GATE DA

Maxima, minima and single-variable optimisation

Maxima and minima locate where a function of one variable turns from rising to falling or the reverse, and single-variable optimisation uses that to find the best value of a quantity. In GATE DA you apply it to find critical points, classify them with the first-, second- and higher-derivative tests, find global extrema on closed, open and unbounded intervals, read convexity and inflection points, and see gradient descent as a repeated step against the derivative.

5 previous-year questions · 5 solved · 6 practice questions

Critical points and monotonicity

Convention for the whole note: $f$ is a real function of one real variable, differentiable where stated. A point $c$ is interior if some open interval around $c$ lies inside the domain (or inside the interval under discussion). $f$ has a local maximum at $c$ if $f (c) \geq f (x)$ for every $x$ in the domain that lies in some open interval around $c$ , and a global maximum on a set $S$ at $c \in S$ if $f (c) \geq f (x)$ for every $x \in S$ . Minima are defined the same way with $\leq$ , and an extremum is a maximum or a minimum. A global extremum attained at an interior point is also a local extremum there.

CRITICAL POINT

f^{'} (c) = 0 or f^{'} (c) does not exist, c interior to the domain

Necessary condition: if $f$ has a local extremum at an interior point $c$ and $f^{'} (c)$ exists, then $f^{'} (c) = 0$ . So every interior local extremum is a critical point. The converse is false: $f (x) = x^{3}$ has $f^{'} (0) = 0$ but no extremum at $0$ .

MONOTONICITY FROM THE SIGN OF THE DERIVATIVE

f^{'} (x) > 0 for all x \in I \Rightarrow f strictly increasing on I, f^{'} (x) < 0 for all x \in I \Rightarrow f strictly decreasing on I

$I$ is an interval, $f$ is continuous on $I$ and differentiable at its interior points. $f^{'} \geq 0$ on $I$ gives non-decreasing; if in addition $f^{'} = 0$ only at isolated points, $f$ is still strictly increasing. The converse of the first implication fails: $x^{3}$ is strictly increasing on $R$ although $f^{'} (0) = 0$ .

EXAM TIP

Check where $f^{'}$ does not exist as well as where it is zero. For $f (x) = ∣ x ∣$ the derivative does not exist at $0$ , and $x = 0$ is the global minimum. Solving only $f^{'} (x) = 0$ misses it.

Classifying critical points: first, second and higher derivative tests

First-derivative test (for a critical point $c$ where $f$ is continuous, and $f$ is differentiable near $c$ except possibly at $c$)

Find a $δ > 0$ such that $f^{'}$ has one constant sign on $(c - δ, c)$ and one constant sign on $(c, c + δ)$ .
If $f^{'}$ changes from positive to negative across $c$ , $c$ is a local maximum.
If $f^{'}$ changes from negative to positive across $c$ , $c$ is a local minimum.
If $f^{'}$ has the same sign on both sides, $c$ is not an extremum.
The test works even where $f^{'} (c)$ does not exist, for example $∣ x ∣$ at $0$ .

SECOND-DERIVATIVE TEST

f^{'} (c) = 0 : f^{''} (c) > 0 \Rightarrow local min, f^{''} (c) < 0 \Rightarrow local max

Needs $f^{'} (c) = 0$ and $f^{''} (c)$ to exist, so it cannot be used at a critical point where $f^{'} (c)$ does not exist. If $f^{''} (c) = 0$ the test is inconclusive and you must use the first-derivative test or higher derivatives.

HIGHER-ORDER DERIVATIVE TEST

f^{'} (c) = \dots = f^{(n - 1)} (c) = 0, f^{(n)} (c) \neq = 0 : n even \Rightarrow {local min if f^{(n)} (c) > 0 local max if f^{(n)} (c) < 0, n odd \Rightarrow no extremum

Here $n \geq 2$ and $f$ is $n$ times differentiable at $c$ . For odd $n \geq 3$ the point $c$ is a point of inflection with a horizontal tangent. Example: $f (x) = x^{4}$ at $0$ has $n = 4$ and $f^{(4)} (0) = 24 > 0$ , so a local minimum; $f (x) = x^{3}$ at $0$ has $n = 3$ , so no extremum.

Function at $x = 0$	$f^{'} (0)$	$f^{''} (0)$	Behaviour at $0$
$x^{2}$	$0$	$2 > 0$	local (and global) minimum
$- x^{2}$	$0$	$- 2 < 0$	local (and global) maximum
$x^{4}$	$0$	$0$	local (and global) minimum; second-derivative test fails
$x^{3}$	$0$	$0$	neither; $f^{'}$ does not change sign (a point of inflection)
$∣ x ∣$	does not exist	does not exist	local (and global) minimum; found only as a non-differentiable critical point

EXAM TIP

When $f^{''} (c) = 0$ do not say "no extremum". $x^{4}$ and $x^{3}$ both have $f^{'} (0) = f^{''} (0) = 0$ , and only one of them has an extremum at $0$ .

Global extrema on an interval

EXTREME VALUE THEOREM

f continuous on [a, b] \Rightarrow f attains a global maximum and a global minimum on [a, b]

Both conditions matter: continuity, and a closed bounded interval. On an open interval an extremum may not exist, for example $f (x) = x$ on $(0, 1)$ has neither. A discontinuous function on a closed interval may also fail, for example $f (x) = x$ for $0 \leq x < 1$ with $f (1) = 0$ has no maximum on $[0, 1]$ .

Procedure for the global maximum and minimum of a continuous $f$ on $[a,b]$

Find all critical points of $f$ in the open interval $(a, b)$ : points where $f^{'} = 0$ and points where $f^{'}$ does not exist.
Evaluate $f$ at each of those critical points and at both endpoints $a$ and $b$ .
The largest of these values is the global maximum on $[a, b]$ and the smallest is the global minimum.
No derivative test is needed for this step; only the comparison of values.

WORKED EXAMPLE

Find the global maximum and global minimum of $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 1$ on $[0, 3]$ , and classify its local extrema.

Solution

$f^{'} (x) = 6 x^{2} - 18 x + 12 = 6 (x - 1) (x - 2)$ , so the critical points are $x = 1$ and $x = 2$ , both inside $(0, 3)$ .
$f^{''} (x) = 12 x - 18$ : $f^{''} (1) = - 6 < 0$ so $x = 1$ is a local maximum; $f^{''} (2) = 6 > 0$ so $x = 2$ is a local minimum.
Values: $f (0) = 1$ , $f (1) = 2 - 9 + 12 + 1 = 6$ , $f (2) = 16 - 36 + 24 + 1 = 5$ , $f (3) = 54 - 81 + 36 + 1 = 10$ .
Largest is $10$ at the endpoint $x = 3$ ; smallest is $1$ at the endpoint $x = 0$ .

Answer: Global maximum $10$ at $x = 3$ , global minimum $1$ at $x = 0$ on $[0, 3]$ . The local maximum at $x = 1$ ( $f = 6$ ) and local minimum at $x = 2$ ( $f = 5$ ) are not global.

Open or unbounded intervals (for continuous $f$ with finitely many critical points, and the limits of $f$ at the excluded or infinite ends existing, possibly $\pm\infty$)

A global maximum or minimum may not exist.
Compute $f$ at every critical point and at every endpoint that belongs to the interval, and the limit of $f$ at every endpoint that does not belong to it (including $\pm \infty$ ).
If the largest attained value is at least every end limit, it is the global maximum; if some end limit is strictly larger, there is no global maximum.
The global minimum is found the same way with smallest in place of largest.

WORKED EXAMPLE

Find the global maximum and global minimum, if they exist, of $f (x) = x e^{- x}$ on $[0, \infty)$ .

Solution

$f^{'} (x) = e^{- x} - x e^{- x} = (1 - x) e^{- x}$ , which is zero only at $x = 1$ .
Values: $f (0) = 0$ and $f (1) = e^{- 1} \approx 0.368$ ; end limit $lim_{x \to \infty} x e^{- x} = 0$ .
The largest attained value $e^{- 1}$ exceeds the end limit $0$ , so it is the global maximum.
The smallest attained value is $0$ at $x = 0$ , and the end limit $0$ is not smaller, so $0$ is the global minimum.

Answer: Global maximum $1/ e$ at $x = 1$ ; global minimum $0$ at $x = 0$ on $[0, \infty)$ .

EXAM TIP

A local maximum can be smaller than the function value at an endpoint. In the first example above the local maximum $6$ loses to the endpoint value $10$ . Always include the endpoints.

Convexity, concavity and points of inflection

CONVEX AND CONCAVE FUNCTIONS (TWICE DIFFERENTIABLE ON AN INTERVAL $I$)

f^{''} (x) \geq 0 for all x \in I ⟺ f convex on I, f^{''} (x) \leq 0 for all x \in I ⟺ f concave on I

Convex (also called concave up) means $f (λ x + (1 - λ) y) \leq λ f (x) + (1 - λ) f (y)$ for all $x, y \in I$ and $λ \in [0, 1]$ . For differentiable $f$ this is equivalent to the graph lying on or above every tangent line: $f (y) \geq f (x) + f^{'} (x) (y - x)$ for all $x, y \in I$ . $f^{''} > 0$ on $I$ makes $f$ strictly convex, but not conversely: $x^{4}$ is strictly convex although $f^{''} (0) = 0$ .

POINT OF INFLECTION

f continuous at c and f^{''} changes sign at c \Rightarrow c is a point of inflection

A point of inflection is a point $c$ where $f$ is continuous and the graph changes from convex to concave or the reverse. If $f^{''} (c)$ exists at an inflection point then $f^{''} (c) = 0$ . The converse is false: $f (x) = x^{4}$ has $f^{''} (0) = 0$ but $f^{''} \geq 0$ on both sides, so $0$ is not an inflection point. An inflection point can also occur where $f^{''}$ does not exist: $f (x) = x ∣ x ∣$ has $f^{''} (x) = - 2$ for $x < 0$ and $2$ for $x > 0$ , with $f^{''} (0)$ undefined.

For a convex function on an interval $I$ , any local minimum is a global minimum on $I$ . If $f$ is also differentiable, every point $c$ with $f^{'} (c) = 0$ is a global minimum on $I$ , because the tangent-line inequality with $f^{'} (c) = 0$ gives $f (y) \geq f (c)$ for all $y \in I$ ; no endpoint comparison is needed there. A strictly convex function has at most one minimiser, and possibly none: $e^{x}$ is strictly convex on $R$ and has no minimum. A concave function behaves the same way with maximum in place of minimum.

EXAM TIP

To count inflection points, list the zeros of $f^{''}$ and the points where $f^{''}$ does not exist (but $f$ is continuous), and keep only those where the sign of $f^{''}$ actually changes. An inflection point need not be a critical point: $f (x) = x^{3} + x$ has an inflection point at $0$ with $f^{'} (0) = 1$ .

Single-variable optimisation problems

Procedure for an optimisation word problem

Name the quantity to be maximised or minimised and write it in terms of the variables.
Use the stated constraint to eliminate all but one variable.
Write the domain of the remaining variable from the physical meaning (lengths positive, and so on).
Differentiate and find the critical points in the domain.
Confirm the optimum is global: compare values with the endpoints (or end limits) of the domain, or use the single-critical-point rule below. A second-derivative test on its own shows only a local extremum.
Answer the question that was asked (the optimal value, or where it occurs).

SINGLE CRITICAL POINT ON AN INTERVAL

f continuous on an interval I, c the only critical point in I, c a local max \Rightarrow c is the global max on I

The same holds with min in place of max. Reason: if some $x_{1} \in I$ had $f (x_{1}) > f (c)$ , $f$ would have a local minimum strictly between $c$ and $x_{1}$ , which would be a second critical point.

WORKED EXAMPLE

A farmer has $80$ m of fencing to enclose a rectangular plot on three sides; the fourth side is an existing wall and needs no fence. Find the largest possible area.

Solution

Let each of the two sides perpendicular to the wall be $x$ m, so the side parallel to the wall is $80 - 2 x$ m, with domain $0 < x < 40$ .
Area $A (x) = x (80 - 2 x) = 80 x - 2 x^{2}$ .
$A^{'} (x) = 80 - 4 x = 0$ gives $x = 20$ , which lies in the domain.
$A^{''} (x) = - 4 < 0$ for every $x$ , so $A$ is concave and the critical point $x = 20$ is the global maximum on the domain.
The parallel side is $80 - 40 = 40$ m, so $A (20) = 20 \times 40 = 800$ .

Answer: Maximum area $800 m^{2}$ with sides $20$ m and $40$ m.

EXAM TIP

A quadratic $a x^{2} + b x + c$ with $a \neq = 0$ has its only extremum, global over all real $x$ , at $x = - b / (2 a)$ , with value $c - b^{2} / (4 a)$ : a minimum if $a > 0$ , a maximum if $a < 0$ . If the domain is restricted and $- b / (2 a)$ lies outside it, the optimum on the domain is at an endpoint.

The one-variable view of gradient descent

Gradient descent looks for a minimum of a differentiable $f$ by repeatedly stepping against the derivative. The sign of $f^{'} (x_{k})$ tells which way $f$ rises, so when $f^{'} (x_{k}) \neq = 0$ a small enough step the opposite way lowers $f$ . Here $η > 0$ is the step size (learning rate).

GRADIENT DESCENT UPDATE (ONE VARIABLE)

x_{k + 1} = x_{k} - η f^{'} (x_{k})

The iterate stays put exactly when $f^{'} (x_{k}) = 0$ , that is at a critical point. That point need not be a minimum if the iterate lands on it exactly. On a non-convex $f$ the iteration can settle at a local minimum that is not global; which one depends on $x_{0}$ and $η$ .

GRADIENT DESCENT ON A QUADRATIC

f (x) = \frac{a}{2} (x - x^{*})^{2}, a > 0 : x_{k + 1} - x^{*} = (1 - η a) (x_{k} - x^{*}) \Rightarrow x_{k} - x^{*} = (1 - η a)^{k} (x_{0} - x^{*})

Converges to $x^{*}$ from every starting point if and only if $∣1 - η a ∣ < 1$ , that is $0 < η < 2/ a$ . With $η = 1/ a$ it reaches $x^{*}$ in one step. For $1/ a < η < 2/ a$ the iterates alternate sides of $x^{*}$ while converging. At $η = 2/ a$ they bounce between $x_{0}$ and $2 x^{*} - x_{0}$ forever, and for $η > 2/ a$ they diverge (whenever $x_{0} \neq = x^{*}$ ). Here $a = f^{''}$ , the constant curvature.

EXAM TIP

For $f (x) = x^{2}$ we have $a = 2$ and the update is $x_{k + 1} = (1 - 2 η) x_{k}$ . With $η = 0.5$ the iterate reaches $0$ in one step; with $η = 1$ it just flips sign forever ( $x_{k + 1} = - x_{k}$ ); with $η > 1$ it diverges (for $x_{0} \neq = 0$ ).

PREVIOUS-YEAR QUESTIONS

5 questions from the DA papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE DA 2026 · GATE DA 2025 · GATE DA 2024

GATE 2026 DA Q.27

GATE 2026 · DAQ.27 · 1 mark · MSQ

Let $f (x) = x^{3} - 3 x^{2} + 2$ be a function defined on $(- 1, 3]$ .

Which of the following statements is/are correct?

(A)
$f (x)$ has exactly two roots in $[- 0.9, 0]$ .
(B)
$f (x)$ has a minimum at 2 only.
(C)
$f (x)$ has maximum at 0 only.
(D)
$f (x)$ has a root at 1.

Answer and solution

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GATE 2025 DA Q.49

GATE 2025 · DAQ.49 · 2 marks · MSQ

Consider the function $f (x) = \frac{x ^{3}}{3} + \frac{7}{2} x^{2} + 10 x + \frac{133}{2}$ , $x \in [- 8, 0]$ . Which of the following statements is/are correct?

(A)
The maximum value of $f$ is attained at $x = - 5$
(B)
The minimum value of $f$ is attained at $x = - 2$
(C)
The maximum value of $f$ is $\frac{133}{2}$
(D)
The minimum value of the derivative of $f$ is attained at $x = - \frac{7}{2}$

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GATE 2025 DA Q.51

GATE 2025 · DAQ.51 · 2 marks · MSQ

Let $f : R \to R$ be a twice-differentiable function and suppose its second derivative satisfies $f^{''} (x) > 0$ for all $x \in R$ . Which of the following statements is/are ALWAYS correct?

(A)
$f$ has a local minima
(B)
There does not exist $x$ and $y$ , $x \neq = y$ , such that $f^{'} (x) = f^{'} (y) = 0$
(C)
$f$ has at most one global minimum
(D)
$f$ has at most one local minimum

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GATE 2024 DA Q.15

GATE 2024 · DAQ.15 · 1 mark · MCQ

For any twice differentiable function $f : R \to R$ , if at some $x^{*} \in R$ , $f^{'} (x^{*}) = 0$ and $f^{''} (x^{*}) > 0$ , then the function $f$ necessarily has a ______ at $x = x^{*}$ .

Note: $R$ denotes the set of real numbers.

(A)
local minimum
(B)
global minimum
(C)
local maximum
(D)
global maximum

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GATE 2024 DA Q.50

GATE 2024 · DAQ.50 · 2 marks · MSQ

Consider the function $f : R \to R$ where $R$ is the set of all real numbers.

f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1

Which of the following statements is/are TRUE?

(A)
$x = 0$ is a local maximum of $f$
(B)
$x = 3$ is a local minimum of $f$
(C)
$x = - 1$ is a local maximum of $f$
(D)
$x = 0$ is a local minimum of $f$

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PRACTICE

6 questions to test yourself

PRACTICE 11 mark · MCQ

Let $f (x) = x^{3} - 6 x^{2} + 9 x + 2$ for real $x$ . What is the value of $f$ at its local minimum?

(A)
$2$
(B)
$6$
(C)
$3$
(D)
$0$

Answer and solution

Answer: (A)

$f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$ , so critical points are $x = 1$ and $x = 3$ .
$f^{''} (x) = 6 x - 12$ : $f^{''} (1) = - 6 < 0$ (local maximum) and $f^{''} (3) = 6 > 0$ (local minimum).
$f (3) = 27 - 54 + 27 + 2 = 2$ . (For reference, the local maximum is $f (1) = 6$ .)

PRACTICE 22 marks · NAT

Let $f (x) = x^{3} - 3 x + 5$ on the closed interval $[- 2, 2]$ . Find (global maximum of $f$ ) minus (global minimum of $f$ ) on this interval.

Answer and solution

Answer: $4$

$f^{'} (x) = 3 x^{2} - 3 = 0$ gives $x = \pm 1$ , both inside $(- 2, 2)$ .
Evaluate: $f (- 2) = - 8 + 6 + 5 = 3$ , $f (- 1) = - 1 + 3 + 5 = 7$ , $f (1) = 1 - 3 + 5 = 3$ , $f (2) = 8 - 6 + 5 = 7$ .
Global maximum is $7$ and global minimum is $3$ , so the difference is $4$ .

PRACTICE 32 marks · MSQ

For real $x$ , consider $f_{1} (x) = x^{4}$ , $f_{2} (x) = x^{3}$ , $f_{3} (x) = ∣ x ∣$ and $f_{4} (x) = - x^{4}$ . For which of these functions does $x = 0$ give a local extremum (a local maximum or a local minimum)?

(A)
$f_{1} (x) = x^{4}$
(B)
$f_{2} (x) = x^{3}$
(C)
$f_{3} (x) = ∣ x ∣$
(D)
$f_{4} (x) = - x^{4}$

Answer and solution

Answer: (A), (C), (D)

$x^{4} \geq 0 = f_{1} (0)$ for all $x$ , so $0$ is a minimum, even though $f_{1}^{''} (0) = 0$ .
$x^{3}$ is negative for $x < 0$ and positive for $x > 0$ , so $f_{2}$ has no extremum at $0$ .
$∣ x ∣ \geq 0 = f_{3} (0)$ , so $0$ is a minimum, although $f_{3}^{'} (0)$ does not exist.
$- x^{4} \leq 0 = f_{4} (0)$ , so $0$ is a maximum.

PRACTICE 42 marks · NAT

Gradient descent is applied to $f (x) = 3 (x - 2)^{2}$ with $x_{0} = 7$ and step size $η = 0.1$ , using $x_{k + 1} = x_{k} - η f^{'} (x_{k})$ . Find $x_{2}$ .

Answer and solution

Answer: $2.8$ (accepted range $\pm 0.001$ )

$f^{'} (x) = 6 (x - 2)$ , so $x_{k + 1} = x_{k} - 0.6 (x_{k} - 2)$ , which gives $x_{k + 1} - 2 = 0.4 (x_{k} - 2)$ .
$x_{0} - 2 = 5$ , so $x_{1} - 2 = 2$ and $x_{2} - 2 = 0.8$ .
Hence $x_{2} = 2.8$ .

PRACTICE 51 mark · MCQ

The number of points of inflection of $f (x) = x^{4} - 4 x^{3}$ on the real line is

(A)
$0$
(B)
$1$
(C)
$2$
(D)
$3$

Answer and solution

Answer: (C)

$f^{'} (x) = 4 x^{3} - 12 x^{2}$ and $f^{''} (x) = 12 x^{2} - 24 x = 12 x (x - 2)$ .
$f^{''}$ is positive for $x < 0$ , negative for $0 < x < 2$ and positive for $x > 2$ , so it changes sign at both $x = 0$ and $x = 2$ .
$f^{''}$ exists everywhere, so there are no other candidates. Hence there are exactly two points of inflection, $x = 0$ and $x = 2$ .

PRACTICE 61 mark · MCQ

Gradient descent $x_{k + 1} = x_{k} - η f^{'} (x_{k})$ with step size $η > 0$ is applied to $f (x) = 5 x^{2}$ . For which values of $η$ does it converge to $0$ from every starting point $x_{0}$ ?

(A)
$0 < η < 0.1$
(B)
$0 < η < 0.2$
(C)
$0 < η < 0.4$
(D)
$0 < η < 1$

Answer and solution

Answer: (B)

$f^{'} (x) = 10 x$ , so $x_{k + 1} = (1 - 10 η) x_{k}$ and $x_{k} = (1 - 10 η)^{k} x_{0}$ .
This tends to $0$ for every $x_{0}$ if and only if $∣1 - 10 η ∣ < 1$ , that is $0 < η < 0.2$ .
At $η = 0.2$ the iterates flip sign forever, and above it they diverge, so the range is $0 < η < 0.2$ .

OFFICIAL GATE 2027 SYLLABUS WORDING

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GATE DA

maxima and minima, optimization involving a single variable

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