Question
Prove that the function f given by f(x) = x2 - x + 1 is neither strictly increasing nor decreasing on (-1, 1).

Answer

It is given that function f(x) = x2 - x + 1
$f'(x) = 2x - 1$
If $f'(x) = 0$, then we get
$~~~ ~~~x=\frac{1}{2}$ 
So, the point x = $\frac{1}{2}$ divides the interval (-1, 1) into two disjoint intervals, $\left(-1, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, 1\right)$ 
In interval $\left(-1, \frac{1}{2}\right)$, we have
    $f'(x) = 2x - 1 < 0$
Therefore, the given function 'f 'is strictly decreasing in interval $\left(-1, \frac{1}{2}\right)$ 
Also, in interval $\left(\frac{1}{2}, 1\right)$ 
    $f'(x) = 2x - 1 > 0$
Hence, the given function 'f ' is strictly increasing in interval for $\left(\frac{1}{2}, 1\right)$.
Therefore, f is neither strictly increasing nor decreasing in interval (-1, 1).

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