Find the absolute maximum and absolute minimum values of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$ in $[1, 5]$.

Find the absolute maximum and absolute minimum values o — CBSE Class 12 Mathematics 2025 PYQ

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f(x) = 2x^3 - 15x^2 + 36x + 1

CBSECBSE Class XII Board 2025Subjective· 5mImportance★★★★★

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Concept understanding — Critical Points Analysis

Critical Points Analysis: Where Functions Change Direction

Hiking a mountain range, you reach peaks (highest spot around), valleys (bottoms), and flat stretches where the ground doesn't slope. These special locations — peaks, valleys, and flat spots — are critical points.

The Intuition

A function's graph is like that trail. At most points it is rising (positive slope) or falling (negative slope). At a critical point something changes: the slope becomes zero, or the slope doesn't exist (a sharp corner).

Throw a ball straight up: at the very top of its arc it stops for an instant before falling. Its velocity — the rate of change of height — is zero at that moment. That's a critical point.

The Precise Definition

x = c

Why Two Conditions?

Derivative equals zero catches the "flat" spots — peaks, valleys, horizontal plateaus — where the tangent line is horizontal.

Derivative does not exist catches sharp corners (like the tip of $|x|$ at $x=0$ ), vertical tangents, and cusps. Even without a zero slope, these can be peaks or valleys.

Watch out

A common mistake: thinking every critical point is a maximum or minimum. Not true. A critical point could be a "saddle point" — flat but neither. For example, $f(x) = x^3$ at $x=0$ has $f'(0)=0$ , yet the function just passes through with no extremum.

How to Find Critical Points

Find the derivative $f'(x)$ .
Solve $f'(x) = 0$ — these are candidates.
Check where $f'(x)$ does not exist — but only if $f(x)$ exists there (the point must be in the domain).
Collect all such $x$ -values.

Example 1: A Simple Polynomial

Let $f(x) = x^3 - 3x^2 + 1$ .

$f'(x) = 3x^2 - 6x = 3x(x - 2)$ .

$f'(x) = 0 \implies x = 0$ or $x = 2$ . Since $f'$ exists everywhere, the critical points are $x = 0$ and $x = 2$ .

Example 2: A Function with a Corner

Let $f(x) = |x|$ . Here $f'(x)$ does not exist at $x = 0$ (left derivative $-1$ , right derivative $+1$ ), and $f'(x) = 0$ has no solutions. So the only critical point is $x = 0$ .

Note

$x=0$ is actually a minimum of $|x|$ — the sharp corner is a valley.

What Critical Points Tell Us …

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The Mean Value Theorem tells us extreme values occur at critical points or endpoints. For $f(x)=2x^3-15x^2+36x+1$ on $[1,5]$ , the absolute maximum is $f(5)=56$ and the absolute minimum is $f(2)=29$ .

We need the absolute (global) maximum and minimum of a continuous function on a closed interval. The Extreme Value Theorem guarantees both exist. The only places they can occur are at critical points (where the derivative is zero or undefined) or at the endpoints of the interval.

Since $f$ is a polynomial, it's differentiable everywhere, so we only need to find where $f'(x)=0$ .

1. Find the derivative.

f'(x) = 6x^2 - 30x + 36

Factor out the common factor:

f'(x) = 6(x^2 - 5x + 6) = 6(x-2)(x-3)

2. Find critical points inside $[1,5]$ .

Set $f'(x)=0$ :

6(x-2)(x-3)=0 \quad\Rightarrow\quad x=2 \text{ or } x=3

Both $2$ and $3$ lie in the interval $[1,5]$ , so both are candidates.

3. Evaluate $f$ at the critical points and endpoints.

We have four $x$ -values to check: $1$ , $2$ , $3$ , $5$ .

$f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 2 - 15 + 36 + 1 = 24$
$f(2) = 2(8) - 15(4) + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$
$f(3) = 2(27) - 15(9) + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$
$f(5) = 2(125) - 15(25) + 36(5) + 1 = 250 - 375 + 180 + 1 = 56$

4. Compare the values. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 65/1/11 markMCQ
$f(x) = x^3 - 12x$
›Reveal solutionSolution
To find the least value of a continuous function on a closed interval, we evaluate the function at its critical points within the interval and at the interval's endpoints. The smallest of these values is the global minimum. For $f(x) = x^3 - 12x$ on $[0, 3]$ , the least value is $\boxed{-16}$ .

When we need to find the maximum or minimum value of a function over a specific interval, especially a closed one, we are looking for its global extrema. For a continuous function on a closed interval, these global extrema are guaranteed to exist. The key insight from calculus is that these extreme values can only occur at two types of points:

Critical points: These are points where the derivative of the function is zero or undefined. At such points, the function might have a local maximum or a local minimum.

Endpoints of the interval: Even if the function is increasing or decreasing throughout the interval, its maximum or minimum value could be at one of the boundaries.

Therefore, our strategy is to find all potential candidates for the minimum (critical points within the interval and the endpoints) and then compare the function's value at each of these candidates. The smallest value will be the least value of the function on the given interval.

Let's apply this method to $f(x) = x^3 - 12x$ on the interval $x \in [0, 3]$ .

Find the derivative of the function.

To locate critical points, we first need the first derivative of $f(x)$ .

$f(x) = x^3 - 12x$

$f'(x) = \frac{d}{dx}(x^3 - 12x) = 3x^2 - 12$

Find the critical points.

Critical points occur where $f'(x) = 0$ or where $f'(x)$ is undefined. Since $f'(x) = 3x^2 - 12$ is a polynomial, it is defined for all real $x$ . So, we only need to set $f'(x) = 0$ :

$3x^2 - 12 = 0$

$3x^2 = 12$

$x^2 = 4$

Solving for $x$ , we get $x = \pm 2$ .

Identify critical points within the given interval.

The given interval is $[0, 3]$ . We must check which of our critical points fall within this range.

$x = 2$ is in $[0, 3]$ . This is a candidate for the minimum.

$x = -2$ is not in $[0, 3]$ . We discard this critical point for this problem, as it's outside our domain of interest. …
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CBSE 2025Set 65/1/11 markMCQ
$f(x) = x^3 - 3x + 2$
›Reveal solutionSolution
To find the absolute maximum of a continuous function on a closed interval, we evaluate the function at its critical points within the interval and at the interval's endpoints, then pick the largest value. For $f(x) = x^3 - 3x + 2$ on $[0, 2]$ , the absolute maximum value is $\boxed{4}$ .

When we need to find the absolute maximum (or minimum) value of a continuous function over a closed interval, we rely on a fundamental concept from calculus called the Extreme Value Theorem. This theorem guarantees that such a maximum and minimum must exist.

The intuition behind finding these extreme values is that they can occur in one of two places:

At a "peak" or "valley" within the interval: These are points where the function changes from increasing to decreasing (local maximum) or decreasing to increasing (local minimum). At such points, if the function is differentiable, its derivative will be zero. These are called critical points.

At the boundaries of the interval: Even if the function is steadily increasing or decreasing throughout the interval, its highest or lowest value might simply be at one of the endpoints.

Therefore, our strategy is to check all these potential locations: the critical points that fall within our interval, and the two endpoints of the interval. We then compare the function values at all these points to find the absolute maximum.

Here's how we apply this to $f(x) = x^3 - 3x + 2$ on the interval $[0, 2]$ :

Find the derivative of the function. The derivative $f'(x)$ tells us about the slope of the tangent line to the function at any point $x$ . Critical points occur where the tangent line is horizontal, meaning $f'(x) = 0$ .

$f(x) = x^3 - 3x + 2$ $f'(x) = \frac{d}{dx}(x^3 - 3x + 2)$ $f'(x) = 3x^2 - 3$

Find the critical points by setting the derivative to zero. We solve $f'(x) = 0$ to find the $x$ -values where the function might have a local maximum or minimum.

$3x^2 - 3 = 0$ $3(x^2 - 1) = 0$ $x^2 - 1 = 0$
This is a difference of squares, which factors as $(x-1)(x+1)=0$. So, the critical points are $x = 1$ and $x = -1$.

3. Identify which critical points lie within the given interval.

The given interval is $[0, 2]$ . We must only consider critical points that are inside or on the boundary of this interval.

* $x = 1$ is in $[0, 2]$ . This is a relevant critical point. …
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CBSE 2024Set A1 mark
$C$
›Reveal solutionSolution

A point where $f'(c)=0$ or $f$ is not differentiable is called a critical point of $f$ .

In the study of maxima/minima (Application of Derivatives), a point $c$ in the domain of $f$ at which either $f'(c)=0$ (a stationary point) or $f$ fails to be differentiable i …

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CBSE 2019Set ANNUAL1 mark
$\phi(x) = f(x) + f(1-x)$
›Reveal solutionSolution

$f''(x)=0$ forces $f$ to be linear, which makes $\phi(x)=f(x)+f(1-x)$ constant on $[0,1]$ — so $x=\tfrac12$ is neither a genuine maximum nor minimum.

$\phi(x) = f(x)+f(1-x)$ .

$\phi'(x) = f'(x) - f'(1-x)$

At $x=\tfrac12$ : $\phi'\left(\tfrac12\right) = f'\left(\tfrac12\right) - f'\left(\tfrac12\right) = 0$ , so $x=\tfrac12$ is a critical point.

$\phi''(x) = f''(x) + f''(1-x)$

We are given $f''(x)=0$ for all $x\in[0,1]$ , so $f''(1-x)=0$ too, giving $\phi''(x) = 0+0 = 0$ for every $x\in[0,1]$ — the second-derivative test is inconclusive.

In fact, since $f''(x)=0$ throughout $[0,1]$ , $f$ must be a linear function, say $f(x)=mx+c$ . Then

$\phi(x) = (mx+c) + \big(m(1-x)+c\big) = mx+c+m-mx+c = m+2c$ ,

…

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CBSE 2019Set ANNUAL1 markMCQ
$f(x) = \frac{\log x}{x}$
(a) $1$
(b) $\frac{2}{e}$
(c) $e$
(d) $\frac{1}{e}$
›Reveal solutionSolution

Maximum value $=\frac{1}{e}$ at $x=e$ .

…

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Q.Find the absolute maximum and absolute minimum values of the function f(x)=2x3−15x2+36x+1f(x) = 2x^3 - 15x^2 + 36x + 1f(x)=2x3−15x2+36x+1 in [1,5][1, 5][1,5].

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Q.Find the absolute maximum and absolute minimum values of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$ in $[1, 5]$ .

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