Let f-R- R be defined by fx-x-2 x- -1 x2 -1-x- 1 2-x x- 1- then the value of f-1-f0-f1 is

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Definition of Functions

Let f:R→ R ...

Question

Let $f : R \to R$ be defined by $f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x + 2 & (x \leq - 1) x^{2} & (- 1 < x < 1) 2 - x & (x \geq 1) \end{matrix}$ then the value of $f (- 1) + f (0) + f (1)$ is

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Solution

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Similar questions

f : R \to

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x + 2 & (x \leq - 1) x^{2} & (- 1 \leq x \leq 1) 2 - x & (x \geq 1) \end{matrix}

f (- 1.75) + f (0.5) + f (1.5)

f : R \to R

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x + 2 & x \leq - 1 x^{2} & - 1 < x < 1 2 - x & x \geq 1 \end{matrix}

f (- 1.75) + f (0.5) + f (1.5)

f : R \to R

f (x) = ⎧ ⎨ ⎩ \begin{matrix} 3 x - 2, & x < 0 1, & x = 0 4 x + 1, & x > 0 \end{matrix}

f (1)

f (- 1)

f (0)

f (2)

f (x) = \{\begin{matrix} 3 x - 2, & x < 0; \\ 1, & x = 0; \\ 4 x + 1, & x > 0 . \end{matrix}

f : R \to R

{lim}_{x \to 0} [\frac{f (1 + x + x^{2}) + x f (1 + x)}{f (1)}]^{\frac{1}{x (x + 1)}}

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