Question

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

• A. $0$
• B. $1$
• C. $3$
• D. $\frac{1}{3}$
• E. $2$

Answer 1

Answer: (D)

$\frac{1}{3}$

Let's find the first derivative of this function

$\frac{(2x-1)(x^2+x+1)-(2x+1)(x^2-x+1)}{{(x^2+x+1)}^2}$

Putting this to zero, we get

$x^2-1=0$

Which gives us $x=\pm 1$

Since we can see that the double derivative is quite difficult to solve, so lets put the value of $x$ in main function and see the result.

So for $x=1$ the function is $\frac{1}{3}$ and for $x=-1$ function is $2$

So minimum value is $\frac{1}{3}$ at $x=1$