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Differentiability and Optimization Notes for GATE CS

Differentiability and Optimization notes for GATE CS: 39 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

differentiability and optimization notes

Chapter Roadmap: Differentiability and Optimization

Chapter Roadmap: Differentiability and Optimization Your journey through the calculus of change and optimization. 1 Differentiability Conditions for Piecewise Functions (Current) Making sharp corners smooth. Matching limits and derivatives at joints. 2 Nondifferentiability of Maximum Functions Analyzing f (x) = max (g (x), h (x)) and finding where the 'winner' changes. 3 Local Extrema and Smoothness First and second derivative tests, identifying peaks and valleys. 4 Mean Value Theorem and Derivative Bounds Bounding function values using derivative limits over an interval. 4 Topics ~5 Core PYQs

Differentiability Conditions for Piecewise Functions

Differentiability Conditions for Piecewise Functions Why this matters: Piecewise functions often have 'joints' where the formula changes. These joints are natural candidates for sharp corners or breaks. To make the function differentiable everywhere, we must carefully stitch these pieces together. What you'll learn here The strict prerequisite of continuity before differentiability. How to compute and equate Left-Hand and Right-Hand Derivatives. A systematic method to find unknown constants in piecewise definitions. Context: Differentiability & Optimization Topic 1 of 4

The Two-Step Rule for Smooth Joints

@keyframes highlightStep { 0% { background: color-mix(in srgb, var(--color-muted) 5%, transparent); border-left-color: var(--color-border); } 50% { background: color-mix(in srgb, var(--color-primary) 15%, transparent); border-left-color: var(--color-primary); } 100% { background: color-mix(in srgb, var(--color-muted) 5%, transparent); border-left-color: var(--color-border); } } .step-highlight-1 { animation: highlightStep 3s infinite 0s; } .step-highlight-2 { animation: highlightStep 3s infinite 1.5s; } The Golden Rule of Differentiability A function f (x) is differentiable at a point x = c if and only if: 1. It is continuous at x = c . 2. Its left-hand derivative equals its right-hand derivative at x = c . Differentiable at c ⟹ Continuous at c If a function has a jump or a break at x = c, it is impossible for it to have a well-defined tangent line there. Therefore, continuity is the mandatory first step . Breaking it Down For a piecewise function with a joint at x = c : Step 1 (Continuity): lim_{x \to c^{-}} f (x) = lim_{x \to c^{+}} f (x) = f (c) Step 2 (Differentiability): f^{'} (c^{-}) = f^{'} (c^{+}) Both conditions must hold simultaneously.

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Question 1

Level 1: Warm-up

Let f (x) = max (x^{2}, x^{4}) for x \in R . Which of the following statements is true regarding the differentiability of f (x) ?

Question 2

Level 1: Warm-up

Consider the piecewise function defined by f (x) = {2 x + 1 x^{2} + c if x < 1 if x \geq 1 where c is a real constant. If f (x) is differentiable at x = 1, what is the value of c ?

Question 3

Level 1: Warm-up

Consider the following assertion and reason: <b>Assertion (A):</b> The function f (x) = max (x^{3}, 0) is differentiable at x = 0 . <b>Reason (R):</b> The curves y = x^{3} and y = 0 intersect at x = 0 and have the same derivative at x = 0 .

Question 4

Level 1: Warm-up

A piecewise function f (x) has exactly one joint at x = c and contains two unknown real constants a and b . If f (x) is given to be differentiable everywhere, what is the minimum number of independent equations you must formulate to uniquely determine the values of a and b ?

Question 5

Level 1: Warm-up

Let f (x) = max (x^{2} - 4, 0) for x \in R . How many points of non-differentiability does f (x) have?

Question 6

Level 1: Warm-up

Consider the function f (x) = ∣ x^{2} - 4∣ . At how many distinct points in the interval [- 5, 5] is f (x) NOT differentiable?

Question 7

Level 1: Warm-up

A function f (x) is differentiable for all real x and has a local maximum at x = 3 . What is the exact value of f^{'} (3) ?

Question 8

Level 1: Warm-up

Let f (x) = x^{2} on the closed interval [- 1, 2] . Which of the following statements about the extrema of f (x) is TRUE?

Question 9

Level 1: Warm-up

Let f (x) = {x^{3} 6 x - 8 if x \leq 2 if x > 2 . What is the maximum of the left-hand derivative and the right-hand derivative of f (x) at the boundary x = 2 ?

Question 10

Level 1: Warm-up

Consider the function f (x) = ∣ x + 2∣ - 1 . Which of the following statements is true regarding its local extrema?

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Differentiability and Optimization Notes for GATE CS

Differentiability and Optimization notes for GATE CS: 39 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Differentiability and Optimization

Your journey through the calculus of change and optimization.

Differentiability Conditions for Piecewise Functions (Current)

Making sharp corners smooth. Matching limits and derivatives at joints.

Nondifferentiability of Maximum Functions

Analyzing

f (x) = max (g (x), h (x))

and finding where the 'winner' changes.

Local Extrema and Smoothness

First and second derivative tests, identifying peaks and valleys.

Mean Value Theorem and Derivative Bounds

Bounding function values using derivative limits over an interval.

4 Topics ~5 Core PYQs

Differentiability Conditions for Piecewise Functions

Why this matters: Piecewise functions often have 'joints' where the formula changes. These joints are natural candidates for sharp corners or breaks. To make the function differentiable everywhere, we must carefully stitch these pieces together.

What you'll learn here

The strict prerequisite of continuity before differentiability.
How to compute and equate Left-Hand and Right-Hand Derivatives.
A systematic method to find unknown constants in piecewise definitions.

Context: Differentiability & Optimization Topic 1 of 4

The Two-Step Rule for Smooth Joints

The Golden Rule of Differentiability

A function $f (x)$ is differentiable at a point $x = c$ if and only if:

1. It is continuous at $x = c$ .

2. Its left-hand derivative equals its right-hand derivative at $x = c$ .

Differentiable at c ⟹ Continuous at c

If a function has a jump or a break at $x = c$ , it is impossible for it to have a well-defined tangent line there. Therefore, continuity is the mandatory first step.

Breaking it Down

For a piecewise function with a joint at

x = c

Step 1 (Continuity): $lim_{x \to c^{-}} f (x) = lim_{x \to c^{+}} f (x) = f (c)$
Step 2 (Differentiability): $f^{'} (c^{-}) = f^{'} (c^{+})$

Both conditions must hold simultaneously.

Computing Left and Right Derivatives

Formal Definitions vs. Practical Shortcuts

Formal Definition

The left-hand derivative (LHD) and right-hand derivative (RHD) at $x = c$ are defined using limits:

LHD = h \to 0^{-} lim \frac{f ( c + h ) - f ( c )}{h}

RHD = h \to 0^{+} lim \frac{f ( c + h ) - f ( c )}{h}

Practical Shortcut for Piecewise Functions

If the function is already known to be continuous at $x = c$ , you do not need to use the limit definition. You can simply differentiate the individual pieces and evaluate them at $x = c$ .

Let $f (x) = {g (x) h (x) x < c x \geq c$

If continuous at $c$ :

$LHD = g^{'} (c)$
$RHD = h^{'} (c)$

For differentiability, we just need $g^{'} (c) = h^{'} (c)$ .

Differentiability and Optimization: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Engineering Mathematics MCQ

Let $f (x) = max (x^{2}, x^{4})$ for $x \in R$ . Which of the following statements is true regarding the differentiability of $f (x)$ ?

A.
$f (x)$ is differentiable everywhere.
B.
$f (x)$ is not differentiable at $x = - 1$ and $x = 1$ .
C.
$f (x)$ is not differentiable at $x = - 1, 0$ , and $x = 1$ .
D.
$f (x)$ is not differentiable only at $x = 0$ .

Correct Answer:

Step-by-Step Solution

Key idea: The function $max (g (x), h (x))$ is non-differentiable only where $g (x) = h (x)$ AND $g^{'} (x) \neq = h^{'} (x)$ .

Step 1: Find the intersection points.

$x^{2} = x^{4} ⟹ x^{4} - x^{2} = 0 ⟹ x^{2} (x^{2} - 1) = 0$ .

The roots are $x = 0, x = 1, x = - 1$ .

Step 2: Check the derivatives at each intersection.

Let $g (x) = x^{2}$ and $h (x) = x^{4}$ .

$g^{'} (x) = 2 x$ and $h^{'} (x) = 4 x^{3}$ .

At $x = 0$ : $g^{'} (0) = 0$ and $h^{'} (0) = 0$ . Since $g^{'} (0) = h^{'} (0)$ , the curves are tangent. $f (x)$ is differentiable at $x = 0$ .
At $x = 1$ : $g^{'} (1) = 2$ and $h^{'} (1) = 4$ . Since $2 \neq = 4$ , $f (x)$ is NOT differentiable at $x = 1$ .
At $x = - 1$ : $g^{'} (- 1) = - 2$ and $h^{'} (- 1) = - 4$ . Since $- 2 \neq = - 4$ , $f (x)$ is NOT differentiable at $x = - 1$ .

Step 3: Conclusion.

The function is not differentiable exactly at $x = - 1$ and $x = 1$ .

Answer: B

Question 2 · Engineering Mathematics MCQ

Consider the piecewise function defined by

$f (x) = {2 x + 1 x^{2} + c if x < 1 if x \geq 1$

where $c$ is a real constant. If $f (x)$ is differentiable at $x = 1$ , what is the value of $c$ ?

A.
0
B.
1
C.
2
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: Differentiability at a joint requires continuity first, then equal left and right derivatives.

Step 1: Check the derivatives at $x = 1$ .

Left-hand derivative (LHD) = $\frac{d}{d x} (2 x + 1) = 2$ .

Right-hand derivative (RHD) = $\frac{d}{d x} (x^{2} + c) = 2 x$ . At $x = 1$ , RHD = $2 (1) = 2$ .

Since LHD = RHD = 2, the derivative condition is satisfied regardless of $c$ .

Step 2: Apply the continuity condition at $x = 1$ .

Left limit = $2 (1) + 1 = 3$ .

Right limit = $(1)^{2} + c = 1 + c$ .

For continuity, Left limit = Right limit $⟹ 3 = 1 + c ⟹ c = 2$ .

Answer: 2

Question 3 · Engineering Mathematics MCQ

Consider the following assertion and reason:

<b>Assertion (A):</b> The function $f (x) = max (x^{3}, 0)$ is differentiable at $x = 0$ .

<b>Reason (R):</b> The curves $y = x^{3}$ and $y = 0$ intersect at $x = 0$ and have the same derivative at $x = 0$ .

A.
Both A and R are true, and R is the correct explanation of A.
B.
Both A and R are true, but R is not the correct explanation of A.
C.
A is true, but R is false.
D.
A is false, but R is true.

Correct Answer:

Step-by-Step Solution

Key idea: Evaluate the truth of the assertion and the reason independently, then check if the reason explains the assertion.

Step 1: Analyze the Assertion (A).

$f (x) = max (x^{3}, 0)$ . At $x = 0$ , the curves $y = x^{3}$ and $y = 0$ intersect.

For $x < 0$ , $x^{3} < 0$ , so $f (x) = 0$ . Derivative is 0.

For $x > 0$ , $x^{3} > 0$ , so $f (x) = x^{3}$ . Derivative is $3 x^{2}$ , which approaches 0 as $x \to 0^{+}$ .

Since the left and right derivatives match (both are 0), $f (x)$ is indeed differentiable at $x = 0$ .

Assertion (A) is True.

Step 2: Analyze the Reason (R).

The curves $y = x^{3}$ and $y = 0$ intersect at $x = 0$ (since $0^{3} = 0$ ).

The derivative of $y = x^{3}$ is $3 x^{2}$ . At $x = 0$ , it is 0.

The derivative of $y = 0$ is 0.

Since both derivatives are 0 at $x = 0$ , they have the same derivative.

Reason (R) is True.

Step 3: Link A and R.

The reason $f (x)$ is differentiable at the intersection is precisely because the derivatives match, making the transition smooth. Thus, R correctly explains A.

Answer: A

Question 4 · Engineering Mathematics MCQ

A piecewise function $f (x)$ has exactly one joint at $x = c$ and contains two unknown real constants $a$ and $b$ . If $f (x)$ is given to be differentiable everywhere, what is the minimum number of independent equations you must formulate to uniquely determine the values of $a$ and $b$ ?

A.
1
B.
2
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: Differentiability at a joint provides exactly two independent constraints: one for continuity and one for smoothness.

Step 1: Analyze the conditions at the joint $x = c$ .

For $f (x)$ to be differentiable at $x = c$ , it must first be continuous at $x = c$ .

Continuity condition: $lim_{x \to c^{-}} f (x) = lim_{x \to c^{+}} f (x)$ . This gives 1 equation involving $a$ and $b$ .

Step 2: Analyze the differentiability condition.

The left-hand derivative must equal the right-hand derivative at $x = c$ .

Differentiability condition: $f^{'} (c^{-}) = f^{'} (c^{+})$ . This gives a 2nd independent equation involving $a$ and $b$ .

Step 3: Determine the minimum number of equations.

Since we have two unknowns ( $a$ and $b$ ), we need exactly 2 independent equations to uniquely determine them. The joint provides exactly these 2 equations.

Answer: 2

Question 5 · Engineering Mathematics MCQ

Let $f (x) = max (x^{2} - 4, 0)$ for $x \in R$ . How many points of non-differentiability does $f (x)$ have?

A.
0
B.
1
C.
2
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: The function $max (g (x), h (x))$ is non-differentiable exactly where $g (x) = h (x)$ and $g^{'} (x) \neq = h^{'} (x)$ .

Step 1: Find the intersection points by setting $g (x) = h (x)$ .

$x^{2} - 4 = 0 ⟹ x^{2} = 4 ⟹ x = 2$ or $x = - 2$ .

Step 2: Check the derivatives at these intersection points.

Let $g (x) = x^{2} - 4$ and $h (x) = 0$ .

$g^{'} (x) = 2 x$ and $h^{'} (x) = 0$ .

At $x = 2$ : $g^{'} (2) = 4$ and $h^{'} (2) = 0$ . Since $4 \neq = 0$ , $f (x)$ is not differentiable at $x = 2$ .

At $x = - 2$ : $g^{'} (- 2) = - 4$ and $h^{'} (- 2) = 0$ . Since $- 4 \neq = 0$ , $f (x)$ is not differentiable at $x = - 2$ .

Step 3: Count the points.

There are exactly 2 points of non-differentiability.

Answer: 2

Question 6 · Engineering Mathematics MCQ

Consider the function $f (x) = ∣ x^{2} - 4∣$ . At how many distinct points in the interval $[- 5, 5]$ is $f (x)$ NOT differentiable?

A.
0
B.
1
C.
2
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: An absolute value function $∣ g (x) ∣$ is non-differentiable exactly where $g (x) = 0$ AND $g^{'} (x) \neq = 0$ .

Step 1: Find the roots of the inside function.

$x^{2} - 4 = 0 ⟹ x^{2} = 4 ⟹ x = 2$ or $x = - 2$ .

Both points are within the interval $[- 5, 5]$ .

Step 2: Check the derivative of the inside function at these roots.

Let $g (x) = x^{2} - 4$ . Then $g^{'} (x) = 2 x$ .

At $x = 2$ , $g^{'} (2) = 4 \neq = 0$ . So $f (x)$ is not differentiable at $x = 2$ .

At $x = - 2$ , $g^{'} (- 2) = - 4 \neq = 0$ . So $f (x)$ is not differentiable at $x = - 2$ .

Step 3: Count the points.

There are exactly 2 points where the function is not differentiable.

Answer: 2

Question 7 · Engineering Mathematics MCQ

A function $f (x)$ is differentiable for all real $x$ and has a local maximum at $x = 3$ . What is the exact value of $f^{'} (3)$ ?

A.
0
B.
1
C.
3
D.
-1

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct application of Fermat's Theorem for stationary points.

Step 1: Recall Fermat's Theorem.

If a function $f (x)$ has a local extremum (maximum or minimum) at a point $x = c$ , and if $f (x)$ is differentiable at $x = c$ , then the derivative at that point must be zero.

Step 2: Apply the theorem to the given information.

We are given that $f (x)$ has a local maximum at $x = 3$ and is differentiable everywhere (including at $x = 3$ ).

Therefore, by Fermat's Theorem, $f^{'} (3)$ must be exactly 0.

Answer: 0

Question 8 · Engineering Mathematics MCQ

Let $f (x) = x^{2}$ on the closed interval $[- 1, 2]$ . Which of the following statements about the extrema of $f (x)$ is TRUE?

A.
The global maximum occurs at $x = 0$ .
B.
The global minimum occurs at $x = 2$ .
C.
The global maximum occurs at $x = 2$ .
D.
The function has no global extrema on this interval.

Correct Answer:

Step-by-Step Solution

Key idea: This is a question about finding global extrema on a closed interval. You must check both critical points and endpoints.

Step 1: Find the critical points inside the interval $(- 1, 2)$ .

$f^{'} (x) = 2 x$ . Setting $f^{'} (x) = 0$ gives $x = 0$ .

Step 2: Evaluate $f (x)$ at the critical point and the endpoints.

At the critical point $x = 0$ : $f (0) = 0^{2} = 0$ .

At the left endpoint $x = - 1$ : $f (- 1) = (- 1)^{2} = 1$ .

At the right endpoint $x = 2$ : $f (2) = 2^{2} = 4$ .

Step 3: Compare the values to find the global extrema.

The values are 0, 1, and 4.

The highest value is 4, which occurs at $x = 2$ . So the global maximum is at $x = 2$ .

The lowest value is 0, which occurs at $x = 0$ . So the global minimum is at $x = 0$ .

Step 4: Match with the options.

Option C correctly states that the global maximum occurs at $x = 2$ .

Answer: C

Question 9 · Engineering Mathematics MCQ

Let $f (x) = {x^{3} 6 x - 8 if x \leq 2 if x > 2$ . What is the maximum of the left-hand derivative and the right-hand derivative of $f (x)$ at the boundary $x = 2$ ?

A.
6
B.
8
C.
12
D.
24

Correct Answer:

Step-by-Step Solution

Key idea: Compute the derivative of each piece separately and evaluate them at the boundary point $x = 2$ .

Step 1: Find the left-hand derivative (LHD) at $x = 2$ .

The left piece is $g (x) = x^{3}$ .

$g^{'} (x) = 3 x^{2}$ .

At $x = 2$ , LHD = $3 (2)^{2} = 3 (4) = 12$ .

Step 2: Find the right-hand derivative (RHD) at $x = 2$ .

The right piece is $h (x) = 6 x - 8$ .

$h^{'} (x) = 6$ .

At $x = 2$ , RHD = 6.

Step 3: Find the maximum of these two values.

$max (12, 6) = 12$ .

Answer: 12

Question 10 · Engineering Mathematics MCQ

Consider the function $f (x) = ∣ x + 2∣ - 1$ . Which of the following statements is true regarding its local extrema?

A.
It has no local minimum because $f^{'} (x)$ is undefined at $x = - 2$ .
B.
It has a local minimum at $x = - 2$ with a value of $- 1$ .
C.
It has a local minimum at $x = 2$ with a value of $1$ .
D.
It has a local maximum at $x = - 2$ with a value of $- 1$ .

Correct Answer:

Step-by-Step Solution

Key idea: This is a question about non-differentiable extrema. Absolute value functions can have extrema at their corners.

Step 1: Analyze the function $f (x) = ∣ x + 2∣ - 1$ .

The absolute value $∣ x + 2∣$ is always greater than or equal to 0.

The minimum value of $∣ x + 2∣$ is 0, which occurs when $x + 2 = 0 ⟹ x = - 2$ .

Step 2: Find the minimum value of $f (x)$ .

At $x = - 2$ , $f (- 2) = ∣0∣ - 1 = - 1$ .

Since $∣ x + 2∣ \geq 0$ for all $x$ , $f (x) \geq - 1$ for all $x$ .

Therefore, $f (x)$ has a local (and global) minimum at $x = - 2$ with a value of $- 1$ .

Step 3: Address the differentiability.

Although $f^{'} (x)$ is undefined at $x = - 2$ (it's a sharp corner), the function still achieves its lowest value there. Extrema do not require differentiability.

Answer: B

differentiability and optimization notes

Chapter Roadmap: Differentiability and Optimization

Chapter Roadmap: Differentiability and Optimization

Differentiability Conditions for Piecewise Functions

Differentiability Conditions for Piecewise Functions

What you'll learn here

The Two-Step Rule for Smooth Joints

The Golden Rule of Differentiability

Try a question

Login to view the complete notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Differentiability and Optimization Notes for GATE CS

Chapter Roadmap: Differentiability and Optimization

Chapter Roadmap: Differentiability and Optimization

Differentiability Conditions for Piecewise Functions

Differentiability Conditions for Piecewise Functions

What you'll learn here

The Two-Step Rule for Smooth Joints

The Golden Rule of Differentiability

Computing Left and Right Derivatives

Formal Definitions vs. Practical Shortcuts

Formal Definition

Practical Shortcut for Piecewise Functions

Differentiability and Optimization: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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