How do you find the domain and range of #f(x) = (2x)/(sqrt(16-8x))#?

Domain:

The domain will be determined by saying that the radical has to be larger than #0#, this being because it's a rational function as well as a radical function.

Consider the function #y = 1/x#.

The domain would be #x !=0#, since if x was equal to 0, then the function would be non-defined.

Now consider the function #y = 1/sqrt(x)#.

Recall that a square root is undefined in the real number system if the number inside the square (the radicand) is less than 0 (#x < 0#). At this point, the domain is #x >= 0#. However, since the radical is in the denominator, and the denominator cannot be equal to #0#, the domain is made #x > 0#, or the #0# is excluded.

Back to our function at hand:

#sqrt(16 - 8x) > 0#

#16 - 8x > 0#

#16 > 8x#

#2 > x#

Hence, the domain is #x< 2, #

The graph of the function, shown in the following image, justifies our answer.

Range:

From as far as I can tell, the range is all the real numbers. I looked at the graph to huge numbers (ex. #2.08 xx 10^4)#, and it looked like the y values were continuing to reduce to #-oo#, indicating the presence of no horizontal asymptotes.

Thus, the range is #y in RR#.

In summary:

For the function #f(x) = (2x)/sqrt(16 - 8x)#:

•The domain is #x < 2#

•The range is #y in RR#

Practice Exercises:

Determine the domain and range of the function #g(x) = (4x)/sqrt(2x^2 + 11x + 12)#.

Hopefully this helps, and good luck!