For all real values of x, the minimum value o | vedclass

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(D) Let $f(x) = \frac{1-x+x^{2}}{1+x+x^{2}}$.Using the quotient rule,$f^{\prime}(x) = \frac{(1+x+x^{2})(-1+2x) - (1-x+x^{2})(1+2x)}{(1+x+x^{2})^{2}}$.

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(D) Let $f(x) = \frac{1-x+x^{2}}{1+x+x^{2}}$.Using the quotient rule,$f^{\prime}(x) = \frac{(1+x+x^{2})(-1+2x) - (1-x+x^{2})(1+2x)}{(1+x+x^{2})^{2}}$.Simplifying the numerator: $(2x^{2}+x-1) - (2x^{3}+x^{2}-x+1) = 2x^{2}-2 = 2(x^{2}-1)$.So,$f^{\prime}(x) = \frac{2(x^{2}-1)}{(1+x+x^{2})^{2}}$.Setting $f^{\prime}(x) = 0$,we get $x^{2}-1 = 0$,which implies $x = 1$ or $x = -1$.Evaluating $f(x)$ at these critical points:$f(1) = \frac{1-1+1}{1+1+1} = \frac{1}{3}$.$f(-1) = \frac{1-(-1)+(-1)^{2}}{1+(-1)+(-1)^{2}} = \frac{1+1+1}{1-1+1} = 3$.Since $f(x) 	o 1$ as $x 	o \pm \infty$,the minimum value is $\frac{1}{3}$.

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

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