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

Question

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

Answer 1

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

or y = 1 – $\frac{2}{\frac{1}{x} + 1 + x}$ = 1 – $\frac{2}{t}$

y is min. when $\frac{2}{t}$ is max.

or t = $\frac{1}{x}$ + 1 + x is min.

$\frac{dt}{dx}$ = 1 – $\frac{1}{x^{2}}$ = 0

\ x = 1, –1 ; $\frac{d^{2}t}{dx^{2}}$ = $\frac{2}{x^{3}}$ = + ive at x = 1 for min.

\ y (min.) = 1 – $\frac{2}{3}$ = $\frac{1}{3}$

Alt. y = $\frac{1}{z}$ where z = $\frac{1 + x + x^{2}}{1 - x + x^{2}}$ . y is min. where z is max.

\ z = 3 \ y = $\frac{1}{3}$ .

Question: For all real x, the minimum value of \(\frac{1 - x + x^{2}}{1 + x + x^{2}}\) is –...