Question

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

Answer 1

Given, $f\left(x\right)=x^2-x+1\Rightarrow f^{\prime }\left(x\right)=2x-1$
On putting f'(x)=0, we get $x=\frac{1}{2}$
$x=\frac{1}{2}$ divides the given interval into two intervals as $(-1,\frac{1}{2})$ and $(\frac{1}{2},1)$.

$\begin{array}{ccc}Intervals& Signof{f}^{\prime }\left(x\right)& Natureoff\left(x\right)\\ (-1,\frac{1}{2})& -ve& Strictlydecreasing\\ \left(\frac{1}{2}1\right)& +ve& Strictlyincreasing\end{array}$
$\therefore$, f'(x) does not have same sign throughtout the interval (-1, 1).
Thus, f(x) is neither increasing nor decreasing strictly in the interval (-1, 1).