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📝 Optimization problems on closed intervals (26 MCQs)

📖 From Calculus • 5. The derivative in Graphing and Applications • 26 questions available

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What is Optimization problems on closed intervals?

Definition:
Optimization on closed intervals $[a, b]$ guarantees absolute extrema by the Extreme Value Theorem. Solve by finding critical points in $(a, b)$ , evaluating $f$ at these points and endpoints $a, b$ , then comparing values to identify the global maximum and minimum.

Example:
Maximize $f(x) = x^3 - 3x$ on $[0, 2]$ . Critical point $x=1$ gives $f(1)=-2$ . Endpoints: $f(0)=0, f(2)=2$ . Absolute max is $2$ , min is $-2$ .

Reason:
Closed intervals ensure boundedness, making the comparison method reliable for finding definitive optimal values in constrained physical or geometric contexts.

Easy

Medium

Hard

📝 All Optimization problems on closed intervals MCQs

Q1. Given the function $f(x)=x^{2}-4x+3$ on the closed interval $[1,5]$ , which point must be evaluated to find the absolute minimum?

A.x=1

B.x=2 ✅

C.x=3

D.x=5

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The derivative $f'(x)=2x-4$ equals zero at $x=2$ , which lies inside $[1,5]$ . According to the Extreme Value Theorem, the absolute minimum occurs either at critical points or endpoints, and evaluating $f(2)$ yields the smallest value, making $x=2$ the required point.

Q2. If a continuous function $g$ attains its maximum at an interior point of $[a,b]$ , what can be inferred about $g'(c)$ at that point?

A.Positive

B.Negative

C.Zero ✅

D.Undefined

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: When a continuous function reaches an interior extremum, Fermat's theorem states that the derivative at that point must be zero (provided the derivative exists). Hence, the derivative $g'(c)$ is zero, indicating a horizontal tangent at the maximum.

Q3. Suppose $h(x)=\sin x$ on $[0,2\pi]$ . Which statement correctly describes the relationship between critical points and endpoints for determining absolute extrema?

A.Only critical points need to be checked

B.Endpoints and critical points both must be checked ✅

C.Only endpoints matter

D.Neither endpoints nor critical points affect extrema

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The Extreme Value Theorem requires evaluating both endpoints and any interior points where the derivative is zero or undefined. For $\sin x$ , the critical points at $\pi/2$ and $3\pi/2$ together with the endpoints $0$ and $2\pi$ determine the absolute maximum and minimum.

Q4. A function $p(x)$ is defined on $[-2,4]$ and is known to be decreasing throughout the interval. Which endpoint gives the absolute maximum?

x=-2

✅

x=4

C.Both endpoints

D.None of the above

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If a function is decreasing on a closed interval, its largest value occurs at the leftmost point. Therefore the absolute maximum is attained at the endpoint $x=-2$ , while the minimum occurs at $x=4$ .

Q5. Let $q(x)=\ln(x)$ on $[1,e^{2}]$ . If the derivative $q'(x)$ is never zero in this interval, what conclusion follows about the location of the absolute minimum?

A.At

x=1

✅

B.At

x=e^{2}

C.No minimum exists

D.Minimum at an interior point

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Since $q'(x)=1/x>0$ for all $x>0$ , the function is strictly increasing. Consequently the smallest value on the interval occurs at the left endpoint $x=1$ , giving the absolute minimum there.

Q6. Consider a piecewise function defined on $[0,3]$ with a discontinuity at $x=2$ . Which statement about the Extreme Value Theorem (EVT) is correct?

A.EVT guarantees both max and min

B.EVT guarantees at least one of them

C.EVT does not apply because the function is not continuous on the whole interval ✅

D.EVT applies regardless of continuity

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The EVT requires continuity on the entire closed interval. A discontinuity at $x=2$ violates this condition, so the theorem cannot guarantee the existence of absolute extrema for the piecewise function.

Q7. Compare the process of finding absolute extrema on $[a,b]$ for a polynomial versus a rational function with vertical asymptotes inside the interval. Which statement is accurate?

A.Both require only endpoints

B.Polynomials need critical points, rational functions need only asymptotes

C.Both require critical points and endpoints, but rational functions also need to examine points of discontinuity ✅

D.Rational functions never have absolute extrema

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Polynomials are continuous, so checking critical points and endpoints suffices. Rational functions may have vertical asymptotes within $[a,b]$ ; those points must also be considered because the function can become unbounded, affecting the existence of extrema.

Q8. Evaluate which of the following intervals guarantees that a continuous function $f$ will have both an absolute maximum and minimum.

(0,1)

[0,1)

(0,1]

[0,1]

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Only a closed and bounded interval, such as $[0,1]$ , satisfies the hypotheses of the Extreme Value Theorem, ensuring that a continuous function attains both its absolute maximum and minimum on that set.

Q9. Given two continuous functions $f$ and $g$ on $[a,b]$ where $f(x)\le g(x)$ for all $x$ in the interval, what can be deduced about their absolute maxima?

f

has a larger maximum

g

has a larger maximum ✅

C.Both have the same maximum

D.No relation can be inferred

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Since $f(x)\le g(x)$ everywhere, the largest value attained by $g$ cannot be smaller than any value of $f$ . Therefore the absolute maximum of $g$ is at least as large as that of $f$ , guaranteeing that $g$ 's maximum is the larger one.

Q10. A function $r(x)=\frac{1}{x-1}$ is defined on $[0,2]$ except at $x=1$ . How does the presence of a vertical asymptote affect the existence of absolute extrema on this interval?

A.Extrema exist at the endpoints

B.No absolute extrema because the function is unbounded ✅

C.Extrema exist at interior points

D.Extrema exist at both endpoints and interior

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The vertical asymptote at $x=1$ causes the function to approach $\pm\infty$ , making it unbounded on the interval. Because the Extreme Value Theorem requires boundedness, the function cannot possess absolute maximum or minimum values on $[0,2]$ .

Q11. For a continuous function $s(x)$ on $[a,b]$ that is strictly convex, which of the following statements about its absolute minimum is true?

A.Minimum occurs at the left endpoint

B.Minimum occurs at the right endpoint

C.Minimum occurs at a unique interior point where the derivative is zero ✅

D.Minimum does not exist

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: A strictly convex function has a single global minimum where its derivative changes sign. Since the function is differentiable, the unique point where $s'(c)=0$ inside $(a,b)$ yields the absolute minimum, regardless of the endpoint values.

Q12. Two functions $u(x)=x^{3}$ and $v(x)=x$ are both continuous on $[-1,2]$ . Which function attains its absolute maximum at an interior point of the interval?

u(x)

only

v(x)

only

C.Both

D.Neither ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Evaluating $u(x)=x^{3}$ gives values $-1$ at $-1$ and $8$ at $2$ ; the interior critical point at $0$ yields $0$ , which is not maximal. For $v(x)=x$ , the maximum is $2$ at the endpoint. Hence neither function reaches its absolute maximum inside the interval.

Q13. Apply the Extreme Value Theorem to explain why a continuous function on a closed interval must attain its absolute maximum.

A.Because its derivative is zero somewhere

B.Because the interval is bounded

C.Because the image of a compact set under a continuous map is compact ✅

D.Because the function is differentiable

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The Extreme Value Theorem relies on the fact that a continuous image of a compact set (the closed interval) is itself compact, meaning it is closed and bounded. Consequently the function's range contains its supremum and infimum, guaranteeing the existence of absolute maximum and minimum values.

Q14. Synthesize the relationship between critical points and endpoints when determining absolute extrema for $f(x)=\sqrt{x}$ on $[0,4]$ .

A.Only critical points matter

B.Only endpoints matter

C.Both must be considered ✅

D.Neither matters

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Although $f'(x)=\frac{1}{2\sqrt{x}}$ is undefined at $x=0$ , the endpoint $x=0$ still needs evaluation. The interior critical point does not exist, so the absolute maximum occurs at the right endpoint $x=4$ and the minimum at the left endpoint. Both types of points are essential.

Q15. Explain why the function $w(x)=|x|$ on $[-2,2]$ has its absolute minimum at $x=0$ even though the derivative does not exist there.

A.Because endpoints have larger values

B.Because the Extreme Value Theorem forces a minimum

C.Because the absolute value attains its smallest value at zero ✅

D.Because derivative zero implies minimum

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The absolute value function reaches its smallest possible output, zero, at $x=0$ . Even though the derivative is undefined at that point, the Extreme Value Theorem guarantees an absolute minimum on a closed interval, and the minimum value is clearly achieved at the origin.

Q16. If a function $y(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with $y'(c)=0$ for some $c\in(a,b)$ , what can be inferred about absolute extrema?

A.Must be an absolute extremum at

c

B.Could be a local extremum ✅

C.Cannot be an extremum

D.Must be an inflection point

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Fermat's theorem states that a zero derivative indicates a possible local extremum, but it does not guarantee an absolute one. Additional evaluation of the function values at endpoints and other critical points is required to determine whether the point $c$ is indeed an absolute maximum or minimum.

Q17. Consider a function $z(x)=e^{-x^{2}}$ on $[-3,3]$ . Using symmetry, determine the location(s) of absolute maximum without evaluating derivatives.

A.At

x=-3

B.At

x=3

C.At

x=0

✅

D.At both endpoints

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The function $e^{-x^{2}}$ is even, so its values are symmetric about the y‑axis. Because the exponent $-x^{2}$ is largest (i.e., least negative) at $x=0$ , the function attains its greatest value $e^{0}=1$ there, making $x=0$ the absolute maximum.

Q18. Given a continuous function $k(x)$ on $[0,5]$ that attains its maximum value at two distinct points, what does this indicate about the function's behavior?

A.The function is constant on the interval

B.The function has a flat top region ✅

C.The function is increasing then decreasing

D.The function is decreasing then increasing

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: If the same maximal value occurs at more than one point, the function must be constant on the segment joining those points, creating a flat region where the function remains at its maximum height. This pattern reflects a plateau rather than a single peak.

Q19. What does the Extreme Value Theorem state for functions on a finite closed interval?

A.Continuous functions have at least one root

B.Continuous functions attain both absolute maximum and minimum ✅

C.Differentiable functions are monotonic

D.All functions are bounded

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The Extreme Value Theorem asserts that any function continuous on a closed, bounded interval $[a,b]$ must achieve a greatest (maximum) and least (minimum) value somewhere in that interval.

Q20. Define a critical point of a function on an interval.

A.Point where the function is undefined

B.Point where the derivative is zero or undefined ✅

C.Endpoint of the interval

D.Point where the function attains its maximum

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A critical point occurs at any interior point where the derivative either equals zero or fails to exist. These points are candidates for local extrema and must be examined when applying the Extreme Value Theorem to locate absolute extrema.

Q21. What is the meaning of 'closed interval' in real analysis?

A.Interval that includes its endpoints ✅

B.Interval that excludes endpoints

C.Interval of infinite length

D.Interval only containing rational numbers

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: A closed interval $[a,b]$ contains both of its endpoints $a$ and $b$ . This property, together with boundedness, is essential for many theorems such as the Extreme Value Theorem, which rely on the interval being compact.

Q22. If a function $f$ is increasing on $[a,b]$ , where is its absolute maximum located?

A.At

a

B.At

b

✅

C.At an interior point

D.Does not exist

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: An increasing function never decreases as $x$ grows, so the greatest value on the interval occurs at the rightmost endpoint $b$ . Hence the absolute maximum is attained at $x=b$ .

Q23. Compare the effect of adding a constant $k$ to a function $f(x)$ on its absolute extrema over $[a,b]$ .

A.Shifts both extrema up by

k

✅

B.Shifts only the maximum

C.Shifts only the minimum

D.No effect

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Adding a constant $k$ to a function translates its entire graph vertically. Every output value, including the absolute maximum and minimum, increases by exactly $k$ . Therefore both extrema are shifted upward by the same amount.

Q24. Given $m(x)=x^{4}-4x^{2}$ on $[-2,2]$ , which points must be evaluated to find the absolute maximum?

A.Only endpoints

B.Only critical points

C.Endpoints and critical points ✅

D.Neither

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: First find critical points: $m'(x)=4x^{3}-8x=4x(x^{2}-2)=0$ gives $x=0,\pm\sqrt{2}$ (the latter lie outside the interval). Evaluate $m$ at the endpoints $-2,2$ and the interior critical point $0$ ; the largest of these values yields the absolute maximum.

Q25. Synthesize why the Extreme Value Theorem does not apply to the function $n(x)=\frac{1}{x}$ on the interval $(0,1]$ .

A.Function not continuous at 0

B.Interval not closed

C.Both A and B ✅

D.Neither

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The EVT requires the domain to be a closed, bounded set and the function to be continuous on that set. The interval $(0,1]$ omits the left endpoint, so it is not closed, and the function $\frac{1}{x}$ is also discontinuous at $x=0$ . Both violations prevent the theorem from applying.

Q26. Apply the concept of compactness to explain why a continuous function on $[a,b]$ attains its bounds.

A.Image of compact set under continuous map is compact ✅

B.Derivative exists

C.Function is monotonic

D.Interval is infinite

💡 Difficulty: medium | ✅ Correct: A

🔗 Related Topics (MCQs)

📝 Increasing and decreasing functions using derivative MCQs 📝 Concavity and inflection points calculus MCQs 📝 Logistic growth curves calculus MCQs 📝 Analysis of Functions II (Relative Extrema) in calculus MCQs 📝 Relative maxima and minima local extrema MCQs

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