Is f(x)=±√(x2+1)f(x)=\pm\sqrt{(x^2+1)} a function from R→R\mathbb{R} \to \mathbb{R}? [on hold]

Hey I just wanted to know if the following is a function from R→R\mathbb{R} \to \mathbb{R}?

f(x)=±√(x2+1)f (x) = \pm\sqrt{(x^2 + 1)}



3 Answers



The way the function is written, f(0)f(0) has two values, 1 and -1. A function cannot map one value onto two values.



Yes thank you very much.
– Shivam Tiwari

Your function actually consists of two functions. Let’s denote two functions f+f_+ and f−f_- by
f_\pm:\mathbb R\to \mathbb R,\quad f_\pm(x) = \pm\sqrt{x^2+1}
For all x∈Rx\in\mathbb R, we have x2∈[0,∞)x^2\in[0,\infty), which implies that
f_+(x) =& \sqrt{x^2+1}\in [1,\infty) \\
f_-(x) =& -\sqrt{x^2+1}\in(-\infty,-1]
Hence you will never reach the interval (-1,1) with either function. You could still say that the functions are from R\mathbb R to R\mathbb R, but they are not what is called surjective in that case.



I’m guessing you meant to say f±(x)=±√x2+1f_\pm (x) = \pm \sqrt{x^2+1}?
– Myridium
2 days ago



Yup, thanks. Edited it.
– Sjors Heefer
2 days ago

x2≥0→x2+1≥0+1√x2+1≥√1→1≤√x2+1<∞=R+−[0,1)→x^2 \geq 0 \to x^2+1 \geq 0+1 \\ \sqrt{x^2+1}\geq \sqrt{1}\\ \to 1\leq \sqrt{x^2+1} < \infty =\mathbb{R^+}-[0,1) \to so this is not R→R\mathbb{R} \to \mathbb{R}      In the expression f:A→Bf : A \rightarrow B, BB is the codomain, not the range. – Myridium 2 days ago      There can be functions from R→R\mathbb{R} \to \mathbb{R} without having R\mathbb{R} as the image (this particular example is not a function, but that has nothing to do with your reasoning). – Morgan Rodgers 2 days ago