Here is the problem of the day.
\[ \textrm{Find the domain of }\textit{f(x) = } \frac{\sqrt{1-ln(x)}}{2-x^{2}} \]
\[ \textrm{Answer:} \left ( 0,\sqrt{2} \right )\cup \left ( \sqrt{2},e \right ] \]
The way to solve the “find the domain” problems can be generalized into a few steps:
- Factor the numerator or denominator of the function if possible and look for all discontinuities –
- find holes: removable discontinuities occur when the same factor exists in numerator and denominator. The zeros of those factors are holes which cannot be in the domain of the function.
- find zeros of the denominator (vertical asymptotes): non-removable discontinuities at the zeros of the factors in the denominator. These zeros cannot be in the domain of the function. In this case, the denominator factors are (x-\(\sqrt{2}\)) and (x+\(\sqrt{2})\), and the zeros are at \( \pm \sqrt{2} \), so these values cannot be in the domain of the function.
- Look for all embedded functions like ln(x) which have limited domains themselves, in this case, the domain of ln(x) is \( \left (0,\infty \right ) \). This will limit the domain of the entire function to be greater than zero, or or whatever the domain of the embedded function happens to be.
- Look for any expressions under radicals. The expression under the radical cannot be less than zero, so set the expression under the radical to be \( \geq 0 \) and solve this equation. The answer to this equation will also limit the domain of the function. In this case, we have 1- ln(x)\( \geq 0 \), and so the value of ln(x) \( \leq 1 \). This means x \( \leq e \).
- Look at all the limitations found above and determine the values of x allowed in the domain. We know that \( \pm \sqrt{2} \) are excluded, we know 0<x<\(\infty\), and x \( \leq e \). This means 0<x \( \leq e \), with \( \pm \sqrt{2} \) excluded, so the domain is \( \left ( 0,\sqrt{2} \right )\cup \left ( \sqrt{2},e \right]\)