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)?
A function fff from R\RR to R\RR is a rule that assigns to each element x∈Rx\in \Rx∈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)=1xf(x)=\dfrac{1}{x}f(x)=x1
x≠0x\not=0x=0
The function fff is undefined at x=0.x=0.x=0.
Therefore a function f(x)=1xf(x)=\dfrac{1}{x}f(x)=x1 is not a function from R\RR to R.\R.R.
b) f(x)=xf(x)=\sqrt{x}f(x)=x
x≥0x\geq0x≥0
The function fff is undefined for x∈R,x<0.x\in \R,x<0.x∈R,x<0.
Therefore a function f(x)=xf(x)=\sqrt{x}f(x)=x is not a function from R\RR to R.\R.R.
c) f(x)=±x2+1f(x)=\pm\sqrt{x^2+1}f(x)=±x2+1
f(0)=±02+1=±1f(0)=\pm\sqrt{0^2+1}=\pm1f(0)=±02+1=±1
A relation f(x)=±x2+1f(x)=\pm\sqrt{x^2+1}f(x)=±x2+1 assigns to x=0x=0x=0 two elements −1-1−1 and 1.1.1.
Therefore f(x)=±x2+1f(x)=\pm\sqrt{x^2+1}f(x)=±x2+1 is not a function from R\RR to R.\R.R.
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