The set of points where fx-x-1- x - is differentiable

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The set of po...

Question

The set of points where $f (x) = \frac{x}{1 + | x |}$ is differentiable

(- \infty, - 1) \cup (- 1, \infty)

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(- \infty, \infty)

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(0, \infty)

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(- \infty, 0) \cup (0, \infty)

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

The set of all points, where the function $f (x) = \frac{x}{1 + | x |}$ is differentiable, is

(- \infty, \infty)

(- \infty, 0) \cup (0, \infty)

(0, \infty)

[0, \infty]

f (x) = e^{x} - x

g (x) = x^{2} - x, \forall x \in R

x \in R

h (x) = (f \circ g) (x)

Solution set of the equation $∣ ∣ \frac{x}{x - 1} ∣ ∣ + | x | = \frac{x^{2}}{| x - 1 |}$ is :

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