The domain of the function f(x) = sqrt{frac{x | vedclass

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Question

(A) For the function $f(x) = \sqrt{\frac{x}{1+x}}$ to be defined, the expression inside the square root must be non-negative:$\frac{x}{1+x} \ge 0$Also

Vedclass · Accepted Answer

$(-\infty, -1) \cup [0, \infty)$

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(A) For the function $f(x) = \sqrt{\frac{x}{1+x}}$ to be defined, the expression inside the square root must be non-negative:$\frac{x}{1+x} \ge 0$Also, the denominator cannot be zero, so $1+x 
eq 0$, which means $x 
eq -1$.We use the sign scheme (Wavy Curve Method) for the inequality $\frac{x}{x+1} \ge 0$:The critical points are $x = 0$ and $x = -1$.Testing intervals:$1$. For $x > 0$, $\frac{x}{1+x} > 0$ (Positive).$2$. For $-1 $3$. For $x  0$ (Positive).Since we need $\frac{x}{1+x} \ge 0$, we include $x=0$ and exclude $x=-1$.Thus, the domain is $(-\infty, -1) \cup [0, \infty)$.

The domain of the function $f(x) = \sqrt{\frac{x}{1+x}}$ is

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