Show that the function $f(x) = x^3 - 6x^2 + 12x + 50$ has neither a local maximum nor a local minimum value.

> [!TLDR] > The derivative is a perfect square $3(x-2)^2\ge 0$, so it never changes sign — the function is non-decreasing and has no local extremum. > [!FORMULA] > A critical point $c$ (where $f'(c)=0$) is a local extremum only if $f'$ **changes sign** across $c$. If $f'$ keeps the same sign, there is no local max or min. 1. Given $f(x)=x^3-6x^2+12x+50$. 2. Differentiate: $f'(x)=3x^2-12x+12=3(x^2-4x+4)=3(x-2)^2$. 3. Set $f'(x)=0\Rightarrow (x-2)^2=0\Rightarrow x=2$ (the only critical point). 4. Examine the sign of $f'(x)=3(x-2)^2$: it is $\ge 0$ for every real $x$ and equals $0$ only at $x=2$. So $f'$ does **not** change sign at $x=2$. …

Show that the function f(x) = x³ - 6x² + 12x + 50 – Class 12 Applied Mathematics

3.5 · Q4

f(x) = x^3 - 6x^2 + 12x + 50

Dnh Dd CbseNCERTSubjective· 3mImportance★★★★★

72% · 63/87 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Monotonic Function Analysis

Monotonic Function Analysis

The Intuition

Imagine you are walking along a path that only ever goes uphill, or only ever goes downhill. You never have to go up and then down, or down and then up. That path is monotonic — it moves in one consistent direction.

A function is monotonic when its output (the $y$ -value) never reverses direction as its input (the $x$ -value) increases. If it always goes up (or stays flat), it is increasing. If it always goes down (or stays flat), it is decreasing. If it does both — rises, then falls — it is not monotonic.

Note

The word "monotonic" comes from Greek monotonos — "one tone." Just as a monotone voice stays on a single pitch, a monotonic function stays on a single trend.

The Precise Definition

Let $f$ be a function defined on an interval $I$ . We say:

$f$ is increasing (or non-decreasing) on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) \leq f(x_2)$ .
$f$ is strictly increasing on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) < f(x_2)$ .
$f$ is decreasing (or non-increasing) on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) \geq f(x_2)$ .
$f$ is strictly decreasing on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) > f(x_2)$ .

If a function is either increasing or decreasing on an interval, it is called monotonic on that interval.

Watch out

| Common Mistake | Correct Understanding |

|----------------|----------------------|

| "Increasing means $f'(x) > 0$ everywhere" | $f'(x) > 0$ implies strictly increasing, but a function can be increasing even where $f'(x) = 0$ at isolated points (e.g., $f(x) = x^3$ at $x=0$ ). |

| "Monotonic means the whole domain" | A function can be monotonic on a sub-interval but not on its entire domain. For example, $f(x) = x^2$ is decreasing on $(-\infty, 0]$ and increasing on $[0, \infty)$ , but not monotonic on $\mathbb{R}$ . |

How to Determine Monotonicity (The Derivative Test)

For differentiable functions, the derivative tells you the direction:

If $f'(x) \geq 0$ for all $x$ in an interval, then $f$ is increasing on that interval.
If $f'(x) \leq 0$ for all $x$ in an interval, then $f$ is decreasing on that interval.
If $f'(x) > 0$ for all $x$ in an interval (except possibly at isolated points), then $f$ is strictly increasing.
If $f'(x) < 0$ for all $x$ in an interval (except possibly at isolated points), then $f$ is strictly decreasing.

f'(x) \geq 0 \implies f \text{ increasing} \qquad f'(x) \leq 0 \implies f \text{ decreasing}

Why It Matters

Monotonic functions are predictable. They have at most one root (if strictly monotonic), they are invertible (if strictly monotonic), and their extreme values occur only at the endpoints of the interval. In exam problems, you will often be asked to:

Find the intervals where a given function is increasing or decreasing.
Use monotonicity to prove inequalities (e.g., show $e^x > 1 + x$ for $x > 0$ ).
Determine the number of real roots of an equation.

A Worked Example …

Start your 14-day free trial to unlock the full solution →

The derivative is a perfect square $3(x-2)^2\ge 0$ , so it never changes sign — the function is non-decreasing and has no local extremum.

c

Given $f(x)=x^3-6x^2+12x+50$ .
Differentiate: $f'(x)=3x^2-12x+12=3(x^2-4x+4)=3(x-2)^2$ .
Set $f'(x)=0\Rightarrow (x-2)^2=0\Rightarrow x=2$ (the only critical point).
Examine the sign of $f'(x)=3(x-2)^2$ : it is $\ge 0$ for every real $x$ and equals $0$ only at $x=2$ . So $f'$ does not change sign at $x=2$ . …

Start your 14-day free trial to unlock the full solution →

←Previous question · 3.11Q3 — Show that of all rectangles with a given perimeter, the square has the…→Next question · 3.11Q5 — The profit function, in rupees, of a firm selling 'x' items (x ≥ 0) pe…

Q.Show that the function f(x)=x3−6x2+12x+50f(x) = x^3 - 6x^2 + 12x + 50f(x)=x3−6x2+12x+50 has neither a local maximum nor a local minimum value.

🎓Unlock everything free for 14 days

Q.Show that the function $f(x) = x^3 - 6x^2 + 12x + 50$ has neither a local maximum nor a local minimum value.

Unlock everything free for 14 days