4
$\begingroup$

According to my understanding, the domain of $\sqrt{x}$ is $x \ge 0$. If this is correct, shouldn't the domain of the function $$\frac{1}{(x^2-5x)^{1/4}}$$ be $x \le 0$ or $x\ge 5$ instead of $x<0$ or $x>5$? What am I getting wrong? (the answer is $x<0$ or $x>5$).

$\endgroup$
  • 1
    $\begingroup$ denominator cannot be zero $\endgroup$ – valer Feb 23 '18 at 22:34
  • $\begingroup$ Can’t read it. Please take the effort to get familiar with MathJax. $\endgroup$ – Lubin Feb 23 '18 at 22:34
3
$\begingroup$

Note that for

$$f(x)=\sqrt{\frac1{g(x)}}$$

we need also that $$g(x)\neq 0$$

Thus in this case we have

$$x^2-5x>0 \implies x(x-5)>0 \implies x <0 \quad x>5$$

$\endgroup$
  • $\begingroup$ thanks. I was dumb for a second :) $\endgroup$ – user1917231 Feb 23 '18 at 22:42
  • $\begingroup$ Gimusi.Huh.You are fast. $\endgroup$ – Peter Szilas Feb 23 '18 at 22:45
  • $\begingroup$ @user1917231 don't worry about that, I'm dumb too very often :) $\endgroup$ – gimusi Feb 23 '18 at 22:47
  • $\begingroup$ @PeterSzilas It was not so difficult here! :) $\endgroup$ – gimusi Feb 23 '18 at 22:47

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.