1
$\begingroup$

I'm having trouble solving this exercise: (find the domain of this function) $$f(x)=\frac{\ln(4-x^2)}{\ln(x+1)}$$ Here is my attempt:

  1. I've found the domain of the natural logarithm in the numerator, which is equal to: $$4-x^2>0$$ (it must be strictly positive),

    1.1 $$-x^2>-4$$

    1.2 $$x^2<4$$

    1.3 (result) $$-2<x<2$$

  2. I've found the domain of the natural logarithm in the denominator, which is equal to: $$x+1>0$$

1.1 (result) $$x>-1$$

Now, I have to find the interval (on the number line) in which each possible solution is acceptable. So, I've plotted each result on a number line, here is the result:

$$-1<x<2$$

but this question has these possible solutions (it is a multiple choice question):

(1) $-1<x<2, except\space0$

(2) $-\infty<x<-1$

(3) $-2<x<-1$

(4) $-\infty<x<0$

The solution is $(1)$, but I don't understand why is not 0 included in the solution's interval.

EDIT (correct answer):

In the denominator I would have had considered the equation involving the denominator: $\ln(x+1)=0$. it mustn't be equal to 0. this error occurs because of distraction.

$\endgroup$
2
  • 1
    $\begingroup$ Notice thta $\ln(x+1)=0\iff x=0$, so you must avoid null denominators. $\endgroup$ Dec 12 '20 at 9:17
  • 1
    $\begingroup$ ah, okay, so it was an error of distraction. Thank you $\endgroup$ Dec 12 '20 at 9:19
4
$\begingroup$

At $x=0$ we have $\ln (1-0) = \ln (1) = 0$ in the denominator, which is undefined.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.