Good time of day,

Need some help with my math homework:

Give each an example of a function $f:D\to\Bbb{R}, D\subset\Bbb{R}$ with the given properties. In addition, sketch the graph of the function you have specified.

(a) $D=[0,1]$, $f(0)=1$, $f(1)=2$, there is no $c\in D$ with $f(c)=\frac{3}{2}$

(b) $D=(0,1)$, $f$ is continuous, the interval $f(D)$ is closed and bounded

(c) $D=[0,\infty)$, $f$ is continuous and $f(D)=\Bbb{R}$

(d) $D=[0,1]$, $f$ is bounded and has neither a maximum nor a minimum

(e) $D=(0,1)$, $f$ is continuous and for every $a\in\Bbb{R}$ the set $\{x\in D\mid f(x)=a\}$ has infinitely many elements.

The original text in German

I think I've got the right answers:

a. Doesn't exist, since $D$ is an interval, thus $f(D)$ is an interval and cannot break at any $f(c), c\in{D}$;

b. Doesn't exist, since $(0,1)$ is an open interval and thus $f(0,1)$ cannot be closed;

c. $x*\sin{\root \of{x}}$;

d. Doesn't exist, since $[0,1]$ is closed, and thus $f[0,1]$ cannot be open;

e. Doesn't exist, since $\Bbb{R}$ is unbounded, which means $f(0,1)$ would have to have infinite asymptotes, which breaks the interval $(0,1)$;

Now the problem is that the question specifically states that I need to provide examples and not prove that functions exist and provide examples if they do, and most of these shouldn't even exist in $\Bbb{R}$ in the first place.

I'd like to know if I'm doing it wrong or if the question itself is poorly stated. Attaching image of the original text to show that it's the question that is the problem and not my understanding of it.

Thanks in advance.


Here's some feedback on your proposed answers.

(a) Wrong. What you're saying is true for continuous functions. But notice that this questions does not say that $f$ continuous.

(b) Wrong. A continuous image of a closed interval has to be closed. But the same is not true for open intervals: a continuous image of an open interval doesn't have to be open. It can be open, closed, or neither.

(c) Correct! As a comment, you don't have to have the square root there: $f(x)=x\sin x$ works just as well.

(d) Wrong. See (a).

(e) Wrong. And to be honest, I don't quite understand your reasoning — what does it have to do with asymptotes? A possible idea for constructing an example: to make sure each output value is taken on infinitely many times, we need something oscillating, like a sine function; but so that it oscillates within a bounded interval, we invert its argument and consider $\sin\frac{1}{x}$. It doesn't quite work yet, but we can modify it to get what we want. (As they say in analysis: if you need a nasty counterexample, use $\sin\frac{1}{x}$).

  • $\begingroup$ Oh, that's absolutely true, I just assumed f was continuous, since the entire rest of the homework deals specifically continuous functions. Thanks a lot! $\endgroup$ – Andrii Kozytskyi Dec 21 '17 at 21:25
  • $\begingroup$ @AndriiKozytskyi: No, that doesn't count because $f(x)=\sqrt{x}$ is NOT defined on $(-1,1)$. I guess you're not seeing the answer because you're thinking about monotone functions. But try other functions. Here's a hint: for example, $f(x)=x^2$ on $D=(-1,1)$ has the image $f(D)=[0,1)$, not an open interval. Now, can you come up with an example with both endpoints of $f(D)$ closed? $\endgroup$ – zipirovich Dec 21 '17 at 21:33
  • $\begingroup$ One more question in general and to (e) & (c) specifically. When I'm given a domain $D$ I assume that the function itself exists only on $D$. For example, "a function on $D=[0;\infty)$ with $f(D)=[0;\infty)$ I'd say $f(x)=\root \of{x}$ and not $f(x)=x$. Is this generally the correct way to do such exercises, or does "on $D=(a;b)$" mean any function that behaves like that on the set $D$ specifically? $\endgroup$ – Andrii Kozytskyi Dec 21 '17 at 21:37
  • 1
    $\begingroup$ @AndriiKozytskyi: The phrase "$f(x)$ is a function defined on the domain $D$" means that this function has to be defined at least on $D$, but there's nothing wrong with using function whose natural domains are larger. In this case, we can say that we restrict the function to the domain $D$. In fact, providing the domain (and the codomain, for that matter) is part of defining a function. For example, "$f(x)=x$ on $D=(-\infty,+\infty)$" and "$f(x)=x$ on $D=[0,+\infty)$" are both meaningful functions -- but they are two different functions. $\endgroup$ – zipirovich Dec 21 '17 at 21:58
  • $\begingroup$ Well, that makes the job a hell of a lot easier, I think I got this right this time. Thanks a ton once again $\endgroup$ – Andrii Kozytskyi Dec 21 '17 at 22:10

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