Question
Show that the function x2 - x + 1 is neither increasing nor decreasing on (0, 1).

Answer

f(x) = x2 - x + 1
f'(x) = 2x - 1
For, f(x) to be increasing, we must have
f'(x) > 0
⇒ 2x - 1 > 0
⇒ 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
⇒ 2x - 1 < 0
⇒ 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).

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