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

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

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

Solution:

Given that $f(x)=x^{2}-x+1$
By differentiating the given equation with respect to $x$ , we obtain
$\Rightarrow \mathrm{t}$ $\mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}$ $\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-1$
Taking different region from $(0,1)$
Suppose $x \in(0,1 / 2)$
$\Rightarrow 2 x-1<0$
$\Rightarrow f^{\prime}(x)<0$
Therefore $f(x)$ is decreasing in $(0,1 / 2)$
$\begin{array}{l} \text { Let } x \in(1 / 2,1) \\ \Rightarrow 2 x-1>0 \\ \Rightarrow f^{\prime}(x)>0 \end{array}$
Therefore $f(x)$ is increasing in $(1 / 2,1)$
Thus, from above condition we need to find that
$\Rightarrow f(x)$ is decreasing in $(0,1 / 2)$ and increasing in $(1 / 2,1)$
As a result, condition for $f(x)$ neither increasing nor decreasing in $(0,1)$