Prove that the function $f(x) = x^2 - x + 1$ is neither increasing nor decreasing in (0, 1).

> [!TLDR] > $f'(x)=2x-1$ is negative on $(0,\tfrac12)$ and positive on $(\tfrac12,1)$; because the derivative changes sign inside $(0,1)$, $f$ is neither increasing nor decreasing on the whole interval. > [!FORMULA] > $f$ is increasing on an interval only if $f'\ge0$ throughout, decreasing only if $f'\le0$ throughout. If $f'$ changes sign, $f$ is neither. 1. Given $f(x)=x^2-x+1$, differentiate: $f'(x)=2x-1$. 2. $f'(x)=0\Rightarrow x=\dfrac12$, which lies in $(0,1)$. 3. On $\left(0,\tfrac12\right)$: take $x=\tfrac14\Rightarrow f'=2(\tfrac14)-1=-\tfrac12 0$, so $f$ is increasing here. …

Prove that the function f(x) = x² - x + 1 is neith – Class 12 Applied Mathematics

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f(x) = x^2 - x + 1

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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 …

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$f'(x)=2x-1$ is negative on $(0,\tfrac12)$ and positive on $(\tfrac12,1)$ ; because the derivative changes sign inside $(0,1)$ , $f$ is neither increasing nor decreasing on the whole interval.

f

Given $f(x)=x^2-x+1$ , differentiate: $f'(x)=2x-1$ .
$f'(x)=0\Rightarrow x=\dfrac12$ , which lies in $(0,1)$ .
On $\left(0,\tfrac12\right)$ : take $x=\tfrac14\Rightarrow f'=2(\tfrac14)-1=-\tfrac12<0$ , so $f$ is decreasing here.
On $\left(\tfrac12,1\right)$ : take $x=\tfrac34\Rightarrow f'=2(\tfrac34)-1=+\tfrac12>0$ , so $f$ is increasing here. …

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Q.Prove that the function f(x)=x2−x+1f(x) = x^2 - x + 1f(x)=x2−x+1 is neither increasing nor decreasing in (0, 1).

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Q.Prove that the function $f(x) = x^2 - x + 1$ is neither increasing nor decreasing in (0, 1).

Unlock everything free for 14 days