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📝 Analysis of Functions II (Relative Extrema) in calculus (25 MCQs)

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What is Analysis of Functions II (Relative Extrema) in calculus?

Definition:
Relative extrema are local maximums or minimums where $f(c)$ is greater or less than nearby values. These occur at critical points where $f'(c) = 0$ or $f'(c)$ is undefined, marking peaks and valleys in the function's graph structure.

Example:
For $f(x) = x^2 - 4x$ , $f'(x) = 2x - 4$ . Setting $2x - 4 = 0$ gives $x=2$ , a relative minimum since the parabola opens upward.

Reason:
Identifying these points helps sketch accurate graphs and understand optimal values within specific local neighborhoods of the domain.

Easy

Medium

Hard

📝 All Analysis of Functions II (Relative Extrema) in calculus MCQs

Q1. Which of the following best describes a relative maximum of a function $f$ at a point $c$ ?

f(c)

is the greatest value of

f

on its entire domain.

B.There exists an interval

(c-\delta,\,c+\delta)

such that

f(c) \ge f(x)

for all

x

in that interval. ✅

f&#039;(c)=0

and

f&#039;&#039;(c)<0

f(c)

is greater than

f(x)

for all

x>c

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A relative maximum is a local peak: within some open interval around $c$ the function’s value at $c$ is not exceeded. This definition captures the “local” nature, unlike a global maximum, and does not require second‑derivative information, which may be inconclusive or unavailable.

Q2. According to the First Derivative Test, what must happen to $f'(x)$ at a point $c$ for $f$ to have a relative minimum there?

f&#039;(x)

changes from negative to positive as

x

passes through

c

. ✅

f&#039;(x)

changes from positive to negative as

x

passes through

c

f&#039;(x)

is zero and remains zero in a neighborhood of

c

f&#039;(x)

does not exist at

c

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The First Derivative Test states that if the derivative switches sign from negative (decreasing function) to positive (increasing function) at a point, the function attains a local low there. The sign change indicates a turning point that is a relative minimum.

Q3. If a differentiable function $f$ satisfies $f'(c)>0$ for all $x<c$ and $f'(c)<0$ for all $x>c$ , what can be inferred about $f$ at $c$ ?

c

is a point of inflection.

c

is a relative maximum. ✅

c

is a relative minimum.

c

is a critical point with no extremum.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When the derivative is positive to the left of $c$ (function rising) and negative to the right (function falling), the graph climbs up to $c$ and then descends, forming a peak. Hence $c$ is a relative maximum by the First Derivative Test.

Q4. Which statement correctly compares the information given by $f'(c)=0$ and $f''(c)=0$ at a point $c$ ?

A.Both guarantee a relative extremum at

c

f&#039;(c)=0

indicates a possible extremum;

f&#039;&#039;(c)=0

gives no definitive conclusion. ✅

f&#039;&#039;(c)=0

guarantees a point of inflection;

f&#039;(c)=0

guarantees a relative extremum.

D.Both are necessary and sufficient conditions for a relative minimum.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A zero first derivative signals a critical point, which may be a maximum, minimum, or saddle point. A zero second derivative eliminates the simple curvature test, leaving the nature of the critical point undecided; further analysis (e.g., higher‑order derivatives) is required.

Q5. For the function $g(x)=x^3-3x$ , on which interval is $g$ increasing?

(- \infty,-1)

(-1,1)

(1,\infty)

✅

(- \infty,0)

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The derivative $g'(x)=3x^2-3=3(x^2-1)$ is positive when $|x|>1$ . Hence $g$ increases on $(- \infty,-1)$ and $(1,\infty)$ . Among the choices, only $(1,\infty)$ is listed as an increasing interval.

Q6. A company’s cost function is $C(x)=5x^2-40x+150$ . Which statement correctly identifies the production level that minimizes cost?

A.The minimum occurs at

x=4

. ✅

B.The minimum occurs at

x= -4

C.The minimum occurs at

x=2

D.The function has no minimum because it is unbounded below.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Setting $C'(x)=10x-40=0$ yields $x=4$ . Since the quadratic coefficient is positive, the parabola opens upward, guaranteeing a global minimum at $x=4$ .

Q7. When applying the Second Derivative Test, which condition guarantees a relative minimum at a critical point $c$ ?

f&#039;(c)=0

and

f&#039;&#039;(c)>0

. ✅

f&#039;(c)=0

and

f&#039;&#039;(c)<0

f&#039;(c)\neq0

and

f&#039;&#039;(c)>0

f&#039;(c)=0

and

f&#039;&#039;(c)=0

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The Second Derivative Test states that if the first derivative vanishes at $c$ (making it a critical point) and the second derivative is positive there, the graph is concave upward, indicating a local trough, i.e., a relative minimum.

Q8. A cost function for drilling is $C(d)=2d^2-12d+20$ , where $d$ is distance in kilometers from shore. Which interval contains the distance that yields the lowest cost?

d\in(0,2)

d\in(2,4)

d\in(4,6)

d\in(6,8)

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Differentiate: $C'(d)=4d-12$ . Setting to zero gives $d=3$ . Since the quadratic opens upward, the minimum occurs at $d=3$ , which lies in the interval $(2,4)$ .

Q9. Consider $f(x)=x^4-8x^2+16$ . Which of the following statements correctly compares its relative extrema?

A.It has one relative maximum at

x=0

and no minima.

B.It has two relative minima at

x=\pm2

and a relative maximum at

x=0

. ✅

C.It has a relative minimum at

x=0

and two relative maxima at

x=\pm2

D.It has no relative extrema because the derivative never changes sign.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Compute $f'(x)=4x^3-16x=4x(x^2-4)$ . Critical points are $x=0,\pm2$ . The second derivative $f''(x)=12x^2-16$ gives $f''(0)=-16<0$ (maximum) and $f''(\pm2)=32>0$ (minima). Hence the described pattern holds.

Q10. A particle moves along a line with position $s(t)=t^3-6t^2+9t$ . Using calculus, which time interval contains a relative maximum of its speed?

t\in(0,1)

t\in(1,3)

✅

t\in(3,4)

t\in(4,5)

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Speed is $|v(t)|$ where $v(t)=s'(t)=3t^2-12t+9$ . Critical points of speed occur where $v(t)=0$ or changes sign. Solving $3t^2-12t+9=0$ gives $t=1$ and $t=3$ . Speed increases on $(1,3)$ and then decreases, so a relative maximum of speed occurs within that interval.

Q11. For $h(x)=\ln(x)-x^2$ on $(0,\infty)$ , which statement correctly describes the use of the Second Derivative Test?

h&#039;&#039;(x)=-\frac{1}{x^2}-2<0

for all

x

, so any critical point is a relative maximum. ✅

h&#039;&#039;(x)=\frac{1}{x^2}-2

changes sign, indicating both maxima and minima.

h&#039;&#039;(x)=-\frac{1}{x^2}-2

is always negative, guaranteeing a relative minimum at the critical point.

D.The test is inconclusive because

h&#039;&#039;(x)

is zero at the critical point.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Differentiating gives $h'(x)=\frac{1}{x}-2x$ and $h''(x)=-\frac{1}{x^2}-2$ , which is strictly negative for all positive $x$ . Thus any critical point where $h'(x)=0$ must be a relative maximum, because the graph is concave down everywhere.

Q12. Why can a relative extremum occur at a point where the derivative does not exist?

A.Because the function must be discontinuous there.

B.Because the sign of the derivative can still change across the point even if the derivative is undefined. ✅

C.Because the second derivative test can be applied instead.

D.It cannot; a relative extremum always requires a defined derivative.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A cusp or corner can cause the derivative to be undefined, yet the function may switch from increasing to decreasing (or vice‑versa) at that point. The First Derivative Test relies on sign changes, not on the existence of the derivative itself.

Q13. Let $f_a(x)=x^3-3ax$ where $a$ is a parameter. How does increasing $a$ affect the location of the relative extrema?

A.Both extrema move farther apart as

a

increases.

B.Both extrema move closer together as

a

increases.

C.One extremum moves left while the other moves right, keeping the midpoint at zero. ✅

D.The locations of the extrema are independent of

a

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Critical points satisfy $f_a'(x)=3x^2-3a=0$ → $x=\pm\sqrt{a}$ . As $a$ grows, the positive root moves right and the negative root moves left, symmetrically about the origin, so the midpoint remains at zero while the distance between them increases.

Q14. For $f(x)=e^{-x^2}$ , which interval correctly describes where the function is concave upward?

(- \infty,-1)

(-1,1)

(1,\infty)

✅

D.The function is never concave upward.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Second derivative: $f''(x)=e^{-x^2}(4x^2-2)$ . Concave upward when $f''(x)>0$ ⇒ $4x^2-2>0$ ⇒ $|x|>\frac{\sqrt{2}}{2}\approx0.707$ . Thus for $x>1$ (and similarly $x<-1$ ), the function is concave upward; choice C matches the positive side.

Q15. When solving $f(x)=0$ with Newton’s method, why is it advantageous to start at a point near a relative extremum of $f$ ?

A.Because the derivative is zero there, speeding convergence.

B.Because the function changes sign rapidly near an extremum.

C.Because the tangent line at an extremum provides a better linear approximation for the root. ✅

D.It is not advantageous; starting near an extremum often slows convergence.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Newton’s iteration uses the tangent line $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ . Near a relative extremum, the function’s curvature is small, making the tangent a closer approximation to the function’s shape, which can lead to faster convergence toward a nearby root.

Q16. Which statement correctly distinguishes a global maximum from a relative maximum?

A.A global maximum must occur at a critical point; a relative maximum does not.

B.Every global maximum is also a relative maximum, but not every relative maximum is global. ✅

C.A relative maximum is always higher than any global maximum.

D.Global maxima can only occur on closed intervals, while relative maxima cannot.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A global maximum is the highest value of the function on its entire domain; if it occurs at an interior point, it is also a relative maximum. However, a relative maximum may be merely a local peak that is not the highest overall value.

Q17. Compare the information provided by the First Derivative Test and the Second Derivative Test for a critical point where $f'(c)=0$ and $f''(c)=0$ .

A.Both tests give the same conclusion about the nature of

c

B.The First Derivative Test can still determine the nature of

c

, while the Second Derivative Test is inconclusive. ✅

C.The Second Derivative Test can determine the nature of

c

, while the First Derivative Test is inconclusive.

D.Both tests are inconclusive; higher‑order derivatives are needed.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When the second derivative vanishes, the curvature test fails, leaving the Second Derivative Test inconclusive. The First Derivative Test, which examines sign changes of $f'$ around $c$ , may still reveal whether the function rises then falls (maximum) or vice‑versa (minimum).

Q18. A manufacturer wants to minimize material cost given the surface area function $S(r)=2\pi r^2+ \frac{V}{r}$ for a fixed volume $V$ . Which sequence of steps correctly applies calculus to find the optimal radius?

A.Differentiate

S(r)

, set

S&#039;(r)=0

, solve for

r

, verify with second derivative. ✅

B.Set

S(r)=0

, solve for

r

, then differentiate to check for minima.

C.Differentiate twice, set

S&#039;&#039;(r)=0

, solve for

r

D.Use trial‑and‑error with values of

r

until cost is minimized.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The procedure follows the standard optimization routine: compute the first derivative, locate critical points by solving $S'(r)=0$ , then apply the Second Derivative Test (or sign analysis) to ensure the critical point yields a minimum of the surface area.

Q19. For the piecewise function $p(x)=\begin{cases}x^2 & x\le 0\\ -x & x>0\end{cases}$ , which interval contains a relative maximum?

(-2,0)

(0,2)

(-1,1)

D.No interval contains a relative maximum. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: On $(-\infty,0]$ the function $x^2$ increases as $x$ approaches 0, reaching a value of 0. For $x>0$ , $-x$ decreases from 0 downward. Thus the highest value in any neighborhood occurs at the cusp $x=0$ , but the function does not attain a higher value on either side, so no relative maximum exists.

Q20. Determine all real values of the parameter $k$ for which the function $f_k(x)=x^3-3kx$ has exactly two distinct relative extrema.

k>0

✅

k<0

k\neq0

D.All real

k

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Critical points satisfy $f_k'(x)=3x^2-3k=0$ → $x=\pm\sqrt{k}$ . Real distinct critical points exist only when $k>0$ . For $k=0$ there is a single critical point at the origin, and for $k<0$ the critical points are imaginary, so exactly two relative extrema occur precisely for positive $k$ .

Q21. Explain why the sign of the second derivative $f''(c)$ at a critical point $c$ reflects the curvature of the graph and thus indicates a relative extremum.

A.Because

f&#039;&#039;(c)

measures the rate of change of the slope, positive curvature (concave up) creates a trough, and negative curvature (concave down) creates a peak.

B.Because

f&#039;&#039;(c)

directly gives the value of the function at

c

C.Because a zero second derivative always implies a maximum. ✅

D.Because curvature is unrelated to extrema; only the first derivative matters.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The second derivative quantifies how the slope itself changes. If $f''(c)>0$ , the slope is increasing, meaning the graph bends upward (concave up) and the point resembles the bottom of a valley—a relative minimum. Conversely, $f''(c)<0$ indicates concave down, forming a peak—a relative maximum.

Q22. Given the implicit relation $x^2+y^2=16$ , find the y‑coordinates of points where the curve has relative extrema with respect to $y$ as a function of $x$ .

y=\pm4

y=0

y=\pm\sqrt{8}

✅

D.No relative extrema exist because the curve is a circle.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Treat $y$ as a function of $x$ via implicit differentiation: $2x+2y\,y'=0$ → $y'=-x/y$ . Critical points occur where $y'=0$ ⇒ $x=0$ . Substituting $x=0$ into the original equation gives $y^2=16$ → $y=\pm4$ . However, these correspond to vertical tangents, not extrema of $y$ with respect to $x$ . The extrema of $y$ occur where $dy/dx$ changes sign, i.e., at $y=\pm\sqrt{8}$ .

Q23. A cost function $C(x,y)=x^2+4y^2$ is to be minimized subject to the constraint $x+2y=10$ . Which method correctly yields the optimal solution?

A.Set up the Lagrangian

\mathcal{L}=x^2+4y^2-\lambda(x+2y-10)

, solve

\nabla\mathcal{L}=0

B.Eliminate

y

using the constraint, differentiate the resulting single‑variable function, and solve for

x

C.Both A and B are valid and will give the same solution. ✅

D.Use trial‑and‑error to test integer pairs satisfying the constraint.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Both the Lagrange multiplier technique and substitution reduce the constrained problem to solving for critical points, and each method leads to the same optimal point $(x,y)=(4,3)$ . Either approach correctly applies calculus to locate the minimum under the linear constraint.

Q24. When combining graphing‑utility output with derivative analysis, how can one verify that a visually identified peak is indeed a relative maximum?

A.Check that the graphing utility’s peak coordinate satisfies

f&#039;(c)=0

and that

f&#039;&#039;(c)<0

. ✅

B.Rely solely on the visual appearance; derivative checks are unnecessary.

C.Confirm that the peak occurs at the endpoint of the domain.

D.Ensure the function value at the peak is greater than all other values shown.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: A graphing utility may display a peak due to scaling or resolution artifacts. To rigorously confirm a relative maximum, compute the derivative at the reported coordinate; it should be zero (or undefined with a sign change). Then evaluate the second derivative: a negative value confirms concave‑down curvature, solidifying the classification.

Q25. Prove that if $f$ is continuous on an interval and $f'(c)=0$ with $f''(c)>0$ , then $c$ is a relative minimum, and discuss whether continuity of $f$ is essential for the conclusion.

A.Continuity is unnecessary; the derivative conditions alone guarantee a minimum.

B.Continuity ensures the derivative exists, making the second‑derivative test valid; without continuity the conclusion may fail.

C.Continuity is required because the Mean Value Theorem underpins the second‑derivative test; without it, the result is not guaranteed. ✅

D.Both continuity and differentiability are irrelevant; only the sign of

f&#039;&#039;(c)

matters.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The proof of the Second Derivative Test relies on the Mean Value Theorem, which demands continuity on a closed interval and differentiability on the open interval. If $f$ were discontinuous at $c$ , the derivative information could be misleading, and the claim that $c$ is a relative minimum would not be assured.

🔗 Related Topics (MCQs)

📝 Increasing and decreasing functions using derivative MCQs 📝 Concavity and inflection points calculus MCQs 📝 Logistic growth curves calculus MCQs 📝 Relative maxima and minima local extrema MCQs 📝 First derivative test for local extrema MCQs

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