Let fx-x-3 - x - -1 - x - - -1 - x - 1 -x- 1 - x - 3 then x-fx- 0 is equal to

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Absolute Value Function

Let fx=[x],...

Question

Let $f (x) = ⎧ ⎨ ⎩ \begin{matrix} [x], - 3 < x \leq - 1 | x |, - 1 < x < 1 [x], 1 \leq x < 3 \end{matrix}$

then ${x : f (x) \geq 0}$ is equal to

(- 1, 3)

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[- 1, 3)

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(- 1, 3]

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Solution

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Similar questions

If 1 $\leq$ $| x |$ $\leq$ 3,then x belongs to the interval

\frac{- 1}{| x | - 2} = 1

x =

f (x) = \sqrt{x - 1} + \sqrt{3 - x}

| x - 2 | \leq 1,

f : [1, 3] \to R

(1, 3)

f^{'} (x) = | f (x) |^{2} + 4

x \in (1, 3)

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