# Why this function is undefined at $x=0$ and $x=5$

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 $x>5$).

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

Note that for

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

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

Thus in this case we have

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

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