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📝 Extreme value theorem absolute extrema (25 MCQs)

📖 From Calculus • 5. The derivative in Graphing and Applications • 25 questions available

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What is Extreme value theorem absolute extrema?

Definition:
The Extreme Value Theorem states that if $f$ is continuous on a closed interval $[a, b]$ , then $f$ attains both an absolute maximum and an absolute minimum on that interval. This guarantees existence but not location, requiring evaluation of critical points and endpoints.

Example:
Since $f(x) = \sin(x)$ is continuous on $[0, \pi]$ , it must have an absolute max ( $1$ at $\pi/2$ ) and min ( $0$ at $0, \pi$ ).

Reason:
This theorem provides the theoretical foundation for optimization problems, ensuring solutions exist for continuous functions on bounded closed domains before calculation begins.

Easy

Medium

Hard

📝 All Extreme value theorem absolute extrema MCQs

Q1. Suppose $f$ is continuous on $[0,2]$ except that it has a removable discontinuity at $x=1$ . Which statement is true regarding the existence of absolute extrema on $[0,2]$ ?

A.Both absolute maximum and minimum must exist

B.At least one of them must exist

C.Neither is guaranteed ✅

D.Exactly one is guaranteed

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Because the Extreme Value Theorem requires continuity on the entire closed interval, a single point of discontinuity invalidates the guarantee. A removable discontinuity means the function is not continuous at that point, so we cannot conclude that absolute extrema must occur; they may or may not exist.

Q2. If a function $g$ is continuous on $[a,b]$ and attains its maximum at an interior point $c\in(a,b)$ , what can be inferred about $c$ ?

c

is a critical point where

g&#039;(c)=0

✅

c

must be a point where

g

is not differentiable

c

is necessarily a local minimum

c

cannot be an endpoint

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: By the theorem, any interior point where an absolute extremum occurs must also be a relative extremum, making it a critical point. Hence the derivative at that point either equals zero or does not exist. Since the function is continuous and differentiable in the interior, the derivative must be zero at $c$ .

Q3. Consider a function $h$ that is continuous on $[ -3,3]$ and differentiable on $(-3,3)$ except at $x=0$ where $h'$ does not exist. Which of the following statements is correct?

h

must have an absolute extremum at

x=0

h

may have an absolute extremum at an endpoint ✅

h

cannot have any absolute extremum

h

must have both absolute extrema at interior points

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Continuity on the closed interval guarantees the existence of absolute extrema, but they need not occur at points where the derivative fails. The extrema can appear at the endpoints $-3$ or $3$ or at critical points where the derivative is zero. Since differentiability fails at $0$ , it is not forced to be an extremum.

Q4. A continuous function $p$ on $[0,1]$ has critical points at $x=0.2$ and $x=0.8$ . Which procedure correctly identifies its absolute maximum?

A.Compare values at the critical points only

B.Compare values at the endpoints only

C.Compare values at both endpoints and critical points ✅

D.Evaluate the second derivative at critical points

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The Extreme Value Theorem states that absolute extrema occur either at endpoints or at interior critical points. Therefore, to locate the absolute maximum, one must evaluate $p$ at $x=0,1,0.2,$ and $0.8$ and then compare these four values. The largest among them is the absolute maximum.

Q5. Let $q(x)=\frac{x^3}{3}-x$ on $[-2,2]$ . Using the Extreme Value Theorem, which statement correctly describes the location of its absolute minimum?

A.It occurs at

x=-2

B.It occurs at

x=2

C.It occurs at a critical point inside

(-2,2)

✅

D.It does not exist because

q

is not bounded

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: First find critical points by setting $q'(x)=x^2-1=0$ , yielding $x=\pm1$ . Evaluate $q$ at $-2,-1,1,2$ . The smallest value occurs at $x=-1$ , an interior critical point, confirming that the absolute minimum is located inside the interval rather than at an endpoint.

Q6. A function $r$ is continuous on $[0,5]$ and differentiable on $(0,5)$ . It has critical points at $x=1$ and $x=4$ . Which of the following guarantees that $r$ attains its absolute maximum at $x=4$ ?

r(4) > r(1)

and

r(4) > r(0), r(5)

✅

r&#039;(4)=0

r

is increasing on

(4,5)

r

is concave down on

(0,5)

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The Extreme Value Theorem ensures an absolute maximum exists, but to guarantee it occurs at $x=4$ we must compare the function values at all possible locations: the endpoints and the critical points. If the value at $x=4$ exceeds those at $x=0,5,1$ , then $x=4$ is the absolute maximum.

Q7. Suppose $s$ is continuous on $[a,b]$ and attains its absolute minimum at an interior point $c$ . Which theorem explains why $c$ must be a critical point?

A.Mean Value Theorem

B.Intermediate Value Theorem

C.Fermat’s Theorem on stationary points ✅

D.Rolle’s Theorem

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Fermat’s Theorem (a consequence of the Extreme Value Theorem) states that if a function has a local extremum at an interior point and is differentiable there, the derivative must be zero, making the point critical. Since an absolute minimum is also a local minimum, the theorem applies, ensuring $c$ is a critical point.

Q8. A function $t(x)=\sin(x)$ is considered on the interval $[0,\pi]$ . Which of the following correctly applies the Extreme Value Theorem to identify its absolute extrema?

A.Absolute max at

x=0

, min at

x=\pi

B.Absolute max at

x=\frac{\pi}{2}

, min at

x=0

C.Absolute max at

x=\frac{\pi}{2}

, min at

x=\frac{3\pi}{2}

D.Absolute max at

x=\frac{\pi}{2}

, min at

x=\pi

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: On $[0,\pi]$ , $\sin(x)$ reaches its highest value $1$ at $x=\frac{\pi}{2}$ and its lowest value $0$ at the endpoint $x=\pi$ . The Extreme Value Theorem guarantees these extrema because the function is continuous on the closed interval.

Q9. If a function $u$ is continuous on $[2,6]$ but not differentiable at $x=4$ , which of the following is true about its absolute extrema?

A.They must occur at

x=4

B.They must occur at the endpoints

C.They may occur at either endpoints or at

x=4

✅

D.They cannot exist

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Continuity on a closed interval guarantees the existence of absolute extrema, but differentiability is not required. Hence the extrema could be located at the endpoints or at any interior point where the derivative fails, such as $x=4$ .

Q10. Consider a function $v$ defined on $(-\infty,\infty)$ that is continuous everywhere and unbounded above. Which statement best reflects the Extreme Value Theorem?

v

has an absolute maximum

v

has an absolute minimum

v

has neither absolute maximum nor minimum ✅

D.The theorem does not apply because the interval is not closed

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The Extreme Value Theorem applies only to finite closed intervals. Since the domain is the entire real line, which is not closed and bounded, the theorem gives no guarantee of absolute extrema. An unbounded function further confirms the absence of an absolute maximum.

Q11. A continuous function $w$ on $[0,4]$ has critical points at $x=1$ and $x=3$ . Its values are: $w(0)=2$ , $w(1)=5$ , $w(3)=1$ , $w(4)=4$ . What is the absolute maximum?

w(0)=2

w(1)=5

✅

w(3)=1

w(4)=4

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: By evaluating the function at all endpoints and critical points, we compare the values: 2, 5, 1, and 4. The largest value is 5 at $x=1$ , making it the absolute maximum according to the Extreme Value Theorem.

Q12. Which of the following intervals fails to satisfy the hypotheses of the Extreme Value Theorem for a continuous function $f$ ?

[0,1]

(0,1]

✅

[0,1)

(0,1)

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The theorem requires the interval to be both closed and bounded. $(0,1]$ is not closed at the left endpoint, so the hypothesis is violated. The other intervals are either closed ( $[0,1]$ ) or both open ( $(0,1)$ ), which also fails, but the question asks for a single example; $(0,1]$ clearly lacks a closed left endpoint.

Q13. A function $z$ is continuous on $[a,b]$ and differentiable on $(a,b)$ . It has exactly one critical point $c$ where $z'(c)=0$ . Which conclusion is always true?

z

attains its absolute maximum at

c

z

attains its absolute minimum at

c

z

attains either its absolute maximum or minimum at

c

z

may have both extrema at the endpoints ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The existence of a single critical point does not guarantee that it is an extremum; the absolute maximum and minimum could both lie at the endpoints. The Extreme Value Theorem only ensures that extrema exist somewhere in $[a,b]$ , not necessarily at the interior critical point.

Q14. If a function $f$ is continuous on $[0,2]$ and its derivative $f'$ is positive on $(0,2)$ except at a single point where $f'$ is zero, what can be deduced about absolute extrema?

f

has an absolute maximum at the point where

f&#039;=0

f

has an absolute minimum at the point where

f&#039;=0

f

is strictly increasing, so the absolute minimum is at

x=0

and maximum at

x=2

✅

f

has no absolute extrema

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A positive derivative throughout an interval (except possibly at isolated points) indicates the function is increasing. Therefore, the smallest value occurs at the left endpoint $x=0$ and the largest at the right endpoint $x=2$ , regardless of the isolated zero of the derivative.

Q15. A function $g(x)=\ln(x)$ is considered on the interval $[1, e]$ . Which statement correctly uses the theorem to describe its extrema?

A.Absolute maximum at

x=1

B.Absolute minimum at

x=e

C.Absolute maximum at

x=e

and minimum at

x=1

✅

D.No absolute extrema exist because the domain is unbounded

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The natural logarithm is continuous on $[1,e]$ . It increases as $x$ increases, so the smallest value occurs at the left endpoint $x=1$ ( $\ln 1 =0$ ) and the largest at the right endpoint $x=e$ ( $\ln e =1$ ). Thus, the absolute minimum is at $x=1$ and the absolute maximum at $x=e$ .

Q16. Consider a piecewise function $h(x)=\begin{cases}x^2 & x\le1\\2-x & x>1\end{cases}$ on $[0,2]$ . Which point must be checked to apply the theorem for absolute extrema?

A.Only the endpoints

0

and

2

B.Only the point

x=1

where the definition changes

C.Both endpoints and

x=1

✅

D.Only the interior critical points of each piece

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The function is continuous on the closed interval, but the point $x=1$ is where the definition changes and could be a candidate for an extremum. Therefore, to locate absolute extrema, we must evaluate the function at the endpoints $0$ and $2$ and at the transition point $1$ .

Q17. A continuous function $p(x)$ on $[ -5,5]$ attains its absolute maximum at $x=-5$ . Which inference is logically valid?

p

has no critical points in

(-5,5)

p

must be decreasing on

(-5,5)

C.The derivative at

x=-5

is zero

D.The absolute maximum can only occur at endpoints if the function is monotonic ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: If the absolute maximum occurs at an endpoint, the function need not be monotonic; it could rise and fall within the interval. However, the Extreme Value Theorem tells us that extrema can appear at endpoints regardless of interior behavior. Thus, statement D correctly captures that endpoints can host absolute extrema without implying monotonicity.

Q18. Which of the following statements best captures the necessity of the closed interval condition in the theorem?

A.Closed intervals guarantee differentiability

B.Closed intervals guarantee boundedness of continuous functions ✅

C.Closed intervals ensure the function is linear

D.Closed intervals make the function periodic

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A continuous function on a closed and bounded interval is guaranteed to be bounded, which is essential for the existence of absolute extrema. The closed nature prevents “escaping” to infinity at the endpoints, unlike open intervals where extrema may not exist.

Q19. A function $f$ is continuous on $[0,3]$ and differentiable on $(0,3)$ except at $x=2$ where $f'$ does not exist. If $f(0)=4$ , $f(2)=7$ , and $f(3)=5$ , what is the absolute maximum?

f(0)=4

f(2)=7

✅

f(3)=5

D.Cannot be determined without more information

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Since the function values at all candidates (endpoints and the interior point where the derivative fails) are known, the largest value is $7$ at $x=2$ . The Extreme Value Theorem assures an absolute maximum exists, and among the evaluated points, $f(2)$ is greatest.

Q20. If a function $k$ is continuous on $[0,1]$ and has no critical points in $(0,1)$ , where must its absolute extrema lie?

A.Both must be at interior points

B.Both must be at endpoints ✅

C.One at an endpoint, one interior

D.Cannot be determined

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Without interior critical points, the only possible locations for absolute extrema are the endpoints of the closed interval, as guaranteed by the Extreme Value Theorem. Hence both the absolute maximum and minimum occur at either $x=0$ or $x=1$ .

Q21. A function $m(x)=x^3-3x$ is examined on $[-2,2]$ . After finding critical points, which of the following correctly identifies the absolute minimum?

A.At

x=-2

B.At

x=2

C.At the critical point

x=-1

✅

D.At the critical point

x=1

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Set $m'(x)=3x^2-3=0$ giving $x=\pm1$ . Evaluate $m$ at $-2,-1,1,2$ : $m(-2)=-2$ , $m(-1)=2$ , $m(1)=-2$ , $m(2)=2$ . The smallest value is $-2$ occurring at both $x=-2$ and $x=1$ ; however, the interior critical point $x=1$ yields the same minimum, satisfying the theorem. Thus $x=1$ is an absolute minimum.

Q22. Which scenario demonstrates that continuity alone is insufficient for guaranteeing absolute extrema on an open interval?

f(x)=x

(0,1)

✅

f(x)=\sin(x)

[0,2\pi]

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

(0,1]

f(x)=x^2

[-1,1]

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: On the open interval $(0,1)$ , the function $f(x)=x$ is continuous but lacks endpoints, so it has no absolute maximum or minimum within the interval; it can approach but never attain the supremum or infimum. This illustrates that continuity without a closed interval does not ensure absolute extrema, unlike the theorem’s requirements.

Q23. A continuous function $n$ on $[0,10]$ has absolute maximum value $20$ at $x=3$ and absolute minimum value $-5$ at $x=7$ . Which of the following must be true?

n&#039;(3)=0

and

n&#039;(7)=0

n

is increasing on

[0,3]

and decreasing on

[7,10]

n

attains its extrema only at critical points

n

may have other critical points besides

3

and

7

✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The theorem guarantees that extrema occur at endpoints or critical points, but it does not restrict the function from having additional critical points elsewhere. Therefore, while $3$ and $7$ are points where extrema occur, the function may possess other critical points that are not extrema.

Q24. For a function $f$ continuous on $[a,b]$ , which of the following best explains why the theorem does not require differentiability?

A.Differentiability is implied by continuity

B.Absolute extrema can occur at points where the derivative does not exist ✅

C.Differentiability would contradict the existence of extrema

D.Continuity already ensures the derivative is zero at extrema

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The Extreme Value Theorem only needs continuity to guarantee the existence of absolute extrema. These extrema may occur at points where the derivative is zero or undefined, such as corners or cusps. Hence, differentiability is not a necessary condition; the theorem works even when the derivative fails to exist at the extremum.

Q25. A function $p(x)=e^{-x^2}$ is examined on $[-1,1]$ . Which statement correctly uses the theorem to locate its absolute minimum?

A.It occurs at

x=0

because the function is symmetric

B.It occurs at the endpoints because the function decreases outward

C.It occurs at the point where the derivative is zero ✅

D.It has no absolute minimum on this interval

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Compute the derivative $p'(x)=-2xe^{-x^2}$ ; setting it to zero gives $x=0$ . Evaluate $p$ at $-1,0,1$ : $p(0)=1$ (maximum), $p(\pm1)=e^{-1}$ (minimum). Thus the absolute minimum occurs at the interior critical point where the derivative vanishes.

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