If a function's domain is an empty set, how can its derivative still exist graphically?

Question

I read in my textbook that if we add two functions (with domain $D_1$ and $D_2$), the domain of resulting function is $\displaystyle{D_1 \cap D_2}$.

Now, if we add $f(x)=\ln(1-x)$ and $ g(x)=\dfrac{1}{\sqrt{x-1}}$. The domain of $p(x)=f(x)+g(x)$ is an empty set, so there is "no graph" for this function $p(x)$. How to interpret this?

However, if I take the derivative of this function, I obtain a graph. How is this possible? The function itself did not exist, but we have a slope(derivative) for that function which is represented graphically?

Answer 1

I read in my textbook that if we add two functions(with domain d1 and d2) the domain of resulting function is the intersection of d1 and d2.

Yes, you cannot add numbers that don't exist. So, for $f(x)+g(x)$ to be defined, you need both $f(x)$ and $g(x)$ to be defined, which means that $x$ is in $d_1=dom(f)$ and in $d_2=dom(g)$. Therefore, the domain of $f+g$ is $d_1\cap d_2=dom(f)\cap dom(g)$.

Now if we add $f(x)=\ln(1−x)$ and $g(x)=\frac{1}{\sqrt{x−1}}$, the domain of $p(x)=f(x)+g(x)$ is a null set.

Yes, since $dom(f)=(-\infty,1)$ and $dom(g)=[1,\infty)$, their intersection is empty, so the domain of $p$ is empty.

But if I take derivative of this function, I get a graph.$

If a function is differentiable at $a$, then it is defined at $a$ (by definition of the derivative). By contraposition, since there is no number $a$ where $p$ is differentiable, then $p$ is not differentiable.

Note that the derivative of $f(x)=\ln(1-x)$ is $f'(x)=\frac{-1}{1-x}$ with $dom(f')=(-\infty, 1)$, not $dom(f')=\mathbb R\setminus \{ 1 \}$.

Answer 2

In this particularity, the domain of the derivate should follow anti-derivate function. Ie. $\big( \ln{(1-x)} \big)' = \dfrac{-1}{1-x}, x < 1$. While, on its own, $\dfrac{-1}{1-x}$, has the domain $x \in \mathbb{R} \backslash \{1\}$, in the question it is specifically bounded as the derivate of another function, so consider it as "child"(output) which follows its parent domain. Think of the reverse, $\displaystyle{\int{\dfrac{-1}{1-x}} = \ln|1-x| + c}$. Now, $\dfrac{-1}{1-x}$ is the "parent" and, hence the output, $\ln|1-x|$, must follow the domain of the integrand (input).

Edit: I figure to say that the derivative in your question should not take on values that were never in the domain of the antiderivative, otherwise the it would defy the antiderivative itself. One needs to be consistent with the domain of original function where the question begins.

Answer 3

This answer is in the context of real analysis.

$f(x)=\ln(1-x)$ and $ g(x)=\dfrac{1}{\sqrt{x-1}}$. The domain of $p(x)=f(x)+g(x)$ is an empty set, so there is "no graph" for this function $p(x)$. How to interpret this?

The function $p$ has an empty domain, so has an empty image, so has "no graph".

However, if I take the derivative of this function, I obtain a graph.

Actually, Desmos returns an empty graph for $p'.$

As $p$ is defined nowhere on $\mathbb R,$ it has a derivative nowhere on $\mathbb R;$ that is, $\boldsymbol{p'}$ has an empty domain, regardless of the fact that it has a mapping rule $$x \mapsto -\frac1{2 {(x - 1)}^{\frac32}}- \frac1{1 - x}$$ that applies nowhere.

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Addendum

$p'$ actually has a graph: https://www.desmos.com/calculator/ti0zlsilix (the black one)

In my previous paragraph, the word "defined" is important: we know that the function $p'$ has an empty domain by derivation from how you have defined $p$ as the sum of $f$ and $g.$

When you got Desmos to graph that mapping rule, it is natural for Desmos to do so with its natural domain (its maximal domain). Referring to your black curve as function $q:$ we say that $\boldsymbol {p'}$ is a restriction of $\boldsymbol q.$

P.S. I shall be preachy: technology is a tool, not a substitute for thinking, judgement and volition.