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 $x>5$).

  • 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

Note that for


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

Thus in this case we have

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

  • $\begingroup$ 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

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