# Combining two sets of domains - how?

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.

Thanks.

• Yes, what you wrote looks good to me. Jan 2, 2019 at 5:28
• 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. Jan 2, 2019 at 8:42
• 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?
– weno
Jan 2, 2019 at 9:24

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'.