How do you find domain for #y = sqrt((4+x) / (1-x) ) #?

Question

Eva Adamson · Accepted Answer

Domain #{x:RR, -4<=x<1}#

Range #{y:RR, y>=0]# For real y, #4+x>=0, x>=-4#; and #1-x>0, or, x<1# The domain is therefore #{x:RR, -4<=x<1}# For range, at x=-4, y is 0 and as x increases from -4, y remains positive and as x approaches 1, #y->+oo#. Hence Range would be #{y:RR, y>=0}#

Ethan Adkins · Answer

undefined To find the domain of $ y = \sqrt{\frac{4+x}{1-x}} $, we need to consider the restrictions on the expression under the square root:

1. The denominator $1-x$ cannot be zero, so $1 - x \neq 0$.
2. The expression inside the square root, $\frac{4+x}{1-x}$, must be non-negative, meaning it cannot be negative or undefined.

Solving the first condition, $1 - x \neq 0$, we get $x \neq 1$.

For the second condition, we need to ensure that the expression $4 + x$ is greater than or equal to zero and that $1 - x$ is greater than zero:

1. $4 + x \geq 0$ implies $x \geq -4$.
2. $1 - x > 0$ implies $x < 1$.

Therefore, the domain of $y$ is $x \in (-\infty, 1)$.

Hazel Anglin · Answer

undefined To find the domain of the function $ y = \sqrt{\frac{4 + x}{1 - x}} $, we need to identify any values of $ x $ that would make the expression under the square root invalid.

The expression under the square root must be greater than or equal to zero to ensure that the square root function is defined. Therefore, we need to solve the inequality:

$$ \frac{4 + x}{1 - x} \geq 0 $$

To solve this inequality, we need to consider two cases:

1. When the numerator and denominator have the same sign.
2. When the numerator and denominator have opposite signs.

After identifying the intervals where the expression is greater than or equal to zero, we can determine the domain of the function by considering the union of these intervals.

Once we have found the valid intervals for $ x $, we can express the domain using interval notation.