Question
Show that the function $x^2 - x + 1$ is neither increasing nor decreasing on $(0, 1).$

Answer

$f(x) = x^2 - x + 1$
$f'(x) = 2x - 1$
For, f(x) to be increasing, we must have
$f'(x) > 0$
$\Rightarrow 2x - 1 > 0$
$\Rightarrow 2x > 1$
$\Rightarrow\text{x}>\frac{1}{2}$
$\Rightarrow\text{x}\in\Big(\frac{1}{2},1\Big)$
So, f(x) is increasing on $\text{x}\in\Big(\frac{1}{2},1\Big).$
For, f(x) to be decreasing, we must have
$f'(x) < 0$
$\Rightarrow 2x - 1 < 0$
$\Rightarrow 2x < 1$
$\Rightarrow\text{x}<\frac{1}{2}$
$\Rightarrow\text{x}\in\Big(0,\frac{1}{2}\Big)$
So, f(x) is increasing on $\Big(0,\frac{1}{2}\Big)$
Since, f(x) is increasing on $\Big(\frac{1}{2},1\Big)$ and decreasing on $\Big(0,\frac{1}{2}\Big),\text{f}(\text{x})$ is neither increasing nor decreasing on $(0, 1).$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions