Show that the function


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


Given:- Function $f(x)=x^{2}-x+1$

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

(i) If $f^{\prime}(x)>0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$


(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

Here we have,


$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}-\mathrm{x}+1\right)$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-1$

Taking different region from $(0,1)$

a) let $x \in\left(0, \frac{1}{2}\right)$

$\Rightarrow 2 x-1<0$

$\Rightarrow f^{\prime}(x)<0$

Thus $f(x)$ is decreasing in $\left(0, \frac{1}{2}\right)$

b) let $x \in\left(\frac{1}{2}, 1\right)$

$\Rightarrow 2 x-1>0$

$\Rightarrow f^{\prime}(x)>0$

Thus $f(x)$ is increasing in $\left(\frac{1}{2}, 1\right)$

Therefore, from above condition we find that

$\Rightarrow \mathrm{f}(\mathrm{x})$ is decreasing in $\left(0, \frac{1}{2}\right)$ and increasing in $\left(\frac{1}{2}, 1\right)$

Hence, condition for $f(x)$ neither increasing nor decreasing in $(0,1)$

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