Find the domain of $h(x) = 1 /\sqrt[4]{x^2-5x}$

Question

I need to find the domain of $$h(x) = \dfrac 1{\sqrt[4]{x^2-5x}} $$

I took $x^2-5x$ and set it $>0$

$x^2 > 5x$

I am stuck at this point

I thought I could factor the x out and got $x(x-5) >0$ but I can't do anything with it in this form either.

What should I do?

score 1 · Answer 1 · edited Apr 13 '17 at 12:19

1

Find out where $x(x-5)=0$, and make a sign chart with marks at those points. Check a point in each region, plot where it is + and -, and that will tell you where $x(x-5)>0$!

If you have not seen sign charts, see the answers to this post: How does one construct a 'sign chart' when solving inequalities?

edited Apr 13 '17 at 12:19

Community

1

answered Sep 16 '14 at 22:34

Clinton Bradford

964

(-infinity,0) U (5, infinity) – user137452 Sep 16 '14 at 22:53
Yep! Remember to accept an answer that answers your question. Also, you can markup your math by using TeX, as in $(-\infty,0) \cup (0,\infty)$ – Clinton Bradford Sep 16 '14 at 23:00

Find the domain of $h(x) = 1 /\sqrt[4]{x^2-5x}$

1 Answers1