Question #231948

Why is f not a function from R to R if

a) f (x) = 1/x?

b) f (x) =√x?

c) f (x) = ±√(x^2+1)?


1
Expert's answer
2021-09-02T00:38:00-0400

A function ff from R\R to R\R is a rule that assigns to each element xRx\in \R exactly one

element, called f(x),f(x)R.f(x), f(x)\in \R. in

a) f(x)=1xf(x)=\dfrac{1}{x}

x0x\not=0

The function ff is undefined at x=0.x=0.

Therefore a function f(x)=1xf(x)=\dfrac{1}{x} is not a function from R\R to R.\R.


b) f(x)=xf(x)=\sqrt{x}

x0x\geq0

The function ff is undefined for xR,x<0.x\in \R,x<0.

Therefore a function f(x)=xf(x)=\sqrt{x} is not a function from R\R to R.\R.


c) f(x)=±x2+1f(x)=\pm\sqrt{x^2+1}

f(0)=±02+1=±1f(0)=\pm\sqrt{0^2+1}=\pm1

A relation f(x)=±x2+1f(x)=\pm\sqrt{x^2+1} assigns to x=0x=0 two elements 1-1 and 1.1.


Therefore f(x)=±x2+1f(x)=\pm\sqrt{x^2+1} is not a function from R\R to R.\R.


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