediate Value Theorem

Practice Intermediate Value Theorem 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 Intermediate Value Theorem, 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 $f(x) = x + \ln(x)$ . For which of the following intervals does the Intermediate Value Theorem guarantee a solution to $f(x) = 2$ ?

$[0, 1]$
$[1, 2]$ (correct answer)
$[2, 3]$
$[e, 3]$

Explanation: First, we must consider the domain of

f(x)

. The natural logarithm

\ln(x)

is defined for

x>0

. The function

f(x)

is continuous on its domain

(0, \infty)

. Choice A is invalid because the function is not defined at

x=0

. For the other intervals, we check if the value 2 is between the function values at the endpoints. B)

f(1) = 1 + \ln(1) = 1+0=1

f(2) = 2 + \ln(2)

. Since

\ln(2) > 0

f(2) > 2

. Since

f(1) < 2 < f(2)

, the IVT guarantees a solution in

(1,2)

. C)

f(2) = 2 + \ln(2) > 2

and

f(3) = 3 + \ln(3) > 2

. D)

f(e) = e + \ln(e) = e+1 > 2

and

f(3) = 3 + \ln(3) > 2

Question 2

Let $f(x) = x^3 - 3x + k$ . The Intermediate Value Theorem guarantees that $f(x)$ has a root in the interval $[0, 2]$ for which of the following values of $k$ ?

$k = 3$
$k = 1$
$k = -1$ (correct answer)
$k = -3$

Explanation: For the Intermediate Value Theorem (IVT) to guarantee a root of

f(x)=0

[0, 2]

, the function must be continuous on that interval (which it is, as a polynomial), and the value 0 must be between

f(0)

and

f(2)

. This occurs if

f(0)

and

f(2)

have opposite signs. We evaluate the function at the endpoints:

f(0) = (0)^3 - 3(0) + k = k

and

f(2) = (2)^3 - 3(2) + k = 8 - 6 + k = 2 + k

. We need

k(2+k) < 0

. This inequality holds if and only if

-2 < k < 0

. Among the given choices, only

k = -1

is in this interval.

Question 3

Consider the function $f(x) = x^4 - 2x^3 - 5$ . The Intermediate Value Theorem guarantees that $f(x)$ has a root on which of the following intervals?

$[-1, 0]$
$[0, 1]$
$[1, 2]$
$[2, 3]$ (correct answer)

Explanation: As a polynomial,

f(x)

is continuous everywhere. To guarantee a root on an interval

[a, b]

, the IVT requires

f(a)

and

f(b)

to have opposite signs. We check each interval: A)

f(-1) = (-1)^4 - 2(-1)^3 - 5 = 1 + 2 - 5 = -2

and

f(0) = -5

. Both are negative. B)

f(0) = -5

and

f(1) = 1 - 2 - 5 = -6

. Both are negative. C)

f(1) = -6

and

f(2) = 2^4 - 2(2)^3 - 5 = 16 - 16 - 5 = -5

. Both are negative. D)

f(2) = -5

and

f(3) = 3^4 - 2(3)^3 - 5 = 81 - 54 - 5 = 22

. The signs are opposite. Therefore, the IVT guarantees a root on

[2, 3]

Question 4

Let $f(x) = e^x - x^2$ . On which of the following closed intervals does the Intermediate Value Theorem guarantee that $f(x)$ has at least one root?

$[0, 1]$
$[-1, 0]$ (correct answer)
$[1, 2]$
$[-2, -1]$

Explanation: The function

f(x) = e^x - x^2

is continuous everywhere. To use the IVT to guarantee a root on an interval

[a, b]

, we need

f(a)

and

f(b)

to have opposite signs. We check each interval: A)

f(0) = e^0 - 0^2 = 1

and

f(1) = e^1 - 1^2 = e-1 > 0

. Signs are the same. B)

f(-1) = e^{-1} - (-1)^2 = 1/e - 1 < 0

and

f(0) = 1 > 0

. Signs are opposite. C)

f(1) = e-1 > 0

and

f(2) = e^2 - 4 > 0

. Signs are the same. D)

f(-2) = e^{-2} - (-2)^2 = 1/e^2 - 4 < 0

and

f(-1) = 1/e - 1 < 0

. Signs are the same. The only interval where the function values at the endpoints have opposite signs is

[-1, 0]

Question 5

A function $g(x)$ is defined on the interval $[0, 10]$ . It is known that $g(0) = 5$ and $g(10) = 25$ . At $x=4$ , it is observed that $g(4) = 15$ . Based on this information, what must be true about the function $g(x)$ ?

The function $g(x)$ must be continuous on the interval $[0, 10]$ because it takes on an intermediate value.
The function $g(x)$ must be increasing on the interval $[0, 10]$ because $g(0) < g(4) < g(10)$ .
The function $g(x)$ is not necessarily continuous on the interval $[0, 10]$ . (correct answer)
There must be exactly one value $c \in (0, 10)$ for which $g(c)=15$ .

Explanation: The Intermediate Value Theorem states that if a function is continuous, then it takes on every intermediate value. The converse is not true. A function taking on an intermediate value does not prove that the function is continuous. The function could have a jump discontinuity, for example, but still happen to pass through the value 15. Therefore, we cannot conclude that

g(x)

must be continuous.

Question 6

In which of the following cases can the Intermediate Value Theorem NOT be used to conclude that the equation has a solution in the given interval?

$\frac{2}{x-3} = -1$ in the interval $[0, 4]$ . (correct answer)
$\cos(x) = x$ in the interval $[0, \pi/2]$ .
$x^5 - 3x - 1 = 0$ in the interval $[1, 2]$ .
$e^x = 4-x^2$ in the interval $[0, 2]$ .

Explanation: The Intermediate Value Theorem (IVT) is a powerful tool for proving that equations have solutions. It states that if a function is continuous on a closed interval

[a,b]

and takes on values

f(a)

and

f(b)

with opposite signs, then there must be some point

c

in the interval where

f(c) = 0

. To apply the IVT, you need to rewrite each equation in the form

f(x) = 0

and check two things: is the function continuous on the given interval, and does it change sign? Choice A is the correct answer because

f(x) = \frac{2}{x-3} + 1

is not continuous on

[0,4]

. The function has a vertical asymptote at

x = 3

, which lies within the interval, creating a discontinuity that violates the IVT's requirements. Choice B works because

f(x) = \cos(x) - x

is continuous on

[0, \pi/2]

, with

f(0) = 1 > 0

and

f(\pi/2) = -\pi/2 < 0

, satisfying both IVT conditions. Choice C works because

f(x) = x^5 - 3x - 1

is a polynomial (continuous everywhere), with

f(1) = -3 < 0

and

f(2) = 25 > 0

. Choice D works because

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

is continuous on

[0,2]

, with

f(0) = -3 < 0

and

f(2) = e^2 > 0

. Study tip: Before applying IVT, always check for discontinuities like vertical asymptotes, holes, or jumps within your interval. Continuity is non-negotiable for the theorem to work.

Question 7

Let $h(x) = x^2 - k$ , where $k$ is a constant. The function is continuous on the interval $[1, 3]$ . The Intermediate Value Theorem guarantees that $h(x)=0$ for some $x \in (1, 3)$ if and only if $k$ is in which interval?

$k > 1$
$k > 9$
$0 < k < 1$
$1 < k < 9$ (correct answer)

Explanation: The Intermediate Value Theorem (IVT) states that if a function is continuous on a closed interval and takes on two different values at the endpoints, then it must take on every value between those endpoints somewhere in the interval. When you see IVT problems asking "when does the function equal zero," you need to check when zero falls between the function values at the endpoints. For

h(x) = x^2 - k

, let's find the endpoint values on

[1,3]

$h(1) = 1^2 - k = 1 - k$
$h(3) = 3^2 - k = 9 - k$

For the IVT to guarantee that

h(x) = 0

for some

x \in (1,3)

, zero must lie strictly between

h(1)

and

h(3)

. Since

h(x)

is increasing (it's a parabola opening upward), we need

h(1) < 0 < h(3)

. This gives us:

$1 - k < 0$ , so $k > 1$
$9 - k > 0$ , so $k < 9$

Therefore,

1 < k < 9

. Looking at the wrong answers: Choice A (

k > 1

) is incomplete—it doesn't ensure

h(3) > 0

. Choice B (

k > 9

) makes both endpoint values negative, so zero isn't between them. Choice C (

0 < k < 1

) makes both endpoint values positive, again failing the IVT requirement. Study tip: For IVT problems, always evaluate the function at both endpoints first, then determine what conditions make zero fall between those values. The key is ensuring the function changes sign across the interval.

Question 8

Let $f$ be a function that is continuous on the open interval $(0, 5)$ . It is known that $\lim_{x\to 0^+} f(x) = -\infty$ and $\lim_{x\to 5^-} f(x) = +\infty$ . What can be concluded about the roots of $f(x)$ ?

The IVT cannot be applied because the function is not defined on a closed interval.
The function $f(x)$ is guaranteed to have at least one root in $(0, 5)$ . (correct answer)
The function $f(x)$ is guaranteed to have exactly one root in $(0, 5)$ because it goes from $-\infty$ to $+\infty$ .
Nothing can be concluded about the roots of $f(x)$ without knowing its values at specific points.

Explanation: Although the Intermediate Value Theorem is formally stated for a closed interval

[a, b]

, we can use the given limit information to find such an interval. Since

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

, there must be a number

a

very close to 0 (with

a > 0

) such that

f(a)

is negative. Similarly, since

\lim_{x\to 5^-} f(x) = +\infty

, there must be a number

b

very close to 5 (with

b < 5

) such that

f(b)

is positive. The function

f

is continuous on the closed interval

[a, b]

(since

[a, b] \subset (0, 5)

). Because

f(a) < 0

and

f(b) > 0

, the IVT guarantees there is a root

c \in (a, b)

, which is also in

(0, 5)

Question 9

Let $f$ be a function defined by $f(x) = \begin{cases} x+3 & \text{if } x \le 1 \\ -x+1 & \text{if } x > 1 \end{cases}$ . The function is considered on the closed interval $[0, 3]$ . Which of the following statements about applying the Intermediate Value Theorem (IVT) for a value $N=2$ is correct?

The IVT guarantees a value $c \in (0, 3)$ such that $f(c) = 2$ because $f(3) < 2 < f(0)$ .
The IVT guarantees a value $c \in (0, 3)$ such that $f(c) = 2$ because a solution exists at $x=-1$ .
The IVT cannot be used to guarantee a value $c \in (0, 3)$ such that $f(c)=2$ because $f$ is not continuous on $[0, 3]$ . (correct answer)
The IVT cannot be used to guarantee a value $c \in (0, 3)$ such that $f(c)=2$ because $N=2$ is not between the function's minimum and maximum values on the interval.

Explanation: The Intermediate Value Theorem requires the function to be continuous on the closed interval. We check for continuity at

x=1

. The limit from the left is

\lim_{x\to 1^-} f(x) = 1+3=4

. The limit from the right is

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

. Since the left-hand and right-hand limits are not equal, the function is not continuous at

x=1

. Because

x=1

is in the interval

[0, 3]

, the IVT cannot be applied.

Question 10

An object's velocity, $v(t)$ , is a continuous function of time $t$ for $t \in [0, 8]$ . At time $t=0$ , the object is moving right with velocity $v(0) = 5$ m/s. At time $t=8$ , it is moving left with velocity $v(8) = -3$ m/s. Also, it is known that the object's velocity is never exactly $1$ m/s. Which of the following statements must be true?

The object must be at rest at some time $t \in (0, 8)$ .
The velocity function $v(t)$ cannot be continuous on $[0, 8]$ . (correct answer)
The object's acceleration must be negative for all $t \in (0, 8)$ .
The object's velocity must be greater than $1$ m/s for all $t \in (0, 8)$ .

Explanation: The Intermediate Value Theorem states that if a function is continuous on a closed interval

[a, b]

, it must take on every value between

f(a)

and

f(b)

. Here, the function is velocity

v(t)

, the interval is

[0, 8]

v(0) = 5

, and

v(8) = -3

. If

v(t)

were continuous, it would have to take on every value between

-3

and

5

. The value

N=1

is between

-3

and

5

. However, the problem states that the velocity is never exactly

1

m/s. This is a direct contradiction to the conclusion of the IVT. Since the conclusion is false, one of the hypotheses must be false. The only hypothesis that can be false is that

v(t)

is continuous on

[0, 8]

Question 11

Let $f(x) = x^3 - 6x + 2$ . The Intermediate Value Theorem can be used to prove that the equation $f(x)=N$ has a solution in the interval $[0, 2]$ . Which of the following intervals represents the largest set of values for $N$ for which a solution is guaranteed?

$[0, 2]$
$[-2, 2]$
$[2-4\sqrt{2}, 2]$ (correct answer)
$[2-4\sqrt{2}, 2+4\sqrt{2}]$

Explanation: The Intermediate Value Theorem guarantees that a continuous function on

[a, b]

takes on all values between its minimum and maximum on that interval. First, find the endpoint values:

f(0) = 2

and

f(2) = 2^3 - 6(2) + 2 = 8 - 12 + 2 = -2

. To find the absolute extrema on

[0, 2]

, we check critical points.

f'(x) = 3x^2 - 6

. Setting

f'(x)=0

gives

x^2=2

, so

x = \sqrt{2}

(since

-\sqrt{2}

is not in

[0,2]

). The value at the critical point is

f(\sqrt{2}) = (\sqrt{2})^3 - 6\sqrt{2} + 2 = 2\sqrt{2} - 6\sqrt{2} + 2 = 2 - 4\sqrt{2}

. The values of the function at the endpoints and critical point are

2

-2

, and

2 - 4\sqrt{2} \approx 2 - 5.656 = -3.656

. The absolute minimum on

[0, 2]

2 - 4\sqrt{2}

and the absolute maximum is

2

. The IVT guarantees a solution for any

N

in the range of the function, which is

[2-4\sqrt{2}, 2]

Question 12

Let $f(x) = x^3 - 3x^2 + 5$ . The Intermediate Value Theorem guarantees that there is at least one solution to $f(x)=3$ in the interval $(-1, 3)$ , since $f(-1)=1$ and $f(3)=5$ . What is a valid conclusion about the number of solutions to $f(x)=3$ in this interval?

There is exactly one solution in $(-1, 3)$ because the function is a polynomial.
There are at least two solutions in $(-1, 3)$ because the function is not monotonic.
There may be no solutions, as the IVT only applies to finding roots (where $f(x)=0$ ).
There is at least one solution in $(-1, 3)$ , but the IVT alone does not determine if there are more. (correct answer)

Explanation: When you encounter questions about the Intermediate Value Theorem (IVT), remember that this theorem tells you about the existence of solutions, not their uniqueness. The IVT states that if a function is continuous on a closed interval and takes on two different values, it must take on every value between them at least once. Here,

f(x) = x^3 - 3x^2 + 5

is continuous (polynomials are always continuous), and we have

f(-1) = 1

and

f(3) = 5

. Since

1 < 3 < 5

, the IVT guarantees that

f(x) = 3

has at least one solution in

(-1, 3)

. However, the IVT alone cannot tell us whether there's exactly one solution or multiple solutions. Choice A is incorrect because being a polynomial doesn't guarantee uniqueness of solutions—polynomials can have multiple solutions to

f(x) = k

for any constant

k

. Choice B incorrectly assumes that non-monotonic behavior automatically means multiple solutions exist, but we'd need to analyze the function's behavior more carefully to confirm this. Choice C shows a fundamental misunderstanding—the IVT applies to any horizontal line

f(x) = k

, not just

f(x) = 0

. Choice D correctly states that while the IVT guarantees at least one solution, it doesn't determine whether additional solutions exist. To find the exact number of solutions, you'd need to use calculus tools like analyzing critical points or graphing. Study tip: Remember that the IVT is an existence theorem, not a uniqueness theorem. It guarantees "at least one" but never "exactly one."

Question 13

Let $f(x)$ be a function defined by $f(x) = \frac{x^2+x-6}{x-2}$ for $x \neq 2$ , and $f(2)=3$ . We evaluate $f(0)=-3$ and $f(3)=6$ . A student concludes that because $0$ is between $-3$ and $6$ , the IVT guarantees a solution to $f(c)=0$ in $(0,3)$ . Why is this conclusion invalid?

The function $f(x)$ is not defined at $x=2$ , which is in the interval $(0,3)$ .
The student made a calculation error; $f(0)$ and $f(3)$ do not have opposite signs.
The function $f(x)$ is not continuous at $x=2$ , so the IVT does not apply on $[0,3]$ . (correct answer)
The IVT only guarantees a solution if the function is monotonic on the interval.

Explanation: The Intermediate Value Theorem requires the function to be continuous on the closed interval. We must check for continuity at

x=2

. The limit as

x \to 2

\lim_{x \to 2} \frac{(x+3)(x-2)}{x-2} = \lim_{x \to 2} (x+3) = 5

. However, the function is defined as

f(2)=3

. Since

\lim_{x \to 2} f(x) \neq f(2)

, the function has a removable discontinuity at

x=2

. Because the function is not continuous on the entire interval

[0,3]

, the IVT cannot be applied, and its conclusion is not guaranteed.

Question 14

Let $f$ be a function that is continuous on $[a, b]$ . The Intermediate Value Theorem guarantees that for any $N$ between $f(a)$ and $f(b)$ , there exists at least one $c \in (a, b)$ such that $f(c) = N$ . Which of the following is NOT guaranteed by the theorem?

The value $c$ is in the open interval $(a, b)$ .
If $f(a) < f(b)$ , then the range of $f$ on $[a,b]$ contains the interval $[f(a), f(b)]$ .
There is only one such value $c$ in the interval $(a, b)$ . (correct answer)
The function $f$ must be defined at every point in $[a, b]$ .

Explanation: The Intermediate Value Theorem is an existence theorem; it guarantees the existence of at least one value

c

but says nothing about its uniqueness. There could be multiple values of

c

(a, b)

for which

f(c) = N

. For example, if

f(x) = \sin(x)

[0, 2\pi]

, the equation

f(x)=0.5

has two solutions. The other statements are part of the theorem's conditions or direct consequences: the theorem specifies

c \in (a, b)

, the continuity condition implies

f

is defined on

[a,b]

, and the theorem implies the range contains all values between

f(a)

and

f(b)

Question 15

Let $f(x)$ and $g(x)$ be functions that are continuous for all real numbers. Let $h(x) = f(g(x))$ . Suppose $g(0)=2$ , $g(1)=5$ , $f(2)=10$ , and $f(5)=-4$ . Which of the following is guaranteed by the Intermediate Value Theorem?

There exists $c \in (0, 1)$ such that $h(c) = 0$ . (correct answer)
There exists $c \in (2, 5)$ such that $h(c) = 0$ .
There exists $c \in (0, 1)$ such that $g(c) = 6$ .
There exists $c \in (2, 5)$ such that $f(c) = 11$ .

Explanation: The composition of continuous functions is continuous, so

h(x)

is continuous. We evaluate

h(x)

at the endpoints of the interval

[0, 1]

h(0) = f(g(0)) = f(2) = 10

h(1) = f(g(1)) = f(5) = -4

. Since

h(x)

is continuous on

[0, 1]

and

0

is between

h(0)=10

and

h(1)=-4

, the IVT guarantees there is a

c \in (0, 1)

such that

h(c) = 0

. Choice B is incorrect because the domain of

h

under consideration is based on the domain of

g

, which is

[0, 1]

. Choices C and D are incorrect applications of the IVT as the target values (6 and 11) are outside the range of the respective functions on the given intervals.

Question 16

Let $f(x)$ be a continuous function that maps the interval $[0, 1]$ to itself, i.e., for any $x \in [0, 1]$ , $f(x) \in [0, 1]$ . The existence of a fixed point $c \in [0, 1]$ such that $f(c) = c$ is guaranteed by applying the Intermediate Value Theorem to which function on $[0, 1]$ ?

$g(x) = f(x)$
$g(x) = f(x) - x$ (correct answer)
$g(x) = f(x) + x$
$g(x) = (f(x))^2 - x^2$

Explanation: To find a fixed point

f(c)=c

, we can look for a root of the auxiliary function

g(x) = f(x) - x

. Since

f(x)

and

x

are continuous,

g(x)

is continuous on

[0, 1]

. We check the values of

g(x)

at the endpoints. At

x=0

g(0) = f(0) - 0 = f(0)

. Since the range of

f

[0, 1]

f(0) \ge 0

, so

g(0) \ge 0

. At

x=1

g(1) = f(1) - 1

. Since the range of

f

[0, 1]

f(1) \le 1

, so

g(1) \le 0

. Since

g(x)

is continuous and its values at the endpoints are on opposite sides of (or on) zero, the IVT guarantees there exists a

c \in [0, 1]

such that

g(c) = 0

, which implies

f(c) - c = 0

, or

f(c) = c

Question 17

Let $f(x) = |x-2| - 1$ . The function is continuous everywhere. We note that $f(0)=1$ and $f(4)=1$ . What can be concluded about the roots of $f(x)$ on the interval $[0,4]$ using the Intermediate Value Theorem?

The IVT is inconclusive because $f(0)$ and $f(4)$ have the same sign. (correct answer)
The IVT guarantees there are no roots in the interval $(0,4)$ .
The IVT guarantees there is at least one root in the interval $(0,4)$ .
The IVT cannot be applied because the absolute value function is not differentiable at $x=2$ .

Explanation: The Intermediate Value Theorem can only guarantee a root on an interval

[a,b]

f(a)

and

f(b)

have opposite signs. In this case,

f(0)=1

and

f(4)=1

, which are both positive. Therefore, the IVT is inconclusive; it cannot confirm or deny the existence of a root in

(0,4)

. Although this particular function does have roots at

x=1

and

x=3

, the IVT, based solely on the endpoint values, cannot be used to find them. The differentiability of the function is a condition for the Mean Value Theorem, not the Intermediate Value Theorem, which only requires continuity.

Question 18

Let $f$ be a continuous function on $[0, 2]$ with $f(0)=5$ and $f(2)=1$ . The Intermediate Value Theorem guarantees the existence of a number $c$ in $(0, 2)$ such that $f(c)=N$ . Which of the following is a possible value for $N$ ?

$N=0$
$N=\pi$ (correct answer)
$N=5$
$N=6$

Explanation: The Intermediate Value Theorem guarantees that for a continuous function

f

[a,b]

, the function takes on every value

N

between

f(a)

and

f(b)

. Here, the interval is

[0, 2]

, and the function values at the endpoints are

f(0)=5

and

f(2)=1

. Therefore, the IVT guarantees a solution for any

N

such that

1 < N < 5

. We check the options: A) 0 is not in

(1,5)

. B)

\pi \approx 3.14159

, which is in

(1,5)

. C) 5 is an endpoint value; the theorem guarantees a

c

in the open interval

(a,b)

for values between the endpoint values. D) 6 is not in

(1,5)

Question 19

Let $f(x) = \frac{1}{x-3}$ . A student computes $f(2) = -1$ and $f(4) = 1$ and concludes that there must be a root in $(2, 4)$ . Which of the following is the most significant error in the student's reasoning?

The student made a calculation error; $f(2)$ and $f(4)$ are both positive.
The IVT requires the interval to be centered at the origin, but $(2, 4)$ is not.
The IVT cannot be applied because the function is not continuous on the interval $[2, 4]$ . (correct answer)
The IVT only proves the existence of a root; it does not guarantee its location in a specific interval.

Explanation: The primary condition for applying the Intermediate Value Theorem is that the function must be continuous on the closed interval. The function

f(x) = \frac{1}{x-3}

has an infinite discontinuity at

x=3

, which lies within the interval

[2, 4]

. Because the function is not continuous on this interval, the IVT is not applicable, and its conclusion cannot be drawn. In fact, the function

f(x) = \frac{1}{x-3}

is never equal to zero.

Question 20

A function $f(x)$ is continuous on $[0, 10]$ . It is known that $f(0) = 5$ , $f(10) = 5$ , and for some $a \in (0, 10)$ , $f(a) = 12$ . Which conclusion is guaranteed by the Intermediate Value Theorem?

The function must have a root in $(0, 10)$ .
There is exactly one solution to $f(x)=8$ in $(0, 10)$ .
There are at least two solutions to $f(x)=8$ in $(0, 10)$ . (correct answer)
The function must have a minimum value of 5 on $[0, 10]$ .

Explanation: We can apply the IVT on two sub-intervals. First, consider the interval

[0, a]

. Since

f

is continuous,

f(0) = 5

and

f(a) = 12

, and 8 is between 5 and 12, there must be a solution

c_1 \in (0, a)

f(x)=8

. Second, consider the interval

[a, 10]

. Since

f

is continuous,

f(a) = 12

and

f(10) = 5

, and 8 is between 5 and 12, there must be another solution

c_2 \in (a, 10)

f(x)=8

. Since

c_1

and

c_2

are in non-overlapping intervals, they are distinct. Thus, there are at least two solutions in

(0, 10)