Let $f(x)=x^{-(1/3)}$ and $A$ denote the area of region bounded by $f(x)$ and the X-axis, when $x$ varies from -1 to 1. Which of the following statements is/are TRUE?
I try to explain :
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The following statements certainly have standard definitions: 1) "$f$ is continuous at $x$" ; 2) "$f$ is continuous". For the function $f(x) = x^{-1/3}$, it's certainly true that "$f$ is continuous" and that "$f$ is not continuous at $0$".
However, what about the statement, "$f$ is continuous on $[a,b]$"? Perhaps there is some confusion or disagreement over the meaning of this statement. I think that the statement "$f$ is continuous on $[a,b]$" means that "if $x \in [a,b]$ then $f$ is continuous at $x$".
By that definition, statement 1) is false because the function $f(x) = x^{-1/3}$ is not continuous at $0$.
So, for $f(x) = x^{-1/3}$, the statement "$f$ is continuous" is true, but the statement "$f$ is continuous on $[-1,1]$" is false.
Note: Although your answers contain some true aspects, your reasoning is not completely satisfying. Most of them is due to the incomplete specification of the function $f$. It's absolutely crucial to be fully aware about the domain and codomain of a function. Otherwise no precise analysis of a problem is possible. You may want to check the example in the section formal definition right at the beginning of the referring wiki-page, which addresses a function similar to your problem.
In order to be able to properly answer your question, we add a domain and codomain of $f$. But note, that other domains/codomains are possible. And this changes the question and also might change the answer.
Question: Let $f:\mathbb{R}\setminus \{0\}\rightarrow \mathbb{R}$, with $f(x)=x^{-\frac{1}{3}}$ and let $A$ denote the area of region bounded by $f(x)$ and the $X$-axis, when $x$ varies from $-1$ to $1$. Which of the following statements is/are TRUE?
$f$ is continuous in $[-1, 1]$
$f$ is not bounded in $[-1, 1]$
$A$ is nonzero and finite
$$ $$
Answer:
The statement 1. is FALSE. Reasoning: In order to be continuous in $[-1,1]$ the function $f$ has to be continuous at each point of $[-1,1]$. Since $f$ is not defined at $x=0$, it is neither continuous nor discontinuous at $0$. But $f$ is continuous in $[-1,1]\setminus{\{0\}}$.
Comment to your reasoning since left limit is not equivalent to right limit:
You are addressing the point $0$, so you should specify it because at all other points in $[-1,1]\setminus{\{0\}}$ the left limit of $f$ is equal to the right limit. The term equivalent has a specific mathematical meaning, so you should instead use the term equal.
The statement 2. is FALSE. Reasoning: In order to be bounded in $[-1,1]$ the function $f$ has to be defined in $[-1,1]$ which is not true for $x=0$. Furthermore $f$ is not bounded in $[-1,1]\setminus{\{0\}}$ since e.g. the limit $lim_{x\rightarrow 0^-}=-\infty$.
Comment to your reasoning since $f(x)$ rises to $infinite$ at $x=0$:
Since the function $f$ is not defined at $x=0$ it is better to write since $f(x)$ rises to $infinite$ when $x$ tends to zero. This formulation considers only values $x$ within the domain of $f$.
The statement 3. is FALSE. Reasoning: It is assumed that $x$ varies within the domain $[-1,1]\setminus{\{0\}}$ of $f$ and we further assume the area $A$ is a signed area. Since integration of the positive part of $A$ results in \begin{align*} \lim_{a\rightarrow 0^+}\int_{a}^{1}x^{-\frac{1}{3}}\,dx&=\frac{3}{2}\lim_{a\rightarrow 0^+}\left. x^{\frac{2}{3}}\right|_a^1=\frac{3}{2} \end{align*} which is finite, we obtain due to symmetry $A=0$.
Comment to your reasoning since we can calculate it by doing integrating the function.
The wording is not precise enough, since we are able to calculate an integral and the result might be $\pm\infty$. We have to state that we get a final value by integration.
Usually when integration is involved, we consider areas as signed. That's why I calculated $A=0$. But maybe, according to your conventions with respect to such questions, you have to consider an unsigned area. In this case we obtain $A=3$ and your answer TRUE is correct.
[2015-09-29] Add-on regarding the term discontinuity
The term continuous function is defined with respect to its domain. Therefore it's crucial to specify the domain of a function, if we want to analyse the function with respect to continuity. Outside of the domain of a function this function is not continuous, since it's not even defined there.
Note that when we talk about discontinuities of a one variable function we classify them as either being a removable discontinuity, a jump discontinuity or an essential resp. infinite discontinuity. The key point here is, that each of these discontinuities is defined with respect to the domain of $f$. We conclude, discontinuities are defined solely within the domain of $f$.
Regrettably we often see different explanations which talk about discontinuities at points where a function is not defined. A typical example is \begin{align*} &f:\mathbb{R}\setminus{\{0\}}\rightarrow \mathbb{R}\\ &f(x)=\frac{1}{x} \end{align*} $f$ is not continuous at $x=0$ since it's not even defined there. But we should not say $f$ is discontinuous at $x=0$, since the term discontinuity is due to it's classification in different types only defined within the domain of $f$. Alas, we can read such statements often.
The situation is completely different when we instead consider \begin{align*} &g:\mathbb{R}\rightarrow \mathbb{R}\\ &g(x)= \begin{cases} \frac{1}{x}&x\neq 0\\ 0&x=0 \end{cases} \end{align*} The function $g$ is also defined at $x=0$. Here we can conclude that $g$ has a discontinuity at $x=0$.
Another example: Let's consider the arithmetical function \begin{align*} &h:\mathbb{N}\rightarrow \mathbb{R}\\ &h(x)=x \end{align*}
This function has domain $\mathbb{N}$ and it's not useful to consider properties of $h$ at points outside of it's domain.
Informally: The domain and codomain specify where the function lives and we can't say anything about the function outside of its region of existence.
Conclusion: Since this terminology is not uniquely used within the community it's presumably helpful to discuss this aspect with a teacher to find a common sense.
