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📝 First derivative test for local extrema (24 MCQs)

📖 From Calculus • 5. The derivative in Graphing and Applications • 24 questions available

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What is First derivative test for local extrema?

Definition:
The first derivative test determines local extrema by checking sign changes of $f'(x)$ around critical points. If $f'$ changes from positive to negative, it is a local max; if negative to positive, it is a local min; no change means no extremum.

Example:
For $f(x) = x^3 - 3x$ , $f'(x) = 3x^2 - 3$ . At $x=-1$ , $f'$ changes from positive to negative, indicating a local maximum at $f(-1)=2$ .

Reason:
This test directly links the direction of the function's increase or decrease to the nature of the critical point, providing a reliable classification method.

Easy

Medium

Hard

📝 All First derivative test for local extrema MCQs

Q1. For the function $f(x)=x^{3}-3x$ , which of the following correctly describes the nature of the point at $x=0$ according to the First Derivative Test?

A.Relative maximum

B.Relative minimum

C.Neither a maximum nor a minimum ✅

D.Cannot be determined from the information given

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The derivative $f'(x)=3x^{2}-3$ evaluates to $-3$ at $x=0$ , which is not zero, so $x=0$ is not a critical point. Since a relative extremum can only occur at a critical point where the derivative changes sign, the point at $x=0$ is neither a maximum nor a minimum.

Q2. According to the First Derivative Test, a relative maximum occurs at a critical point where the derivative changes from ___ to ___.

A.positive to negative ✅

B.negative to positive

C.positive to positive

D.negative to negative

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The theorem states that if the derivative is positive on an interval to the left of the critical point and negative on an interval to the right, the function attains a relative maximum at that point. This corresponds to a change from a positive to a negative derivative.

Q3. If a differentiable function has a critical point at $x=c$ and the derivative is positive on both sides of $c$ , what does the First Derivative Test conclude?

x=c

is a relative maximum

x=c

is a relative minimum

x=c

is an inflection point

x=c

is not a relative extremum ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: When the sign of the derivative does not change across a critical point—remaining positive on both sides—the First Derivative Test tells us that the point cannot be a relative extremum. The function continues to increase through the point, so no maximum or minimum occurs there.

Q4. Consider $f(x)=x^{4}$ and $g(x)=x^{3}$ . Both have a critical point at $x=0$ . Which statement correctly applies the First Derivative Test to these points?

A.Both functions have a relative minimum at

x=0

f

has a relative minimum and

g

has no relative extremum at

x=0

. ✅

g

has a relative maximum and

f

has no relative extremum at

x=0

D.Both functions have no relative extremum at

x=0

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For $f(x)=x^{4}$ , $f'(x)=4x^{3}$ changes sign from negative to positive, giving a relative minimum at $x=0$ . For $g(x)=x^{3}$ , $g'(x)=3x^{2}$ is non‑negative on both sides, so the derivative does not change sign; thus, $x=0$ is not a relative extremum for $g$ .}

Q5. A function $h$ satisfies $h'(x)>0$ for $x<2$ and $h'(x)<0$ for $x>2$ . Which of the following must be true at $x=2$ ?

x=2

is a relative maximum ✅

x=2

is a relative minimum

x=2

is a point of inflection

x=2

is not a critical point

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The sign pattern (+ left, – right) matches part (a) of the First Derivative Test, indicating that the function climbs up to $x=2$ and then descends, which characterizes a relative maximum at that point.

Q6. Suppose $p(x)$ is continuous and differentiable except at $x=5$ , where $p'(5)=0$ . If $p'(x)<0$ for $x<5$ and $p'(x)>0$ for $x>5$ , what does the First Derivative Test say about $x=5$ ?

A.Relative maximum

B.Relative minimum ✅

C.Inflection point

D.No relative extremum

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The derivative switches from negative on the left to positive on the right, satisfying part (b) of the theorem. This sign change indicates that the function decreases before $x=5$ and increases after, giving a relative minimum at that point.

Q7. A graph of a function shows the derivative curve crossing the x‑axis at $x=-1$ from negative to positive, and again at $x=3$ from positive to negative. How many relative extrema does the original function have?

A.One relative maximum and one relative minimum ✅

B.Two relative maxima

C.Two relative minima

D.None

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Each crossing of the derivative from negative to positive yields a relative minimum, while a crossing from positive to negative yields a relative maximum. Therefore, the function has one relative minimum at $x=-1$ and one relative maximum at $x=3$ .}

Q8. Which of the following scenarios cannot occur at a critical point according to the First Derivative Test?

A.Derivative changes from positive to negative

B.Derivative changes from negative to positive

C.Derivative remains positive on both sides ✅

D.Derivative is undefined on one side only

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The test requires the derivative to have a sign change to produce a relative extremum. If the derivative stays positive on both sides, the point is not a relative extremum, but such a scenario is still possible. The only impossible case among the options is a derivative that is undefined on only one side while still being a critical point, which violates the continuity hypothesis.

Q9. A function $q(x)$ has a critical point at $x=0$ . Near this point, the sign of $q'(x)$ is negative for $x<0$ and also negative for $x>0$ . Which conclusion follows from the First Derivative Test?

x=0

is a relative maximum

x=0

is a relative minimum

x=0

is a point of inflection

x=0

is not a relative extremum ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Since the derivative does not change sign—remaining negative on both sides—the First Derivative Test indicates that no relative extremum occurs at the point. The function continues to decrease through $x=0$ .

Q10. Consider the function $r(x)=x^{5}-5x^{3}$ . At which critical point does the First Derivative Test guarantee a relative extremum?

x=-\sqrt{3}

x=0

x=\sqrt{3}

D.Both

x=-\sqrt{3}

and

x=\sqrt{3}

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The derivative $r'(x)=5x^{4}-15x^{2}=5x^{2}(x^{2}-3)$ is zero at $x=0,\pm\sqrt{3}$ . The sign of $r'$ changes from positive to negative at $-\sqrt{3}$ and from negative to positive at $\sqrt{3}$ , giving a relative maximum at $-\sqrt{3}$ and a relative minimum at $\sqrt{3}$ . Thus both points guarantee relative extrema.

Q11. A continuous function has a critical point at $x=c$ where the derivative exists and equals zero. If the second derivative $f''(c)>0$ , what does the First Derivative Test imply about the nature of $x=c$ ?

A.Relative maximum

B.Relative minimum

C.Point of inflection ✅

D.Cannot be determined from the First Derivative Test alone

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: When $f''(c)>0$ , the function is concave upward near $c$ , and the derivative changes from negative to positive, satisfying part (b) of the First Derivative Test. Hence, $x=c$ is a relative minimum.

Q12. Which statement best captures the logical relationship between the sign of $f'(x)$ and the existence of a relative extremum at a critical point?

A.A sign change from positive to negative guarantees a relative minimum.

B.A sign change from negative to positive guarantees a relative maximum.

C.A sign change in any direction guarantees some relative extremum. ✅

D.No sign change guarantees a relative extremum.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The theorem specifies that a sign change—whether from positive to negative (yielding a maximum) or from negative to positive (yielding a minimum)—ensures a relative extremum. Thus, any sign change across a critical point guarantees that some type of relative extremum exists.

Q13. Given $s(x)=\frac{x}{1+x^{2}}$ , determine the nature of the critical point at $x=0$ using the First Derivative Test.

A.Relative maximum

B.Relative minimum

C.Neither a maximum nor a minimum ✅

D.Inflection point

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The derivative $s'(x)=\frac{1-x^{2}}{(1+x^{2})^{2}}$ evaluates to $1$ at $x=0$ , which is not zero; therefore, $x=0$ is not a critical point. Since relative extrema can only occur at critical points, the point is neither a maximum nor a minimum.

Q14. A function $f$ is differentiable everywhere and satisfies $f'(x)>0$ for all $x\neq2$ while $f'(2)=0$ . Which conclusion follows from the First Derivative Test?

x=2

is a relative maximum

x=2

is a relative minimum

x=2

is a point of inflection ✅

x=2

is not a relative extremum

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Even though the derivative is zero at $x=2$ , the sign of the derivative does not change (it stays positive on both sides). According to part (c) of the theorem, this means there is no relative extremum at that point.

Q15. Which of the following graphs best illustrates a scenario where the First Derivative Test predicts a relative maximum?

A.[Graph A: derivative positive then negative] ✅

B.[Graph B: derivative negative then positive]

C.[Graph C: derivative positive on both sides]

D.[Graph D: derivative negative on both sides]

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A relative maximum occurs when the derivative is positive to the left of the critical point and negative to the right. Graph A displays exactly that sign pattern, making it the correct illustration.

Q16. If a function’s derivative changes sign from negative to positive at a critical point, which of the following must be true about the original function near that point?

A.It is decreasing then increasing ✅

B.It is increasing then decreasing

C.It has a horizontal inflection point

D.It is constant

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A change from negative (decreasing) to positive (increasing) indicates that the function was decreasing before the point and starts increasing after, which is precisely the behavior described in option A.

Q17. Consider the piecewise function $f(x)=\begin{cases}x^{2}&x\le 1\\2x-1&x>1\end{cases}$ . Using the First Derivative Test, what can be concluded about the point $x=1$ ?

A.Relative maximum

B.Relative minimum

C.Neither a maximum nor a minimum ✅

D.Not a critical point

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The left-hand derivative at $x=1$ is $2$ and the right-hand derivative is $2$ ; the derivative does not change sign, and the function is continuous. Hence, $x=1$ is not a relative extremum.

Q18. A function $g$ satisfies $g'(x)<0$ for $x<0$ and $g'(x)>0$ for $x>0$ . Which of the following best describes the shape of the graph of $g$ near $x=0$ ?

A.A valley-shaped curve ✅

B.A hill-shaped curve

C.A straight line

D.A horizontal inflection

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The derivative being negative to the left (function decreasing) and positive to the right (function increasing) creates a valley shape, indicating a relative minimum at $x=0$ .

Q19. Which of the following is a necessary condition for the First Derivative Test to be applied at a point $x=c$ ?

f

must be continuous at

x=c

✅

f&#039;

must be continuous at

x=c

f&#039;&#039;

must exist at

x=c

f

must be differentiable everywhere

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The theorem explicitly requires the function to be continuous at the critical point. Continuity of the derivative or existence of the second derivative is not required for the basic First Derivative Test.

Q20. A function $h$ has critical points at $x=-2,0,2$ . Its derivative signs are: negative left of $-2$ , positive between $-2$ and $0$ , negative between $0$ and $2$ , and positive right of $2$ . How many relative extrema does $h$ have?

A.One relative maximum and one relative minimum ✅

B.Two relative maxima and one relative minimum

C.Two relative minima and one relative maximum

D.None

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The sign pattern shows a change from negative to positive at $-2$ (relative minimum) and from positive to negative at $0$ (relative maximum). The change from negative to positive at $2$ gives another relative minimum, totaling two minima and one maximum.

Q21. If a function’s derivative is undefined at a point but the function is continuous there, can the First Derivative Test still be used to determine a relative extremum at that point?

A.Yes, as long as the derivative exists on both sides

B.No, the test requires the derivative to exist at the point

C.Yes, but only if the one‑sided limits of the derivative have opposite signs ✅

D.No, continuity alone is sufficient

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The test can be applied when the derivative does not exist at the point provided the one‑sided limits of the derivative exist and have opposite signs, indicating a sign change across the point.

Q22. A differentiable function satisfies $f'(x)=0$ only at $x=4$ . Additionally, $f'(x)>0$ for $x<4$ and $f'(x)<0$ for $x>4$ . What does the First Derivative Test conclude?

x=4

is a relative maximum ✅

x=4

is a relative minimum

x=4

is a point of inflection

D.Cannot be determined without second derivative

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The derivative changes from positive on the left to negative on the right, which matches part (a) of the theorem, indicating a relative maximum at $x=4$ .

Q23. Which of the following best explains why a critical point where the derivative does not change sign cannot be a relative extremum?

A.The function is constant near that point

B.The function continues to increase or decrease through the point ✅

C.The second derivative is zero

D.The function is not continuous at that point

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: If the derivative retains the same sign on both sides, the function either keeps increasing (positive sign) or keeps decreasing (negative sign) through the point, precluding a local maximum or minimum. Hence, no relative extremum can exist.

Q24. A function $p(x)$ has a critical point at $x=1$ . The derivative satisfies $p'(x)>0$ for $0<x<1$ and $p'(x)<0$ for $1<x<2$ . According to the First Derivative Test, what is the nature of $x=1$ ?

A.Relative maximum ✅

B.Relative minimum

C.Inflection point

D.Not a critical point

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The derivative switches from positive (increasing) to negative (decreasing) as $x$ passes through $1$ . This sign change aligns with part (a) of the First Derivative Test, confirming a relative maximum at that point.

🔗 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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