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)}

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3 Answers

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No.

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

yesterday

Your function actually consists of two functions. Let’s denote two functions f+f_+ and f−f_- by

f±:R→R,f±(x)=±√x2+1\begin{align}

f_\pm:\mathbb R\to \mathbb R,\quad f_\pm(x) = \pm\sqrt{x^2+1}

\end{align}

For all x∈Rx\in\mathbb R, we have x2∈[0,∞)x^2\in[0,\infty), which implies that

f+(x)=√x2+1∈[1,∞)f−(x)=−√x2+1∈(−∞,−1]\begin{align}

f_+(x) =& \sqrt{x^2+1}\in [1,\infty) \\

f_-(x) =& -\sqrt{x^2+1}\in(-\infty,-1]

\end{align}

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