Finding domain of a function with square root in bottom of fraction\infty

[tmt1]

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

So, I understand that I need to set

$$x^2 -5x > 0$$

from that I get

$$ x(x-5) > 0$$
and

$$ x > 5$$

However, the answer in the textbook is given:

$$ ( \infty, 0) \cup (5, \infty)$$

Which mean that the graph has a domain including x values of less than 0 and greater? Can someone explain this to me?

 

[I like Serena]

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

So, I understand that I need to set

$$x^2 -5x > 0$$

from that I get

$$ x(x-5) > 0$$
and

$$ x > 5$$

However, the answer in the textbook is given:

$$ ( \infty, 0) \cup (5, \infty)$$

Which mean that the graph has a domain including x values of less than 0 and greater? Can someone explain this to me?

Hi tmt,

The product of 2 numbers is positive if either both numbers are positive, or if both numbers are negative.

It appears you have only considered the case where both are positive.
What do you get if both numbers would be negative?

 

[mathbalarka]

You have almost made it :

$$x(x - 5) > 0$$

As ILS said, both of the factors must be either positive or negative. If both are positive, then $x > 0$ and $x > 5$ and pick the larger of these two ($x > 5$) to make the product positive. If both are negative, $x < 0$ and $x < 5$ and you'd have to pick smaller of these two $(x < 0)$ to make this work. (Why?)

 

[MarkFL]

You have correctly determined:

$$x(x-5)>0$$

There are two points on the number line over which the expression on the left will change sign, and those two critical points are:

$$x=0,5$$

So, divide the real number line in the three intervals:

$$(-\infty,0),\,(0,5),\,(5,\infty)$$

Because all of the roots of the expression on the left are of odd multiplicity (they occur an odd number of times...once in this case), we know the sign of the expression will alternate across the three intervals.

So, pick a test value in any of the intervals, and check to see what the sign of the expression is for that test value. The let the signs of the other two intervals alternate with respect to your test interval. And finally, all intervals for which the sign is positive will be a part of your solution.