Question

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

Answer 1

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.

& \Rightarrow \dfrac{d}{dx}\left[ {{x}^{2}}-x+1 \right]=0 \\\ & \Rightarrow 2x-1=0 \\\ & \Rightarrow x=\dfrac{1}{2} \\\ \end{aligned}$$ So, the function $${{x}^{2}}-x+1$$ has a minima for $$x=\dfrac{1}{2}$$. $$\therefore $$ Maximum value of $${{x}^{2}}-x+1$$ is, $$\begin{aligned} & ={{\left( \dfrac{1}{2} \right)}^{2}}-\dfrac{1}{2}+1 \\\ & =\dfrac{3}{4} \\\ \end{aligned}$$ So, $$\dfrac{3}{4}\le {{x}^{2}}-x+1\le \infty $$ Taking reciprocal we get, $$\begin{aligned} & \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{aligned}$$ **Hence, range of $$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.