The minimum value of {x^2} + frac{1}{1 + {x^2 | vedclass

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(A) Let $f(x) = x^2 + \frac{1}{1 + x^2}$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f'(x) = 2x - \frac{1}{(1 + x^2)^2} \

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(A) Let $f(x) = x^2 + \frac{1}{1 + x^2}$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f'(x) = 2x - \frac{1}{(1 + x^2)^2} \cdot (2x) = 2x \left( 1 - \frac{1}{(1 + x^2)^2} ight)$.Setting $f'(x) = 0$,we get $2x = 0$ or $1 - \frac{1}{(1 + x^2)^2} = 0$.From $2x = 0$,we find $x = 0$.From $1 - \frac{1}{(1 + x^2)^2} = 0$,we get $(1 + x^2)^2 = 1$,which implies $1 + x^2 = 1$ (since $1 + x^2 > 0$),so $x^2 = 0$,which again gives $x = 0$.Evaluating the function at $x = 0$,$f(0) = 0^2 + \frac{1}{1 + 0^2} = 1$.Since $x^2 \ge 0$,let $u = x^2$ where $u \ge 0$. Then $g(u) = u + \frac{1}{1+u} = \frac{u^2 + u + 1}{1+u} = \frac{u^2 - 1 + u + 2}{1+u} = u - 1 + \frac{2}{1+u}$.At $x=0$,$u=0$,$f(0)=1$. For $x 
eq 0$,$x^2 > 0$,the function value increases. Thus,the minimum value is at $x = 0$.

The minimum value of ${x^2} + \frac{1}{1 + {x^2}}$ is at $x=$ ..........

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