📝 Derivatives at endpoints one sided derivatives (13 MCQs)

📖 From Calculus • 3. The Derivation • 13 questions available

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What is Derivatives at endpoints one sided derivatives?

Definition:
At endpoints of a closed interval, derivatives are defined using one-sided limits, where the right-hand derivative applies at the left endpoint and the left-hand derivative applies at the right endpoint, since the function is not defined beyond these boundaries.

Example:
For $f(x) = \sqrt{x}$ on $[0,1]$ , the right-hand derivative at $x=0$ is $\lim_{h \to 0^+} \frac{\sqrt{h}}{h} = \infty$ , so it is not differentiable at zero.

Reason:
Understanding one-sided derivatives is essential for analyzing functions on restricted domains, ensuring correct application of theorems like Rolle's Theorem and Mean Value Theorem which require differentiability on open intervals.

Easy

Medium

Hard

📝 All Derivatives at endpoints one sided derivatives MCQs

Q1. Which of the following correctly expresses the right‑hand derivative $f'_+(a)$ at the left endpoint $a$ of a closed interval $[a,b]$ ?

f&#039;_+(a)=\displaystyle\lim_{h\to0^-}\frac{f(a+h)-f(a)}{h}

f&#039;_+(a)=\displaystyle\lim_{h\to0^+}\frac{f(a+h)-f(a)}{h}

✅

f&#039;_+(a)=\displaystyle\lim_{h\to0}\frac{f(a+h)-f(a)}{h}

f&#039;_+(a)=\displaystyle\lim_{x\to a}\frac{f(x)-f(a)}{x-a}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The right‑hand derivative uses increments that approach zero from the positive side, so the limit is taken as $h\to0^{+}$ . Option B matches this definition, while the other choices either use the wrong sign or a two‑sided limit, which does not define a one‑sided derivative.

Q2. For the function $f(x)=x^{2}$ defined only on $[0,2]$ , which statement about the left‑hand derivative at $x=2$ is correct?

A.It does not exist because the function is not defined beyond 2.

B.It exists and equals 4. ✅

C.It exists and equals 2.

D.It exists and equals 0.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Even though the function is not defined past $x=2$ , the left‑hand derivative uses values approaching 2 from the left. Computing $\lim_{h\to0^-}\frac{(2+h)^{2}-4}{h}=4$ shows the derivative exists and equals 4, confirming option B.

Q3. Consider $f(x)=\begin{cases}x^{2},&x\le 1\\2x+1,&x>1\end{cases}$ on $[0,2]$ . Which statement is true about the one‑sided derivatives at $x=1$ ?

A.Both

f&#039;_-(1)

and

f&#039;_+(1)

exist and are equal. ✅

B.Only

f&#039;_-(1)

exists;

f&#039;_+(1)

does not exist.

C.Only

f&#039;_+(1)

exists;

f&#039;_-(1)

does not exist.

D.Neither one‑sided derivative exists.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For $x\le1$ , the derivative of $x^{2}$ at 1 is $2$ . For $x>1$ , the derivative of $2x+1$ is also $2$ . Both limits exist and coincide, so the correct choice is A.

Q4. If a function is continuous on $[a,b]$ , its right‑hand derivative at $a$ is positive, and its left‑hand derivative at $b$ is negative, what can be inferred?

A.No interior point must have zero derivative.

B.At least one interior point has derivative zero.

C.Exactly one interior point has derivative zero. ✅

D.Derivative zero can occur only at the endpoints.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: By the Mean Value Theorem, continuity on $[a,b]$ and differentiability on $(a,b)$ guarantee a point where the instantaneous rate equals the average rate. The sign change of the endpoint derivatives forces the average rate to be zero, ensuring at least one $c\in(a,b)$ with $f'(c)=0$ .

Q5. How does the definition of differentiability on a closed interval $[a,b]$ differ from that on an open interval $(a,b)$ ?

A.There is no difference; both require two‑sided derivatives everywhere.

B.Closed intervals require one‑sided derivatives at the endpoints. ✅

C.Open intervals require one‑sided derivatives at the endpoints.

D.Closed intervals do not require continuity at the endpoints.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: On a closed interval the function must be differentiable on the interior and possess the appropriate one‑sided derivatives at the included endpoints, whereas an open interval only demands two‑sided derivatives throughout its interior.

Q6. Which of the following functions is differentiable on the half‑open interval $[0,3)$ ?

f(x)=\sqrt{x}

f(x)=|x-2|

f(x)=\ln(x+1)

✅

f(x)=\dfrac{1}{x}

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: $\ln(x+1)$ is defined and smooth for all $x\ge0$ ; its derivative $1/(x+1)$ exists at every point, including the right‑hand limit at $0$ . The other options either have undefined points or nondifferentiable cusps, making C the only correct choice.

Q7. Which statement best describes the geometric meaning of $f'_+(a)$ and $f'_-(b)$ for a function defined on $[a,b]$ ?

A.Both represent slopes of tangent lines at interior points.

f&#039;_+(a)

is the slope of a tangent line from the left,

f&#039;_-(b)

from the right.

f&#039;_+(a)

is the limit of secant slopes as

x

approaches

a

from the right, and

f&#039;_-(b)

is the limit of secant slopes as

x

approaches

b

from the left.

D.They have no geometric interpretation. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The one‑sided derivatives are precisely the limits of the slopes of secant lines as the second point approaches the endpoint from the interior side. This interpretation links the algebraic limit definitions to the visual notion of a tangent line approaching the endpoint.

Q8. Given that a function $f$ is continuous on $[0,1]$ , its left‑hand derivative at $1$ equals 5, and its right‑hand derivative at $0$ does not exist, which statement must be true?

f

is differentiable on

(0,1)

f

cannot be differentiable at any interior point.

f

may be differentiable on

(0,1)

but need not be. ✅

f

is not continuous at

0

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The existence of a one‑sided derivative at an endpoint does not impose any restriction on differentiability inside the interval. Thus $f$ could be differentiable on $(0,1)$ or not; the only certainty is that interior differentiability is independent of the missing right‑hand derivative at 0.

Q9. If a function has a right‑hand derivative at $a$ , what can be said about its continuity at $a$ ?

A.The function must be continuous from the right at

a

. ✅

B.The function must be continuous from the left at

a

C.The function must be continuous on both sides of

a

D.Continuity is not guaranteed at all.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The existence of the limit defining $f'_+(a)$ requires that the numerator $f(a+h)-f(a)$ approach zero as $h\to0^{+}$ . Hence the function cannot jump from the right, guaranteeing right‑hand continuity at $a$ .

Q10. Which principle states that if a function is differentiable on $(a,b)$ and has appropriate one‑sided derivatives at $a$ and $b$ , then it is continuous on $[a,b]$ ?

A.Differentiability implies continuity ✅

B.Mean Value Theorem

C.Intermediate Value Theorem

D.Rolle’s Theorem

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If a function possesses a right‑hand derivative at $a$ and a left‑hand derivative at $b$ , those limits enforce continuity from the respective sides. Combined with interior differentiability, which always yields continuity, the function is continuous on the entire closed interval.

Q11. Why does the existence of a left‑hand derivative at $b$ guarantee continuity from the left at $b$ ?

A.Because the limit defining the derivative forces the function values to approach

f(b)

x\to b^{-}

B.Because the derivative equals the function value at

b

. ✅

C.Because continuity implies the derivative exists, not the reverse.

D.It does not guarantee any form of continuity.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When the left‑hand derivative $\displaystyle\lim_{h\to0^{-}}\frac{f(b+h)-f(b)}{h}$ exists, the numerator must tend to zero as $h\to0^{-}$ . This forces $f(b+h)$ to approach $f(b)$ from the left, establishing left‑hand continuity.

Q12. Outline a proof that a continuous function on $[a,b]$ with $f'_+(a)>0$ and $f'_-(b)<0$ must have some $c\in(a,b)$ where $f'(c)=0$ . Which theorem is central and what are the key steps?

A.Use the Intermediate Value Theorem on

f

directly.

B.Apply Rolle’s Theorem after constructing an auxiliary function that meets the hypotheses. ✅

C.Employ the Mean Value Theorem after showing the average slope is zero; then conclude a point with zero derivative exists.

D.Invoke the Extreme Value Theorem to locate maxima and minima.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The proof begins by noting the average rate of change $\frac{f(b)-f(a)}{b-a}$ must lie between the positive right‑hand derivative at $a$ and the negative left‑hand derivative at $b$ , forcing it to be zero. The Mean Value Theorem then guarantees a point $c$ where the instantaneous derivative equals this average, i.e., $f'(c)=0$ .

Q13. For a function on $[-1,2]$ with $f(-1)=3$ , $f(2)=1$ , $f'_+(-1)=-2$ and $f'_-(2)=4$ , what can be inferred about the average rate of change over the interval relative to the endpoint derivatives?

A.The average rate equals

-\frac{2}{3}

, which lies between

-2

and

4

. ✅

B.The average rate equals

-\frac{2}{3}

, which is less than both endpoint derivatives.

C.The average rate equals

-\frac{2}{3}

, which is greater than both endpoint derivatives.

D.No relationship can be deduced from the given information.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The average rate of change is $\frac{f(2)-f(-1)}{2-(-1)}=\frac{1-3}{3}=-\frac{2}{3}$ . This number is greater than $-2$ and less than $4$ , so it indeed lies between the two one‑sided endpoint derivatives, confirming option A.

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