Board 2024 Class 12 Maths Q22 Solution: Show that 4x³ − 18x² + 27x − 7 has neither maxima nor minima

The question

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

Exam answer

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

\begin{aligned} f'(x) &= 12x^2 - 36x + 27 \\ &= 3(4x^2 - 12x + 9) \\ &= 3(2x - 3)^2 \end{aligned}

Why?

Maxima and minima are decided by $f'(x)$ ⁠: a local maximum or minimum needs $f'(x) = 0$ and a change of sign of $f'$ there.

Differentiate term by term with $\frac{d}{dx}x^n = nx^{n - 1}$ ⁠; the constant $-7$ gives 0.

\begin{aligned} \tfrac{d}{dx}(4x^3) &= 4 \times 3x^2 = 12x^2 \\ \tfrac{d}{dx}(18x^2) &= 18 \times 2x = 36x \\ \tfrac{d}{dx}(27x) &= 27 \times 1 = 27 \\ \tfrac{d}{dx}(7) &= 0 \end{aligned}

Each of 12, 36 and 27 divides by 3: $12 \div 3 = 4$ ⁠, $36 \div 3 = 12$ ⁠, $27 \div 3 = 9$ ⁠.

After taking out the common factor 3, the bracket is a perfect square:

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

Check with $(a - b)^2 = a^2 - 2ab + b^2$ ⁠, $a = 2x$ ⁠, $b = 3$ ⁠:

\begin{aligned} &(2x)^2 - 2(2x)(3) + (3)^2 \\ &= 4x^2 - 12x + 9 \end{aligned}

So $f'(x) = 3(2x - 3)^2 \ge 0$ for all $x \in R$ ⁠, and $f'(x) = 0$ only at $x = \frac32$ ⁠, where $f'$ does not change sign.

Why?

A square is never negative, so $f'(x) \ge 0$ everywhere.

$f'(x) = 0$ needs the square to be 0:

\begin{aligned} 2x - 3 &= 0 \\ 2x &= 3 \\ x &= \tfrac32 \end{aligned}

At $x = \frac32$ ⁠, $f' = 0$ ⁠, but $f'$ is positive on both sides, for example $f'(1) = 3$ and $f'(2) = 3$ ⁠. So $x = \frac32$ is not a turning point.

\begin{aligned} f'(1) &= 3(2(1) - 3)^2 \\ &= 3(-1)^2 = 3(1) = 3 \\[4pt] f'(2) &= 3(2(2) - 3)^2 \\ &= 3(1)^2 = 3(1) = 3 \end{aligned}

Watch out: $f'(x) = 0$ alone does not give a maximum or minimum. The sign of $f'$ must change.

$\therefore$ f is increasing on R.

Why?

$f'(x) \ge 0$ everywhere, with $f' = 0$ only at a single point, so f keeps rising. A rising function has no point that is higher, or lower, than all the points near it.

Hence $f(x)$ has neither maxima nor minima.

Where this comes from

Chapter 6, Application of Derivatives: Increasing function with no maxima or minima.

Question type: Show that a function is increasing everywhere + When the second derivative test fails (combines 2 types)

How to spot it: click to see

Type 4. Show that a function is increasing everywhere

Spot it: "Show that (or prove that) $f$ is increasing on $\R$ ⁠."

Here: First, $f'(x) = 12x^2 - 36x + 27 = 3(2x - 3)^2$ is 3 times a square, so $f'(x) \ge 0$ and $f$ is increasing on $R$ ⁠.

Type 8. When the second derivative test fails: the first derivative test

Spot it: $f''(c) = 0$ at a critical point, or $f$ has a corner there (like $|x|$ at 0), or the question says "first derivative test".

Here: $f'$ is 0 only at $x = \frac32$ ⁠, and the square keeps it positive on both sides. No change of sign, so there is no local maximum or minimum there.

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Q22: Show that 4x3−18x2+27x−74x^3 - 18x^2 + 27x - 74x3−18x2+27x−7 has neither maxima nor minima

Q22: Show that $4x^3 - 18x^2 + 27x - 7$ has neither maxima nor minima