The function f(x) = frac{x^2 - 2}{sqrt{1 + x^ | vedclass

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(C) Given $f(x) = \frac{x^2 - 2}{\sqrt{1 + x^2}}$.To find the critical points,we calculate the derivative $f'(x)$ using the quotient rule:$f'(x) = \fr

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has exactly one point of minima

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(C) Given $f(x) = \frac{x^2 - 2}{\sqrt{1 + x^2}}$.To find the critical points,we calculate the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{\sqrt{1+x^2}(2x) - (x^2-2)\frac{1}{2\sqrt{1+x^2}}(2x)}{1+x^2}$$f'(x) = \frac{2x(1+x^2) - x(x^2-2)}{(1+x^2)^{3/2}}$$f'(x) = \frac{2x + 2x^3 - x^3 + 2x}{(1+x^2)^{3/2}} = \frac{x^3 + 4x}{(1+x^2)^{3/2}} = \frac{x(x^2 + 4)}{(1+x^2)^{3/2}}$.Since $x^2 + 4 > 0$ and $(1+x^2)^{3/2} > 0$ for all $x \in \mathbb{R}$,the sign of $f'(x)$ depends only on $x$.For $x For $x > 0$,$f'(x) > 0$ (function is increasing).At $x = 0$,$f'(x) = 0$ and the sign changes from negative to positive,which indicates that $x = 0$ is a point of local minima.Thus,the function has exactly one point of minima.

The function $f(x) = \frac{x^2 - 2}{\sqrt{1 + x^2}}$:

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