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📝 Concavity and inflection points calculus (19 MCQs)

📖 From Calculus • 5. The derivative in Graphing and Applications • 19 questions available

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What is Concavity and inflection points calculus?

Definition:
Concavity describes the curvature of a graph where $f''(x) > 0$ implies concave up and $f''(x) < 0$ implies concave down. An inflection point occurs where concavity changes, requiring $f''(x) = 0$ or undefined with a sign change.

Example:
For $f(x) = x^3$ , $f''(x) = 6x$ . At $x=0$ , concavity changes from down to up, so $(0,0)$ is an inflection point.

Reason:
The second derivative measures the rate of change of the slope, revealing whether the graph bends upward like a cup or downward like a frown.

Easy

Medium

Hard

📝 All Concavity and inflection points calculus MCQs

Q1. What is the definition of an inflection point for a twice‑differentiable function $f$ ?

A.A point where

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

and the concavity changes ✅

B.A point where

f&#039;(x)=0

C.A point where

f

is not continuous

D.A point where

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

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: An inflection point occurs where the second derivative is zero (or undefined) and the sign of $f''$ changes, indicating a switch from concave up to concave down or vice‑versa. This captures both the algebraic condition and the geometric change in curvature.

Q2. Which statement correctly describes the second‑derivative test for concavity?

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

implies concave down

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

implies concave up

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

implies concave up ✅

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

always indicates an inflection point

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The second‑derivative test states that if the second derivative is positive on an interval, the graph is curving upward, i.e., it is concave up. Conversely, a negative second derivative indicates concave down. Equality to zero alone does not guarantee an inflection point without a sign change.

Q3. If the first derivative $f'(x)$ is increasing on an interval, what can be said about the second derivative $f''(x)$ on that interval?

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

✅

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

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

D.No information can be deduced

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When $f'(x)$ is increasing, its rate of change is positive, which means the second derivative, the derivative of $f'$ , must be positive throughout that interval. This directly links monotonic behavior of the first derivative to the sign of the second derivative.

Q4. Given that $f''(x)>0$ for all $x$ in $(1,4)$ and $f'(2)=5$ , what can be inferred about $f'(4)$ ?

f&#039;(4) > 5

✅

f&#039;(4) = 5

f&#039;(4) < 5

D.Cannot be determined from the information given

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Since $f''(x)>0$ means the first derivative is strictly increasing, any value of $f'$ taken at a larger $x$ must be larger than the value at a smaller $x$ . Hence $f'(4)$ must exceed the known value $f'(2)=5$ .

Q5. Suppose a function $g$ has a point $c$ where $g''(c)=0$ and $g'''(c)\neq 0$ . Which inference about concavity near $c$ is most justified?

A.Concavity changes sign at

c

✅

B.Concavity remains the same on both sides of

c

C.No inflection point occurs at

c

D.Insufficient information to decide

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A non‑zero third derivative indicates that the second derivative crosses zero with non‑zero slope, causing a sign change. Therefore the concavity switches from up to down or vice‑versa, confirming that $c$ is an inflection point.

Q6. If a function $h$ is concave up on $(a,b)$ and has a local maximum at $x=m$ within that interval, what must be true about $m$ ?

m

must be an endpoint of the interval

h&#039;&#039;(m)<0

C.Such a maximum cannot exist ✅

h&#039;(m)=0

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A concave‑up graph curves upward, so any interior point cannot be a local maximum; the function is rising then falling only at endpoints. Therefore a local maximum inside a region of concave up is impossible, forcing the conclusion that the premise cannot occur.

Q7. A function $p$ satisfies $p''(x)=6x-4$ . Determine the interval(s) where $p$ is concave down and justify your answer.

(-\infty,\frac{2}{3})

✅

(\frac{2}{3},\infty)

(-\infty,0)

D.No interval of concave down exists

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Concave down corresponds to a negative second derivative. Solving $6x-4<0$ yields $x<\frac{2}{3}$ . Hence every $x$ less than $\frac{2}{3}$ lies in a region where the graph bends downward, giving the interval $(-\infty,\frac{2}{3})$ .

Q8. For the function $q(x)=\ln(x^2+1)$ , the second derivative is $q''(x)=\frac{2(1-x^2)}{(x^2+1)^2}$ . At which of the following points does the concavity change?

x=-1

✅

x=0

x=1

D.No change in concavity

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Concavity changes where the second derivative passes through zero. Setting $1-x^2=0$ gives $x=\pm1$ . Both points cause a sign change, but the question asks for a single choice; selecting $-1$ satisfies the condition and demonstrates the concept.

Q9. If a function $r$ has $r''(x)>0$ for $x<2$ and $r''(x)<0$ for $x>2$ , what can be deduced about the nature of $x=2$ ?

x=2

is a local maximum of

r&#039;

x=2

is a local minimum of

r&#039;

x=2

is an inflection point of

r

✅

x=2

has no special significance

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The sign of the second derivative switches from positive to negative at $x=2$ , indicating a change from concave up to concave down. This sign change is precisely the definition of an inflection point for the original function $r$ .}

Q10. A cubic function $s(x)=ax^3+bx^2+cx+d$ has an inflection point at $x=1$ . Which relationship among the coefficients must hold?

a+b+c=0

3a+2b+c=0

6a+2b=0

✅

a=0

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The second derivative of a cubic is $s''(x)=6ax+2b$ . Setting $s''(1)=0$ gives $6a+2b=0$ . This linear relation between $a$ and $b$ is necessary for the graph to change concavity at $x=1$ .

Q11. Consider a function $t$ with $t''(x) < 0$ for all $x$ . Which of the following statements is necessarily true?

t

is increasing

t

is decreasing

t

is concave down ✅

t

has an inflection point

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A uniformly negative second derivative means the graph bends downward everywhere, which is the definition of concave down. The sign of the first derivative is unrelated to this condition, so only the concavity statement must hold.

Q12. Compare the concavity of $f(x)=x^4$ and $g(x)=x^3$ on the interval $(-1,1)$ .

A.Both are concave up

B.Both are concave down

f

is concave up while

g

changes concavity ✅

f

changes concavity while

g

is concave down

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: For $f(x)=x^4$ , $f''(x)=12x^2$ is always positive, so the graph is concave up throughout. For $g(x)=x^3$ , $g''(x)=6x$ changes sign at $x=0$ , producing both concave up and concave down portions within $(-1,1)$ .

Q13. Evaluate which function has a larger region of concave up: $h_1(x)=\sin x$ or $h_2(x)=\ln x$ on their domains.

h_1

(0,\pi)

✅

h_2

(1,\infty)

C.Both have infinite concave‑up intervals

D.Neither has a concave‑up region

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The second derivative of $\sin x$ is $-\sin x$ ; it is positive when $\sin x<0$ , which occurs on intervals such as $(\pi,2\pi)$ . In contrast, $\ln x$ has second derivative $-1/x^2<0$ everywhere, so it never exhibits concave up. Hence $h_1$ possesses the larger concave‑up region.

Q14. Given $f(x)=e^{x^2}$ , determine all intervals where $f$ is concave down.

A.No interval of concave down exists ✅

(-\infty,0)

(0,\infty)

(-\infty,\infty)

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Computing derivatives yields $f''(x)=e^{x^2}(4x^2+2)$ , which is strictly positive for every real $x$ . Since the second derivative never becomes negative, the function is never concave down, so the correct answer is that no such interval exists.

Q15. Differentiate between an inflection point and a point of non‑differentiability. Which statement is accurate?

A.Both require

f&#039;&#039;

to exist

B.An inflection point may have

f&#039;&#039;=0

while non‑differentiability means

f&#039;

does not exist ✅

C.Non‑differentiability implies an inflection point

D.They are the same concept

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: An inflection point is characterized by a change in the sign of the second derivative, which can occur even when $f''=0$ but exists. A point of non‑differentiability, however, is where the first derivative fails to exist, unrelated to curvature change. Thus the second statement correctly distinguishes the two ideas.

Q16. A function $k$ is defined piecewise: $k(x)=x^2$ for $x\le0$ and $k(x)=-x^2$ for $x>0$ . Determine whether $x=0$ is an inflection point, justifying with derivatives.

A.Yes, because concavity changes

B.No, because

k&#039;

is discontinuous

C.Yes, because

k&#039;&#039;

exists and changes sign

D.No, because

k

is not twice differentiable at 0 ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: To be an inflection point, the function must be at least twice differentiable near the point so that the sign of the second derivative can be examined. Here, the left‑hand second derivative is $2$ and the right‑hand second derivative is $-2$ ; however, the second derivative does not exist at $0$ because the first derivative is not continuous there. Hence $x=0$ is not an inflection point.

Q17. For the function $m(x)=x^3-3x$ , find the x‑coordinate(s) of inflection points and explain why they are inflection points.

x=0

✅

x=\pm1

x=\pm\sqrt{1}

D.No inflection points

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The second derivative of $m$ is $m''(x)=6x$ . Setting this equal to zero gives $x=0$ . On either side of zero, $m''$ changes sign (negative for $x<0$ , positive for $x>0$ ), confirming a change in concavity and thus an inflection point at $x=0$ .

Q18. Which statement correctly compares the curvature of $p(x)=x^2$ and $q(x)=x^4$ at $x=0$ ?

A.Both have zero curvature

p

has greater curvature ✅

q

has greater curvature

D.Curvature is undefined for both

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Curvature at a point depends on the second derivative. At $x=0$ , $p''(0)=2$ while $q''(0)=0$ . A larger second derivative indicates stronger bending, so the parabola $p$ exhibits greater curvature than the flatter quartic $q$ at the origin.

Q19. Given $f(x)=\ln(x)$ , determine the intervals where the function is concave up.

(0,1)

(1,\infty)

C.No interval of concave up ✅

D.All positive

x

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The second derivative of $\ln x$ is $-1/x^{2}$ , which is negative for every $x>0$ . Because concave up requires a positive second derivative, the function is never concave up on its domain, making the correct choice

🔗 Related Topics (MCQs)

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

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