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

Given: $f\left(x\right)=x^2-x+1$

Differentiating w.r.t $x,$

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

Putting $f^{\prime }\left(x\right)=0$

$\Rightarrow 2x-1=0$

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

Since $xϵ(-1,1)$

So, plotting point on real number line,

Intervals	Sign of $f^{\prime }\left(x\right)=2x-1$	Nature of $f\left(x\right)$
$(-1,\frac{1}{2})$	$f^{\prime }\left(x\right)<0$	Strictly decreasing
$(\frac{1}{2},1)$	$f^{\prime }\left(x\right)>0$	Strictly increasing

$f\left(x\right)$ is strictly decreasing for

$xϵ(-1,\frac{1}{2})\&f\left(x\right)$

is strictly increasing for

$xϵ(\frac{1}{2},1)$

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