Find the domain of the function h where h(x)=(x(x-12))^1/2

Question

Find the domain of the function h where h(x)=(x(x-12))^1/2

Answer 1

Hi Virginia:

 

The function h(x) is defined as the square root of x(x-12).  The domain of the square root function, in your case, is defined as the values of x for which x(x-12) is greater or equal to zero.  So, let's see how to find out where that is.

 

First, let's find out where x(x-12)=0.  This is true if x=0 or if x=12.  These two points on the x axis are where x(x-12) changes sign.  So, we have to check the sign of x(x-12) for x less than 0, for x between 0 and 12, and for x greater than 12.

 

For x less than 0 let's try x=-1.  We find that x(x-12)=(-1)(-1-12)=+13, so the domain includes the interval x<0.

 

For x between 0 and 12, let's try x=1.  We find that x(x-12)=1(1-12)=-11, so the domain does not include the interval (0,12).

 

For x greater than 12, let's try x=13.  We find that x(x-12)=13(13-12)=+13, so the domain includes the interval x>12.

 

Combining these results, the domain is x<0 and x>12.

 

 

Answer 2

You have some function to the 1/2 power. The 1/2 power is the same thing as the square root.

Can you think of anything that we can't take the square root of? Negative numbers should be the thing that pops into your head anytime you see a square root.

 

So we can't have anything inside the square root that is negative. This means that anything inside the square root needs to be positive. In math-speak, that is...

 

x(x-12) > 0

 

Now we should figure out when the argument of the square root is positive.

 

Finding the roots of this equation tells us that x = 0 and x = 12

 

When is x(x - 12) > 0? You can figure this out a few ways. If you have a graphing calculator, plot this function. If you don't we will have to go with the tried and trued method of picking a value for x.

 

When x is less than 0 (one of the roots), what is the sign of the function? Try -1 and see what happens:

 

(-1) * (-1 - 12) = (-1) * (-13) = 13, which is positive.

 

Now pick a value for x that is in the middle of the roots...perhaps x = 1?

 

(1) * ( 1 - 12)  = 1 * (-11) = -11, a negative number (which is bad for the square root).

 

Using a number greater than 12 for x yields a positive number again, which you can figure out on your own.

 

So, the domain of the original function is (-∞,0) and (12, +∞).

 

I hope this helps! Please ask any questions if you have them

 

 

Answer 3

Hi Virginia,

 

h(x) = √(x(x-12))

The argument of the square root cannot be negative

x(x -12) = 0

x = 0, x = 12 are excluded from the domain of h(x)

x(x - 12) < 0  on the interval (0, 12)

The domain of h(x)  is (-∞, 0) and (12, ∞)