← Back to 5. The derivative in Graphing and Applications

📝 Second derivative test for local extrema (15 MCQs)

📖 From Calculus • 5. The derivative in Graphing and Applications • 15 questions available

←Previous Topic:First derivative test for local extrema📝 Second derivative test for local extremaMultiplicity and graph behavior:Next Topic→

What is Second derivative test for local extrema?

Definition:
The second derivative test uses $f''(c)$ to classify critical points where $f'(c)=0$ . If $f''(c) > 0$ , $f$ has a local minimum; if $f''(c) < 0$ , $f$ has a local maximum; if $f''(c)=0$ , the test is inconclusive and requires further analysis.

Example:
For $f(x) = x^2$ , $f'(0)=0$ and $f''(0)=2 > 0$ , confirming a local minimum at $x=0$ with value $f(0)=0$ .

Reason:
Concavity provides immediate insight into whether a critical point is a peak or valley, often simplifying calculations compared to analyzing sign charts of the first derivative.

Easy

Medium

Hard

📝 All Second derivative test for local extrema MCQs

Q1. According to the Second Derivative Test, which condition on $f''(x_0)$ guarantees a relative minimum at $x_0$ ?

f&#039;&#039;(x_0) > 0

✅

f&#039;&#039;(x_0) < 0

f&#039;&#039;(x_0) = 0

D.No condition on

f&#039;&#039;

is needed

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: If the first derivative is zero at $x_0$ and the second derivative is positive, the graph is concave up near that point. Concave‑up behavior forces the function to lie above the tangent line on both sides, establishing a local minimum. This matches part (a) of the Second Derivative Test.

Q2. What does the Second Derivative Test conclude when $f''(x_0)=0$ and $f'(x_0)=0$ ?

A.The point is a relative maximum

B.The point is a relative minimum ✅

C.The test is inconclusive

D.The function is not differentiable at

x_0

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When both the first and second derivatives vanish, the curvature test cannot determine the sign of the concavity. Consequently, the test provides no information about extrema; the point may be a maximum, minimum, or neither, depending on higher‑order terms. Hence the test is labeled inconclusive.

Q3. Given $f(x)=x^{3}$ , what does the Second Derivative Test reveal about $x=0$ ?

A.Relative maximum

B.Relative minimum

C.Neither maximum nor minimum ✅

D.Inconclusive because

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

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: For $f(x)=x^{3}$ we have $f'(0)=0$ and $f''(0)=0$ . The test is inconclusive, and examining the function shows it increases through the origin, so $x=0$ is neither a maximum nor a minimum. This illustrates the need for higher‑order analysis.

Q4. Which of the following functions has a relative minimum at the origin according to the Second Derivative Test?

f(x)=-x^{4}

f(x)=x^{4}+x^{2}

f(x)=x^{4}

f(x)=x^{3}

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The function $f(x)=x^{4}$ satisfies $f'(0)=0$ and $f''(0)=0$ but higher‑order terms are positive, giving a local minimum. The other choices either have negative curvature or are odd functions, which do not produce a minimum at the origin.

Q5. For $f(x)=x^{2}-4x+3$ , what does the Second Derivative Test say about the critical point at $x=2$ ?

A.Relative maximum ✅

B.Relative minimum

C.Inconclusive

D.No critical point exists

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Compute $f'(x)=2x-4$ , giving a critical point at $x=2$ . The second derivative $f''(x)=2$ is positive, so the graph is concave up there. By the test, $x=2$ is a relative minimum.

Q6. If $f''(c)>0$ for $c$ in an interval, what can be said about $f'(x)$ for $x<c$ and $x>c$ close to $c$ ?

f&#039;(x)<0

left,

f&#039;(x)>0

right

f&#039;(x)>0

left,

f&#039;(x)<0

right

f&#039;(x)=0

on both sides ✅

D.No information about

f&#039;

can be deduced

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: When the second derivative is positive, the first derivative is increasing. Near a stationary point where $f'(c)=0$ , the increase forces $f'$ to be negative to the left and positive to the right, indicating a local minimum.

Q7. Which statement correctly compares the first and second derivative tests for locating relative extrema?

A.The second test works at any critical point, the first does not

B.The first test requires continuity, the second does not

C.The second test is easier when the second derivative exists, but the first works even when it does not ✅

D.Both tests give identical conclusions for all functions

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The second derivative test is simpler when a second derivative exists because it uses curvature, yet it only applies at stationary points where $f''$ exists. The first derivative test, however, works at any critical point of a continuous function, making it more generally applicable.

Q8. How does concavity relate to the type of relative extremum at a stationary point?

A.Concave down implies a relative minimum

B.Concave up implies a relative maximum

C.Concave down implies a relative maximum

D.Concavity gives no information about extrema ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: If the graph is concave down ( $f''<0$ ) near a stationary point, the function bends downward, creating a peak, i.e., a relative maximum. Conversely, concave up ( $f''>0$ ) produces a trough, a relative minimum. This is the geometric basis of the Second Derivative Test.

Q9. Given $f'(x)=(x-2)(x+3)^{2}$ , at which critical points can the Second Derivative Test definitively classify the extremum?

A.At

x=2

only ✅

B.At

x=-3

only

C.At both

x=2

and

x=-3

D.At neither point

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Compute $f''(x)= (x+3)^{2}+2(x-2)(x+3)$ . Evaluating yields $f''(2)=25>0$ (minimum) and $f''(-3)=0$ (inconclusive). Thus the test definitively classifies the extremum at $x=2$ but not at $x=-3$ .

Q10. If $f'(c)=0$ but $f''(c)>0$ does not hold, what can be concluded about the nature of $c$ ?

c

must be a relative maximum

c

must be a relative minimum

C.No conclusion can be drawn from the Second Derivative Test alone ✅

c

is a point of inflection

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: When $f''(c)$ is not positive (it could be zero or negative), the Second Derivative Test cannot determine the extremum type. Additional analysis, such as higher‑order derivatives or the first derivative test, is required to draw any conclusion.

Q11. Suppose $f''$ changes sign at a stationary point $c$ . What does the Second Derivative Test indicate?

A.A relative maximum at

c

B.A relative minimum at

c

C.The test is inconclusive because

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

D.Both a maximum and a minimum simultaneously ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: If $f''$ changes sign, then $f''(c)$ must be zero, making the test inconclusive. The sign change suggests a possible inflection point, but the Second Derivative Test does not provide a definitive classification, so option D is correct.

Q12. Does the Second Derivative Test provide any information for $f(x)=x^{5}$ at its only stationary point?

A.Yes, it shows a relative maximum

B.Yes, it shows a relative minimum

C.No, because

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

makes the test inconclusive ✅

D.It indicates a point of inflection

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: $f'(x)=5x^{4}$ gives a stationary point at $x=0$ . The second derivative $f''(x)=20x^{3}$ also vanishes at the origin, so the test is inconclusive. Higher‑order analysis shows the function is increasing through the point, confirming no extremum.

Q13. Why does the Second Derivative Test fail to classify the extremum of $f(x)=x^{3}$ at $x=0$ ?

A.Because \(f'(0)\neq0

B.Because

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

and higher‑order terms dominate ✅

C.Because the function is not continuous

D.Because the test only works for polynomials of even degree

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For $f(x)=x^{3}$ , both the first and second derivatives vanish at the origin. The second derivative test therefore gives no curvature information. The cubic term dominates, causing the function to increase through the point, so no extremum exists. The failure stems from the zero second derivative.

Q14. For $f(x)=e^{x^{2}}$ , what does the Second Derivative Test reveal at $x=0$ ?

A.Relative maximum

B.Relative minimum

C.Inconclusive

D.No stationary point exists ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: $f'(x)=2xe^{x^{2}}$ gives a stationary point at $x=0$ . The second derivative $f''(x)=2e^{x^{2}}+4x^{2}e^{x^{2}}$ evaluates to $2>0$ at the origin, indicating concave up and thus a relative minimum.

Q15. Can the Second Derivative Test be applied to a piecewise function that has a cusp at the critical point?

A.Yes, if each piece is twice differentiable

B.No, because the second derivative does not exist at the cusp

C.Yes, but only the first derivative test is reliable ✅

D.It depends on the continuity of the first derivative

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: At a cusp the derivative is not defined or is not continuous, so the second derivative does not exist. Consequently, the Second Derivative Test cannot be used, and one must rely on the First Derivative Test or direct analysis.

🔗 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

←Previous Topic:First derivative test for local extrema📝 Second derivative test for local extremaMultiplicity and graph behavior:Next Topic→