A function f f f from R \R R to R \R R is a rule that assigns to each element x ∈ R x\in \R x ∈ R exactly one
element, called f ( x ) , f ( x ) ∈ R . f(x), f(x)\in \R. f ( x ) , f ( x ) ∈ R . in
a) f ( x ) = 1 x f(x)=\dfrac{1}{x} f ( x ) = x 1
x ≠ 0 x\not=0 x = 0
The function f f f is undefined at x = 0. x=0. x = 0.
Therefore a function f ( x ) = 1 x f(x)=\dfrac{1}{x} f ( x ) = x 1 is not a function from R \R R to R . \R. R .
b) f ( x ) = x f(x)=\sqrt{x} f ( x ) = x
x ≥ 0 x\geq0 x ≥ 0
The function f f f is undefined for x ∈ R , x < 0. x\in \R,x<0. x ∈ R , x < 0.
Therefore a function f ( x ) = x f(x)=\sqrt{x} f ( x ) = x is not a function from R \R R to R . \R. R .
c) f ( x ) = ± x 2 + 1 f(x)=\pm\sqrt{x^2+1} f ( x ) = ± x 2 + 1
f ( 0 ) = ± 0 2 + 1 = ± 1 f(0)=\pm\sqrt{0^2+1}=\pm1 f ( 0 ) = ± 0 2 + 1 = ± 1
A relation f ( x ) = ± x 2 + 1 f(x)=\pm\sqrt{x^2+1} f ( x ) = ± x 2 + 1 assigns to x = 0 x=0 x = 0 two elements − 1 -1 − 1 and 1. 1. 1.
Therefore f ( x ) = ± x 2 + 1 f(x)=\pm\sqrt{x^2+1} f ( x ) = ± x 2 + 1 is not a function from R \R R to R . \R. R .
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