For all real x,the minimum value of the funct | vedclass

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Question

(A) We have $f(x)=\frac{1-x+x^2}{1+x+x^2}$.To find the critical points,we calculate the derivative $f^{\prime}(x)$ using the quotient rule:$f^{\prime}

Vedclass · Accepted Answer

$\frac{1}{3}$

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(A) We have $f(x)=\frac{1-x+x^2}{1+x+x^2}$.To find the critical points,we calculate the derivative $f^{\prime}(x)$ using the quotient rule:$f^{\prime}(x)=\frac{(1+x+x^2)(2x-1)-(1-x+x^2)(2x+1)}{(1+x+x^2)^2}$.Expanding the numerator:$f^{\prime}(x)=\frac{(2x-1+2x^2-x+2x^3-x^2)-(2x+1-2x^2-x+2x^3+x^2)}{(1+x+x^2)^2}$.$f^{\prime}(x)=\frac{(2x^3+x^2+x-1)-(2x^3-x^2+x+1)}{(1+x+x^2)^2}$.$f^{\prime}(x)=\frac{2x^2-2}{(1+x+x^2)^2} = \frac{2(x^2-1)}{(1+x+x^2)^2}$.Setting $f^{\prime}(x)=0$,we get $x^2-1=0$,which implies $x=1$ or $x=-1$.Evaluating the function at these points:For $x=1$,$f(1)=\frac{1-1+1}{1+1+1} = \frac{1}{3}$.For $x=-1$,$f(-1)=\frac{1-(-1)+(-1)^2}{1+(-1)+(-1)^2} = \frac{1+1+1}{1-1+1} = 3$.Comparing the values,the minimum value of $f(x)$ is $\frac{1}{3}$.

For all real $x$,the minimum value of the function $f(x)=\frac{1-x+x^2}{1+x+x^2}$ is

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