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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 $x>5$).

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1 Answer

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Note that for

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

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

Thus in this case we have

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

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