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The question asks "Decide whether the following is a function or not; justify your answer." Then they give the following piecewise "function": $$ f(x) = \begin{cases} x+1, & \text{if } x < 0\\ \sqrt{x+3}, & \text{if } x > -3 \end{cases} $$

Now so far I have the following sketch: enter image description here

What I want to know is that in the case of $\sqrt{x+3}$ does it need to be drawn from $x>-3$ (indicated with red) or from $\sqrt{3}$.

I know that, if the part I have indicated in red should not be included, it will be a function. But if it stays it is not a function.

gt6989b
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pabhp
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  • You may want to take a look at this:https://en.wikipedia.org/wiki/Vertical_line_test – ashK Feb 14 '18 at 17:07

2 Answers2

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Your relation is not a function, since given any point $x \in (-3,0)$, the quantity $f(x)$ is not uniquely defined.

You are right, if you exclude this interval from any one of the branches, it will become a function.

The way the problem is formulated, to draw its graph you need to indeed draw both branches the way you showed.

gt6989b
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  • gt69 what is a map? And this is a relation right? Thank you, liked. – King Tut Feb 14 '18 at 15:59
  • @KingTut it is definitely a relation, but as a relation, you are not mapping $\mathbb{R} \to \mathbb{R}$. Instead, you are mapping $\mathbb{R}$ to $\mathbb{R} \times \mathbb{R}$ over the interval $(-3,0)$... – gt6989b Feb 14 '18 at 16:01
  • @KingTut looked up and there are conflicting definitions. Here is one link to explain the difference: https://math.stackexchange.com/q/95741/16192. Will change the answer. – gt6989b Feb 14 '18 at 16:08
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Why shouldn't the part in red be included? That graphs exactly what happens on the second leg of the proposed definition when $x > -3$! So yes, the combined clauses give two values to the "function" between $-3$ and $0$. So, on modern definitions, it isn't a function.

It is always worth noting though, as a footnote, that the insistence that functions (properly so-called) be single-valued is a relative late-comer. G.H. Hardy, for example, in his once-canonical A Course in Pure Mathematics explicitly and emphatically says that the characteristic of taking just one value for a given argument "is by no means involved in the general idea of a function".

Peter Smith
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