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

Jul 1, 2016

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 $y = \frac{1}{x}$.

The domain would be $x \ne 0$, since if x was equal to 0, then the function would be non-defined.

Now consider the function $y = \frac{1}{\sqrt{x}}$.

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 $x \ge 0$. 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:

$\sqrt{16 - 8 x} > 0$

$16 - 8 x > 0$

$16 > 8 x$

$2 > x$

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 $- \infty$, indicating the presence of no horizontal asymptotes.

Thus, the range is $y \in \mathbb{R}$.

In summary:

For the function $f \left(x\right) = \frac{2 x}{\sqrt{16 - 8 x}}$:

•The domain is $x < 2$

•The range is $y \in \mathbb{R}$

Practice Exercises:

Determine the domain and range of the function $g \left(x\right) = \frac{4 x}{\sqrt{2 {x}^{2} + 11 x + 12}}$.

Hopefully this helps, and good luck!