Question

Do anyone know how to solve this problem

Answer 1

Answer:

The domain is $\boxed{D_f = [0, \infty)}$.

Step-by-step explanation:

Let $f(x) = \dfrac{\sqrt{2x}}{x+1}$.

Since we have a fraction, we need to make sure that its denominator isn't zero, since division by zero isn't defined. This means that:

$x+1 \neq 0 \iff x \neq -1.$

On the other hand, there's also a square root, whose argument can't be negative. This means that:

$2x \geq 0 \iff x \geq 0.$

So the domain is:

$\boxed{D_f =\{x\in\mathbb{R}: x\geq 0 \textrm{ and } x \neq -1\} = [0, \infty)}.$