The minimum of fx-1-x-x2-1-x-x2 occurs at x -

The correct option is B 1 f(x)=1+2x1 x+x^2 f^1(x)=2(1 x+x^2) 2x(2x 1)(1 x+x^2)^2 0= 2x^2+2(1 x+x^2)^2 x=±1 f^1(x) gt;0 If 1 ...

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The minimum o...

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The minimum of $f (x) = \frac{1 + x + x^{2}}{1 - x + x^{2}}$ occurs at x =

-1

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-2

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f (x) = \frac{x}{1 + x^{2}}

f (x) = {cos}^{- 1} (\frac{2 x}{1 + x^{2}})

f (x)

f (x) = {cos}^{- 1} (\frac{2 x}{1 + x^{2}})

f^{'} (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} \frac{- 2}{1 + x^{2}}, & | x | < 1 \end{matrix} \begin{matrix} \frac{2}{1 + x^{2}}, & | x | > 1 \end{matrix} \end{matrix}

f (x) = \frac{x^{2} - 1}{x^{2} + 1}

x ϵ R

x

f (x)

f (x)

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{x^{2} - 1}{x^{2} - 2 | x - 1 | - 1}, & x \neq 1 \frac{1}{2}, & x = 1 \end{matrix}

x = 1

f (x) = x^{2} + \frac{1}{x^{2}} and g (x) = x - \frac{1}{x}, x ϵ R - {- 1, 0, 1}, and h (x) = \frac{f (x)}{g (x)}

h (x)

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