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. $\frac{1}{3}$

Answer 1

Answer: (D)

$\frac{1}{3}$

Answer 2

$Let ƒ(x)=\frac{1-x+x^{2}}{1+x+x^{2}.}$

$∴ƒ'(x)=\frac{(1+x+x^{2})(-1+2x)-(1-x+x^{2})(1+2x)}{(1+x+x^{2})^{2}}$

$=\frac{-1+2x-x+2x^{2}-x^{2}+2x^{3}-1-2x+x+2x^{2}-x^{2}-2x^{3}}{(1+x+x^{2})^{2}}$

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

$∴ƒ'(x)=0⇒x^{2}=1⇒x=\pm1$

Now$,ƒ''(x)=\frac{2[(1+x+x^{2})^{2}(2x)-(x^{2}-1)(2)(1+x+x^{2})(1+2x)]}{(1+x+x^{2})^{4}}$

$=\frac{4(1+x+x^{2})[(1+x+x^{2})x-(x^{2}-1)(1+2x)]}{(1+x+x^{2})^{4}}$

$=\frac{4[x+x^{2}+x^{3}-x^{2}-2x^{3}+1+2x]}{(1+x+x^{2})^{3}}$

$=\frac{4(1+3x-x^{3})}{(1+x+x^{2})^{3}}$

And,$ƒ''(1)=\frac{4(1+3-1)}{(1+1+1)^{3}}=\frac{4(3)}{(3)^{3}}=\frac{4}{9}>0$

Also,$ƒ''(-1)=\frac{4(1-3+1)}{(1-1+3)^{3}}=4(-1)=-4<0$

⧠ By second derivative test,f is the minimum at $x=1$ and the minimum value is given

by$ƒ(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$

The correct answer is D.