Question

Prove that the function $f$ given by $f\left(x\right)=x^2-x+1$ is neither strictly increasing nor strictly decreasing on $(-1,1)$.

Answer 1

Let $f\left(x\right)=x^2-x+1$ for $x\in (-1,1)$

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

$\Rightarrow f^{\prime }\left(x\right)=0\Rightarrow 2x-1=0$

$\Rightarrow x=\frac{1}{2}$

Since $x\in (-1,1)$.The point $x=\frac{1}{2}$ divide the intervals $(-1,1)$ into two disjoint intervals.

$(-1,\frac{1}{2})$ and $(\frac{1}{2},1)$

Let $-1<x<\frac{1}{2}\Rightarrow x=0\in (-1,\frac{1}{2})$

$\Rightarrow f^{\prime }\left(0\right)=2\left(0\right)-1=-1<0$

$\therefore f\left(x\right)$ is strictly decreasing.

$(\frac{1}{2},1)$ and $(\frac{1}{2},1)$

Let $-1<x<\frac{1}{2}\Rightarrow x=0.7\in (\frac{1}{2},1)$

$\Rightarrow f^{\prime }\left(0.7\right)=2\left(0.7\right)-1=0.4>0$

$\therefore f\left(x\right)$ is strictly increasing.

Hence $f^{\prime }\left(x\right)<0$ for $x\in (-1,\frac{1}{2})$

and $f^{\prime }\left(x\right)>0$ for $x\in (\frac{1}{2},1)$

Hence $f\left(x\right)$ is neither decreasing nor increasing on $(-1,1)$

Hence proved.