Show that the function $f(x) = 4x^3 - 18x^2 + 27x - 7$ has neither maximum nor minimum value.

Show that the function f(x) = 4x³ - 18x² + 27x - 7 has — CBSE Class 12 Mathematics 2024 PYQ

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f(x) = 4x^3 - 18x^2 + 27x - 7

CBSECBSE Class XII Board 2024Subjective· 2mImportance★★★★★

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Concept understanding — Monotonic Function Analysis

Monotonic Function Analysis

A function is monotonic on an interval when it moves in a single direction across that interval — either always rising or always falling, never doubling back. Derivatives give us a clean, mechanical way to detect this, which is why monotonicity is one of the first applications of the derivative.

Increasing, Decreasing, Monotonic

On an interval $I$ , a function $f$ is:

increasing if $x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$ ,
strictly increasing if $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ ,
decreasing if $x_1 < x_2 \Rightarrow f(x_1) \ge f(x_2)$ ,
strictly decreasing if $x_1 < x_2 \Rightarrow f(x_1) > f(x_2)$ .

A function that is either increasing throughout $I$ or decreasing throughout $I$ is called monotonic on $I$ .

The Derivative Test

The slope of the tangent tells you the direction of travel. If $f$ is differentiable on an open interval $I$ :

f'(x) > 0 \text{ on } I \implies f \text{ is strictly increasing on } I

The idea is intuitive: a positive slope means the graph climbs as you move right, a negative slope means it falls.

How to Analyse Monotonicity

Compute $f'(x)$ .
Solve $f'(x) = 0$ (and note where $f'$ is undefined). These critical points split the domain into intervals.
Test the sign of $f'$ in each interval.
Read off where $f$ increases ( $f' > 0$ ) and decreases ( $f' < 0$ ).

Example. For $f(x) = x^2 - 4x + 1$ , $f'(x) = 2x - 4$ . So $f'(x) < 0$ for $x < 2$ and $f'(x) > 0$ for $x > 2$ : the function decreases on $(-\infty, 2)$ and increases on $(2, \infty)$ . …

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The function $f(x) = 4x^3 - 18x^2 + 27x - 7$ is strictly increasing for all real $x$ because its derivative $f'(x) = 12(x - 1.5)^2$ is always non-negative and zero only at a single point (a point of inflection). A strictly monotonic function cannot have local maxima or minima, so it has neither.

Why this approach works

The standard way to find maxima and minima of a differentiable function is to examine its derivative. At a local maximum or minimum, the derivative must be zero (a critical point). But the converse is not true: a zero derivative could also indicate a point of inflection — where the function keeps increasing (or decreasing) but momentarily flattens out.

So the real question is: does $f'(x)$ change sign around its zeros? If it does, we have an extremum. If it doesn't, the function is monotonic and has no turning points.

Let's check.

Step-by-step solution

1. Compute the first derivative.

f'(x) = \frac{d}{dx}(4x^3 - 18x^2 + 27x - 7) = 12x^2 - 36x + 27

2. Factor the derivative completely.

Factor out the common factor 3:

f'(x) = 3(4x^2 - 12x + 9)

Notice that $4x^2 - 12x + 9 = (2x - 3)^2$ . So:

f'(x) = 3(2x - 3)^2

f'(x) = 12\left(x - \frac{3}{2}\right)^2

3. Analyse the sign of $f'(x)$ .

Since a square is always $\ge 0$ , we have:

f'(x) \ge 0 \quad \text{for all } x \in \mathbb{R}

The derivative is zero only when $x = \frac{3}{2}$ .

4. Interpret what this means.

If $f'(x) > 0$ everywhere except at a single isolated point where it is zero, the function is strictly increasing on the whole real line. The point $x = \frac{3}{2}$ is not a turning point — it's a point where the graph momentarily has a horizontal tangent but continues rising on both sides.

Watch out

A common mistake is to see $f'(c) = 0$ and immediately conclude $c$ is a maximum or minimum. That's only true if the derivative changes sign. Here it doesn't — it stays positive on both sides of $x = \frac{3}{2}$ .

5. Confirm with the second derivative test (optional).

$f''(x) = 24x - 36 = 12(2x - 3)$ …

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

CBSE 2024Set ANNUAL1 markMCQ
$Q. The interval in which the function f(x) = [ILLEGIBLE — page defect: heavy scan grain/noise obscures the printed function and range], 0 \leq x \leq [ILLEGIBLE] is strictly [ILLEGIBLE — increasing or decreasing, not legible] :$
(a) [ILLEGIBLE — page defect]
(b) [ILLEGIBLE — page defect]
(c) [ILLEGIBLE — page defect]
(d) [ILLEGIBLE — page defect]
›Reveal solutionSolution

This question cannot be honestly answered: the printed function, its domain range, whether it asks for increasing or decreasing, and all four options are lost to a heavy scan-grain/noise defect on the source PDF.

Note on the source scan: the question stem (which function f(x), the exact interval endpoint(s), and whether it asks about strictly increasing or strictly decreasing) and all four multiple-choice options are obscured by heavy scan grain/noise on the original page. Rather than invent a plausible-looking function and answer, we report this honestly.

…

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CBSE 2020Set ANNUAL1 mark
$(-1, 1)$
›Reveal solutionSolution

The sign of $f'(x)$ changes at $x=\tfrac12$ , which lies inside $(-1,1)$ , so $f$ is not monotonic on the whole interval.

Given $f(x)=x^2-x+1$ , so $f'(x)=2x-1$ .

Set $f'(x)=0$ : $2x-1=0 \Rightarrow x=\tfrac12$ , which lies inside $(-1,1)$ .

For $-1<x<\tfrac12$ : $f'(x)=2x-1<0$ , so $f$ is decreasing on $\left(-1,\tfrac12\right)$ .

For $\tfrac12<x<1$ : $f'(x)=2x-1>0$ , so $f$ is increasing on $\left(\tfrac12,1\right)$ .

…

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Q.Show that the function f(x)=4x3−18x2+27x−7f(x) = 4x^3 - 18x^2 + 27x - 7f(x)=4x3−18x2+27x−7 has neither maximum nor minimum value.

Why this approach works

Step-by-step solution

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Q.Show that the function $f(x) = 4x^3 - 18x^2 + 27x - 7$ has neither maximum nor minimum value.

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