Question

For all real $x$, the minimum value of $\dfrac{1-x+{{x}^{2}}}{1+x+{{x}^{2}}}$ is:

Answer 1

Hint: Here, to find the minimum value of the function, $\dfrac{1-x+{{x}^{2}}}{1+x+{{x}^{2}}}$ , first find $f'(x)$ and then substitute it to zero. i.e. put $f'(x)=0$ to find the value of $x$, and then substitute that value in$\dfrac{1-x+{{x}^{2}}}{1+x+{{x}^{2}}}$, to obtain the minimum value. So, to find $f'(x)$ we have to use the quotient rule:

Complete step-by-step answer:
${{\left( \dfrac{u}{v} \right)}^{'}}=\dfrac{u'v-uv'}{{{v}^{2}}}$

Here, for all real $x$, we have to find the minimum value of $\dfrac{1-x+{{x}^{2}}}{1+x+{{x}^{2}}}$.

Let us consider,

 $f(x)=\dfrac{1-x+{{x}^{2}}}{1+x+{{x}^{2}}}\text{ }.....\text{ (1)}$

Now, to find the minimum value, we have to find the value of $x$ when $f'(x)=0$.

First, let us calculate $f'(x)$.

$f'(x)=\dfrac{d}{dx}\left( \dfrac{1-x+{{x}^{2}}}{1+x+{{x}^{2}}} \right)$

Here, $f(x)$ is of the form $\dfrac{u}{v}$. Therefore to find $f'(x)$ we have to apply the quotient rule. The quotient rule is given by:

${{\left( \dfrac{u}{v} \right)}^{'}}=\dfrac{u'v-uv'}{{{v}^{2}}}$

 So, from equation (1) we have $u=1-x+{{x}^{2}}$ and $v=1+x+{{x}^{2}}$.

By applying the quotient rule in equation (1) we get:

$\Rightarrow$ $f'(x)=\dfrac{\dfrac{d}{dx}\left( 1-x+{{x}^{2}} \right)\left( 1+x+{{x}^{2}} \right)-\left( 1-x+{{x}^{2}} \right)\dfrac{d}{dx}\left( 1+x+{{x}^{2}} \right)}{{{\left( 1+x+{{x}^{2}} \right)}^{2}}}\text{ }.....\text{ (2)}$


 Now, let us find:

$\begin{align}

  & \dfrac{d}{dx}\left( 1-x+{{x}^{2}}

\right)=\dfrac{d}{dx}(1)+\dfrac{d}{dx}(-x)+\dfrac{d}{dx}\left( {{x}^{2}} \right) \\

 & \dfrac{d}{dx}\left( 1-x+{{x}^{2}} \right)=\dfrac{d}{dx}(1)-\dfrac{d}{dx}(x)+\dfrac{d}{dx}\left(

{{x}^{2}} \right) \\

 & \dfrac{d}{dx}\left( 1-x+{{x}^{2}} \right)=0-1+2x \\

 & \dfrac{d}{dx}\left( 1-x+{{x}^{2}} \right)=-1+2x\text{ }.......\text{ (3)} \\

\end{align}$

$\begin{align}

  & \dfrac{d}{dx}\left( 1+x+{{x}^{2}}

\right)=\dfrac{d}{dx}(1)+\dfrac{d}{dx}(x)+\dfrac{d}{dx}\left( {{x}^{2}} \right) \\

 & \dfrac{d}{dx}\left( 1+x+{{x}^{2}} \right)=0+1+2x \\

 & \dfrac{d}{dx}\left( 1+x+{{x}^{2}} \right)=1+2x\text{ }......\text{ (4)} \\

\end{align}$

Next, by substituting equation (3) and equation (4) in equation (2) we obtain:

$\Rightarrow$ $f'(x)=\dfrac{(-1+2x)\left( 1+x+{{x}^{2}} \right)-\left( 1-x+{{x}^{2}} \right)(1+2x)}{{{\left( 1+x+{{x}^{2}} \right)}^{2}}}\text{ }......\text{ (5)}$

In the next step let us find $(-1+2x)\left( 1+x+{{x}^{2}} \right)$ i.e. multiply each term of the first part by each term of the second part.

$\begin{align}

  & (-1+2x)\left( 1+x+{{x}^{2}} \right)=-1\times 1+-1\times x+-1\times {{x}^{2}}+2x\times 1+2x\times x+2x\times {{x}^{2}} \\

 & (-1+2x)\left( 1+x+{{x}^{2}} \right)=-1-x-{{x}^{2}}+2x+2{{x}^{2}}+2{{x}^{3}} \\

\end{align}$

Now, by adding the coefficients of similar terms we get:

$\Rightarrow$ $(-1+2x)\left( 1+x+{{x}^{2}} \right)=-1+x+{{x}^{2}}+2{{x}^{3}}\text{ }.......\text{ (6)}$

Now, find the value of $\left( 1-x+{{x}^{2}} \right)(1+2x)$ i.e. multiply each term of the first part by each term of the second part.

$\begin{align}

  & \left( 1-x+{{x}^{2}} \right)(1+2x)=1\times 1+1\times 2x-x\times 1-x\times 2x+{{x}^{2}}\times 1+{{x}^{2}}\times 2x \\

 & \left( 1-x+{{x}^{2}} \right)(1+2x)=1+2x-x-2{{x}^{2}}+{{x}^{2}}+2{{x}^{3}} \\
\end{align}$

Now, by adding the coefficients of similar terms we get:

$\left( 1-x+{{x}^{2}} \right)(1+2x)=1+x-{{x}^{2}}+2{{x}^{3}}\text{ }.......\text{ (7)}$

So, by substituting equation (6) and equation (7) in equation (5) we obtain:

$\begin{align}

  & f'(x)=\dfrac{-1+x+{{x}^{2}}+2{{x}^{3}}-\left( 1+x-{{x}^{2}}+2{{x}^{3}} \right)}{{{\left( 1+x+{{x}^{2}} \right)}^{2}}} \\

 & f'(x)=\dfrac{-1+x+{{x}^{2}}+2{{x}^{3}}-1-x+{{x}^{2}}-2{{x}^{3}}}{{{\left( 1+x+{{x}^{2}} \right)}^{2}}} \\

\end{align}$

Next, by adding the coefficients of the similar terms we get:

$f'(x)=\dfrac{-2+2{{x}^{2}}}{{{\left( 1+x+{{x}^{2}} \right)}^{2}}}$

 Now, we have to put $f'(x)=0$.i.e.

$\begin{align}

  & f'(x)=\dfrac{-2+2{{x}^{2}}}{{{\left( 1+x+{{x}^{2}} \right)}^{2}}}=0 \\

 & \Rightarrow \dfrac{-2+2{{x}^{2}}}{{{\left( 1+x+{{x}^{2}} \right)}^{2}}}=0 \\

\end{align}$

In the next step, we have to do cross multiplication. Then we obtain:

$-2+2{{x}^{2}}=0$

Now, by taking -2 to the right side it becomes 2. i.e. we get:

$2{{x}^{2}}=2$

 Next, by cross multiplication we get:

$\begin{align}

  & {{x}^{2}}=\dfrac{2}{2} \\

 & {{x}^{2}}=1 \\

\end{align}$

By taking the square root we get:

$x=\pm 1$

 Now we have two values for $x$ i.e. $x=1$ or $x=-1$, they are called the critical points.
 So, to find the minimum value we have to substitute $x=\pm 1$ in equation (1) and identify the smallest value.

In the next step put $x=1$ in equation (1), we obtain:

$\begin{align}

  & f(1)=\dfrac{1-1+{{1}^{2}}}{1+1+{{1}^{2}}} \\

 & f(1)=\dfrac{0+1}{2+1} \\

 & f(1)=\dfrac{1}{3} \\

\end{align}$

 Now, put $x=-1$ in equation (1), we get:

$\begin{align}

  & f(-1)=\dfrac{-1-(-1)+{{(-1)}^{2}}}{1+(-1)+{{(-1)}^{2}}} \\

 & f(-1)=\dfrac{1+1+1}{1-1+1} \\

 & f(-1)=\dfrac{2+1}{0+1} \\

 & f(-1)=\dfrac{3}{1} \\

 & f(-1)=3 \\

\end{align}$

Hence, at $x=1$ we have $f(1)=\dfrac{1}{3}$ and at $x=-1$ we have $f(-1)=3$.
Therefore, we can say that the minimum value of $\dfrac{1-x+{{x}^{2}}}{1+x+{{x}^{2}}}$ is $\dfrac{1}{3}$.

Note: Here, we got the values of $x$ as, $x=\pm 1$, they are the critical points, not the minimum value. To find the minimum value you have to put $x=\pm 1$ in $\dfrac{1-x+{{x}^{2}}}{1+x+{{x}^{2}}}$ and then identify which one is smaller, that will be the corresponding minimum value.