Question

The maximum value of $\frac{x^2-x+1}{x^2+x+1}$ for real values of $x$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D

$3$

Explanation for the correct option:

Step 1: Find the critical points of the given function.

A function $f\left(x\right)=\frac{x^2-x+1}{x^2+x+1}$ is given.

Rewrite the given function as follows:

$$\begin{gathered} f\left(x\right)=\frac{x^2-x+1+x-x}{x^2+x+1} \\ \Rightarrow f\left(x\right)=1-\frac{2x}{x^2+x+1} \end{gathered}$$

Differentiate both sides with respect to $x$.

$$\begin{gathered} f\text{'}\left(x\right)=0-\frac{2\left(x^2+x+1\right)-2x(2x+1)}{{\left(x^2+x+1\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{2x^2+2x+2-4x^2-2x}{{\left(x^2+x+1\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{-2x^2+2}{{\left(x^2+x+1\right)}^2} \end{gathered}$$

Put $f\text{'}\left(x\right)$ equal to zero to find the critical points.

$$\begin{gathered} \frac{-2x^2+2}{{\left(x^2+x+1\right)}^2}=0 \\ \Rightarrow -2x^2+2=0 \\ \Rightarrow x^2=1 \\ \Rightarrow x=\pm 1 \end{gathered}$$

So, the critical points are $1,-1$.

Step 2: Find the maximum value of the given function.

Since, the critical points are $1,-1$.

Evaluate $f\left(x\right)$ for $x=1$ as follows:

$$\begin{gathered} f\left(1\right)=\frac{1^2-1+1}{1^2+1+1} \\ \Rightarrow f\left(1\right)=\frac{1}{3} \end{gathered}$$

So, the value of $f\left(x\right)$ for $x=1$ is $\frac{1}{3}$.

Similarly, Evaluate $f\left(x\right)$ for $x=-1$ as follows:

$$\begin{gathered} f(-1)=\frac{{\left(-1\right)}^2-\left(-1\right)+\left(1\right)}{{\left(-1\right)}^2+\left(-1\right)+1} \\ \Rightarrow f(-1)=3 \end{gathered}$$

So, the value of $f\left(x\right)$ for $x=-1$ is $3$.

Therefore, the maximum value of the given function is $3$.

Hence, option $D$ is the correct answer.