Prove that the function$f$ defined by$f\left( x \right) = {x^2} - x + 1$ is neither increasing nor decreasing in $\left( { - 1,1} \right)$ . Hence find the interval in which $f\left( x \right)$ is:
(a) Strictly increasing
(b) Strictly decreasing

Hint: Try to figure out the points where the function is changing its behaviour first. Plot them on a number line for a better understanding of behaviour. Check whether the intervals made by that point are increasing or decreasing.

Complete step-by-step answer:
We are given with $f\left( x \right) = {x^2} - x + 1$and our motive is to check the behaviour of this function in $\left( { - 1,1} \right)$
Let’s start with finding out the Critical Points, i.e. point at which the function changes its behaviour of increasing and decreasing.
At these points, the first derivative of a function is zero. So, we’ll use this information to find them.
$f'\left( x \right) = \dfrac{{d\left( {{x^2} - x + 1} \right)}}{{dx}} = 2x - 1 + 0$
For a critical point, $f'\left( x \right) = 0$
$f'\left( x \right) = 0 \Rightarrow 2x - 1 = 0$
Therefore, $x = \dfrac{1}{2}$
So, this shows that at $x = \dfrac{1}{2}$, the function changes its behaviour of increasing and decreasing.
But $x = \dfrac{1}{2}$lie inside the interval$\left( { - 1,1} \right)$, making two disjoint intervals, i.e. $\left( { - 1,\dfrac{1}{2}} \right)$and $\left( {\dfrac{1}{2},1} \right)$
To check behaviour in the interval $\left( { - 1,1} \right)$we should check for these two small intervals also. This can be determined by the sign of $f'\left( x \right)$,i.e. the increasing interval will have a positive sign of $f'\left( x \right)$and negative for decreasing.
For $x \in \left( { - 1,\dfrac{1}{2}} \right)$; let $x = 0 \in \left( { - 1,\dfrac{1}{2}} \right)$ then $f'\left( 0 \right) = 2 \times 0 - 1 = - 1 < 0$
For $x \in \left( {\dfrac{1}{2},1} \right)$; let $x = \dfrac{3}{4} \in \left( { - 1,\dfrac{1}{2}} \right)$ then $f'\left( {\dfrac{3}{4}} \right) = 2 \times \dfrac{3}{4} - 1 = 1.5 - 1 = 0.5 > 0$

So the nature of $f\left( x \right)$ in interval $\left( { - 1,\dfrac{1}{2}} \right)$ is strictly decreasing and in the interval $\left( {\dfrac{1}{2},1} \right)$is strictly increasing.
Hence, we can concur that $f\left( x \right) = {x^2} - x + 1$ in the interval $\left( { - 1,1} \right)$ is neither increasing nor decreasing due to its changing point in between the interval.
And since $f\left( x \right) = {x^2} - x + 1$ is a quadratic polynomial with domain$ \in \left( { - \infty ,\infty } \right)$ and one critical point at $x = \dfrac{1}{2}$
We can say $f\left( x \right)$is strictly decreasing in the interval $\left( { - \infty ,\dfrac{1}{2}} \right)$ and strictly increasing in the interval $\left( {\dfrac{1}{2},\infty } \right)$

Note: Don’t get confused while deciding intervals. Draw a rough curve with a number line for understanding better. An alternative approach can be that of finding intervals of strictly increasing and decreasing behaviour then check for the interval $\left( { - 1,1} \right)$to prove as asked in the question.