Question

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

a) zero

Correct answer: a) zero

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Answer

Let $$ y = \frac{1 - x + x^2}{1 + x + x^2} $$

First, we differentiate with respect to $$ x $$: $$ \frac{dy}{dx} = \frac{(1 + x + x^2)(-1 + 2x) - (1 - x + x^2)(1 + 2x)}{(1 + x + x^2)^2} $$

Simplifying the numerator: $$ \frac{dy}{dx} = \frac{-1 + 2x + x^2 + 2x^3 - 1 - 2x + x + 2x^2 - x^2 - 2x^3}{(1 + x + x^2)^2} $$ $$ = \frac{-2 + 2x^2 - 2x^2 + 2x^2}{(1 + x + x^2)^2} $$ $$ = \frac{2(x^2 - 1)}{(1 + x + x^2)^2} $$

Set $$ \frac{dy}{dx} = 0 $$ to find critical points: $$ \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 $$

Now, find the second derivative: $$ \frac{d^2y}{dx^2} = \frac{2(1 + x + x^2)^2 \cdot (2x) - (x^2 - 1) \cdot 2(1 + x + x^2)(1 + 2x)}{(1 + x + x^2)^4} $$ $$ = \frac{4(1 + x + x^2)(1 + x + x^2) \cdot x - (x^2 - 1)(1 + 2x)}{(1 + x + x^2)^4} $$ $$ = \frac{4[x + x^2 + x^3 - x^2 - 2x^3 + 1 + 2x]}{(1 + x + x^2)^3} $$ $$ = \frac{4(1 + 3x - x^3)}{(1 + x + x^2)^3} $$

Evaluate the second derivative at $$ x = 1 $$ and $$ x = -1 $$: $$ \left(\frac{d^2y}{dx^2}\right)_{x=1} = \frac{4(1 + 3(1) - 1^3)}{(1 + 1 + 1)^3} $$ $$ = \frac{4(3)}{3^3} $$ $$ = \frac{4}{9} > 0 $$

For $$ x = -1 $$: $$ \left(\frac{d^2y}{dx^2}\right)_{x=-1} = \frac{4(1 + 3(-1) + (-1))}{(1 - 1 + 1)^3} $$ $$ = \frac{4(1 - 3 - 1)}{1^3} $$ $$ = 4(-1) $$ $$ = -4 < 0 $$

Thus, by the second derivative test, the function reaches a minimum at $$ x = 1 $$.

The minimum value is: $$ y = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3} $$

Hence, the correct option is (d) not (a).


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