Finding the Minimum Value of the Given Expression

We are asked to find the minimum value of the expression \( \frac{{1 - x + {x^2}}}{{1 + x + {x^2}}} \) for all real x. This is a rational expression.

Let the given expression be \( y = \frac{{x^2 - x + 1}}{{x^2 + x + 1}} \). We need to find the range of this expression to determine its minimum value.

Notice that the denominator \( x^2 + x + 1 \) has a discriminant \( \Delta = 1^2 - 4(1)(1) = -3 \). Since the discriminant is negative and the coefficient of \( x^2 \) is positive (1), the quadratic \( x^2 + x + 1 \) is always positive for all real x. Similarly, the numerator \( x^2 - x + 1 \) has a discriminant \( \Delta = (-1)^2 - 4(1)(1) = -3 \), which is also negative. Since the coefficient of \( x^2 \) is positive, \( x^2 - x + 1 \) is always positive for all real x. Therefore, the expression \( y \) is always positive.

Method 1: Using Algebraic Method (Discriminant)

We can treat this problem by rearranging the expression into a quadratic equation in terms of x and using the discriminant condition for real x.

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

Cross-multiply:

\( y(x^2 + x + 1) = x^2 - x + 1 \)

\( yx^2 + yx + y = x^2 - x + 1 \)

Rearrange into a standard quadratic equation form \( Ax^2 + Bx + C = 0 \):

\( (y - 1)x^2 + (y + 1)x + (y - 1) = 0 \)

For real x to exist, the discriminant of this quadratic equation must be greater than or equal to zero (\( \Delta \ge 0 \)).

The discriminant is \( \Delta = B^2 - 4AC \). Here, \( A = y-1 \), \( B = y+1 \), \( C = y-1 \).

\( \Delta = (y+1)^2 - 4(y-1)(y-1) \)

\( \Delta = (y+1)^2 - 4(y-1)^2 \)

Expand the squares:

\( \Delta = (y^2 + 2y + 1) - 4(y^2 - 2y + 1) \)

\( \Delta = y^2 + 2y + 1 - 4y^2 + 8y - 4 \)

\( \Delta = -3y^2 + 10y - 3 \)

For real x, we require \( \Delta \ge 0 \):

\( -3y^2 + 10y - 3 \ge 0 \)

Multiply by -1 and reverse the inequality sign:

\( 3y^2 - 10y + 3 \le 0 \)

Now, we need to find the roots of the quadratic \( 3y^2 - 10y + 3 = 0 \). We can use the quadratic equation formula or factorization.

Using factorization: \( 3y^2 - 9y - y + 3 = 0 \)

\( 3y(y - 3) - 1(y - 3) = 0 \)

\( (3y - 1)(y - 3) = 0 \)

The roots are \( y = \frac{{1}}{{3}} \) and \( y = 3 \).

The inequality \( 3y^2 - 10y + 3 \le 0 \) represents the values of y for which the parabola \( 3y^2 - 10y + 3 \) (which opens upwards as the coefficient of \( y^2 \) is positive) is below or on the x-axis. This occurs between the roots.

So, the range of y is \( \frac{{1}}{{3}} \le y \le 3 \).

This means the minimum value of the expression is \( \frac{{1}}{{3}} \) and the maximum value is 3.

Note: The case \( y-1=0 \) (i.e., \( y=1 \)) was implicitly handled because it leads to \( 2x=0 \), \( x=0 \), which is a real solution, and \( y=1 \) is within the range \( [1/3, 3] \).

Method 2: Using Calculus (Differentiation)

Let \( f(x) = \frac{{x^2 - x + 1}}{{x^2 + x + 1}} \). To find the minimum value, we can use differentiation. We find the derivative \( f'(x) \) and set it to zero to find critical points.

Using the quotient rule, \( f'(x) = \frac{{(2x-1)(x^2+x+1) - (x^2-x+1)(2x+1)}}{{(x^2+x+1)^2}} \)

Let's simplify the numerator:

\( (2x-1)(x^2+x+1) = 2x(x^2+x+1) - 1(x^2+x+1) = 2x^3 + 2x^2 + 2x - x^2 - x - 1 = 2x^3 + x^2 + x - 1 \)

\( (x^2-x+1)(2x+1) = x^2(2x+1) - x(2x+1) + 1(2x+1) = 2x^3 + x^2 - 2x^2 - x + 2x + 1 = 2x^3 - x^2 + x + 1 \)

Numerator of \( f'(x) \) is \( (2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1) \)

\( = 2x^3 + x^2 + x - 1 - 2x^3 + x^2 - x - 1 \)

\( = 2x^2 - 2 \)

So, \( f'(x) = \frac{{2x^2 - 2}}{{(x^2+x+1)^2}} \)

To find critical points, set \( f'(x) = 0 \):

\( \frac{{2x^2 - 2}}{{(x^2+x+1)^2}} = 0 \)

Since the denominator is always positive, we only need the numerator to be zero:

\( 2x^2 - 2 = 0 \)

\( 2x^2 = 2 \)

\( x^2 = 1 \)

\( x = \pm 1 \)

These are the critical points where the minimum value or maximum value might occur. Evaluate the original expression at these critical points:

For \( x = 1 \): \( f(1) = \frac{{1^2 - 1 + 1}}{{1^2 + 1 + 1}} = \frac{{1 - 1 + 1}}{{1 + 1 + 1}} = \frac{{1}}{{3}} \)

For \( x = -1 \): \( f(-1) = \frac{{(-1)^2 - (-1) + 1}}{{(-1)^2 + (-1) + 1}} = \frac{{1 + 1 + 1}}{{1 - 1 + 1}} = \frac{{3}}{{1}} = 3 \)

We should also consider the limit of the rational expression as \( x \to \pm \infty \).

\( \lim_{{x \to \pm \infty}} \frac{{x^2 - x + 1}}{{x^2 + x + 1}} = \lim_{{x \to \pm \infty}} \frac{{x^2(1 - \frac{1}{x} + \frac{1}{x^2})}}{{x^2(1 + \frac{1}{x} + \frac{1}{x^2})}} = \lim_{{x \to \pm \infty}} \frac{{1 - \frac{1}{x} + \frac{1}{x^2}}}{{1 + \frac{1}{x} + \frac{1}{x^2}}} = \frac{{1 - 0 + 0}}{{1 + 0 + 0}} = 1 \)

The function approaches 1 as \( x \) goes to positive or negative infinity. Comparing the values at critical points (1/3 and 3) and the limit (1), the smallest value is 1/3.

Conclusion on Minimum Value

Both the algebraic method using the discriminant and the calculus method using differentiation show that the range of the expression is \( [1/3, 3] \). Therefore, the minimum value of the expression \( \frac{{1 - x + {x^2}}}{{1 + x + {x^2}}} \) for all real x is \( \frac{{1}}{{3}} \).