I'm doing the exercises related to finding monotonicity/extreme values.

I have given function: $f(x) = -3x + \ln x$

Domain of $f$: $D_{f} = (0, \infty)$

Derivative: $f'(x) = -3 + \frac{1}{x}$

Domain of the derivative (teacher requires this): $D_{f'} = (-\infty,0)\cup(0,\infty)$

Now, how can I combine both domains? I need to write the mutual part of both domains. How to do it in a good math-fashioned style without "syntax mistakes"?

Something like that should do the job?

$\begin{cases} D_{f} = (0, \infty) \\ D_{f'} = (-\infty,0)\cup(0,\infty) \\ \end{cases}$

$\Rightarrow D_{f} \cap D_{f'} = (0, \infty)$

Is this correct?

I am asking this because if the solution(s) of $f'(x) = 0$ don't belong to the domain of $D_{f'}$, then I do not take them into account.


  • $\begingroup$ Yes, what you wrote looks good to me. $\endgroup$
    – zipirovich
    Jan 2, 2019 at 5:28
  • $\begingroup$ The teacher is wrong with her naive definition of function. The domain of -3 + 1/x is not the domain of f'. They are two different functions. $\endgroup$ Jan 2, 2019 at 8:42
  • $\begingroup$ Why finding domain of derivative is wrong? If I get solution that belongs to domain of function, but does not belong to domain of derivative, it will be invalid solution. Right? $\endgroup$
    – weno
    Jan 2, 2019 at 9:24

2 Answers 2


I used to use the example of f(x) = ln(x), x > 0, as one in which the derivative, f '(x)=1/x SEEMS to have a "larger" domain, namely the set of all nonzero reals. What has to be emphasized is the so-called "natural domain convention" (domain is implied by formula) does NOT apply for the derivative.

In fact, the domain of f ' is ALWAYS a subset of the domain of f, from the very definition of f '(x) being the limit of the difference quotient, in which appears f(x) itself. So, to say f '(x) makes sense requires that the f(x) that is part of its definition must make sense first.


f:(0,oo) -> R, x -> -3x + ln x.
f':(0,oo) -> R, x -> -1 + 1/x.

g:R\{0} -> R, x -> -1 + 1/x.
g and f' are different because they are defined on different domains.
The domain of f' cannot have points outside the domain of f
because f'(x) cannot be calculated when f(x) is not defined.
However, as you were considering, g restriced to (0,oo) = f'.

The teacher is wrong.
As f'(-1) is not defined, -1 cannot be in domain f'.
g is not f'.


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