If fx - 2x-1- x-0 x2- x-0- then the values of f-1-f1 are

The correct option is A 1,1 f(x)=2x+1 f( 1)=2( 1)+1 ; since x lt;0 #160; #160; #160; #160; #160; = 1 f(1)=(1)^2=1 ...

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Global Maxima

If fx = 2...

Question

If $f (x)$ $= {\begin{matrix} 2 x + 1; x \leq 0 x^{2}; x > 0 \end{matrix}$ , then the values of $f (- 1), f (1)$ are

- 1, 1

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1, - 1

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- 1, - 1

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1, 1

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f (x)

= {\begin{matrix} 2 x + 1; x \leq 0 x^{2}; x > 0 \end{matrix}

f (\frac{1}{2}), f (\frac{- 1}{2})

f : (- \infty, + 1] \to R, g : [- 1, \infty) \to R

f (x) = \sqrt{1 - x}

g (x) = \sqrt{1 + x}

f (x) + \frac{1}{g (x)}

x \in

f : R \to (- 1, 1)

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

f^{- 1} (x)

f (x) = 2 x - 1

x > 1

= x^{2} + 1

- 1 \leq x \leq 1

\frac{f (1) + f (3) + f (0)}{f (2) + f (- 1) + f (\frac{1}{2})}

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