Find the range of the function \[f\left( x \right)=\dfrac{1}{{{x}^{2}}-x+1}\].

Hint: First find the range of the function \[{{x}^{2}}-x+1\]. See the co – efficient of \[{{x}^{2}}\] and find the value of the determinant of \[{{x}^{2}}-x+1\]. If the value of determinant is negative and co – efficient of \[{{x}^{2}}\] is positive then the value of quadratic equation is always positive and if co – efficient of \[{{x}^{2}}\] is negative then the value of quadratic equation is always negative. Once the range of \[{{x}^{2}}-x+1\] is found, take its reciprocal to get the answer.

Complete step by step answer:
Here, we have been provided with the function: - \[f\left( x \right)=\dfrac{1}{{{x}^{2}}-x+1}\].
Let us first find the range of \[{{x}^{2}}-x+1\].
Clearly, we can see that co – efficient of \[{{x}^{2}}=1\], which is positive. Now,
Determinant = \[{{b}^{2}}-4ac\].
Here, b = -1, a = 1, c = 1.
\[\Rightarrow \] Determinant = \[{{\left( -1 \right)}^{2}}-4\times 1\times 1=1-4=-3\], which is negative.
Hence, the value of \[{{x}^{2}}-x+1\] is positive for all values of x. The value of \[{{x}^{2}}-x+1\] can extend upto infinity. So, the maximum value of \[{{x}^{2}}-x+1\] is infinite.
Now, to determine the minimum value of \[{{x}^{2}}-x+1\], let us differentiate and substitute it equal to 0 to find the value of x.
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left[ {{x}^{2}}-x+1 \right]=0 \\
 & \Rightarrow 2x-1=0 \\
 & \Rightarrow x=\dfrac{1}{2} \\
\end{align}\]
So, the function \[{{x}^{2}}-x+1\] has a minima for \[x=\dfrac{1}{2}\].
\[\therefore \] Maximum value of \[{{x}^{2}}-x+1\] is,
\[\begin{align}
  & ={{\left( \dfrac{1}{2} \right)}^{2}}-\dfrac{1}{2}+1 \\
 & =\dfrac{3}{4} \\
\end{align}\]
So, \[\dfrac{3}{4}\le {{x}^{2}}-x+1\le \infty \]
Taking reciprocal we get,
\[\begin{align}
  & \dfrac{4}{3}\ge \dfrac{1}{{{x}^{2}}-x+1}\ge \dfrac{1}{\infty } \\
 & \Rightarrow 0\le \dfrac{1}{{{x}^{2}}-x+1}\le \dfrac{4}{3} \\
\end{align}\]

Hence, range of \[f\left( x \right)=\dfrac{1}{{{x}^{2}}-x+1}\] is \[\left[ 0,\dfrac{4}{3} \right]\].

Note: One may note that when we have taken the reciprocal of \[{{x}^{2}}-x+1\], the sign of inequality changes its direction. This is an important property of inequality. You must note that, since the function is a quadratic equation therefore on differentiating we will get only one value of x which will be either a maxima or minima point.