Question

Find the domain of the real valued function: f(x) = $\frac{1}{x + |x|}$

Answer 1

Answer:      Domain = ( 0 , + ∞ )

The given function is

$f(x)=\frac{1}{x+|x|}$

By Definition

|x| = x   for   x ≥ 0

|x| = - x   for   x < 0

⇒ The given function does not hold for |x| = - x   for   x < 0 Because

$f(x)=\frac{1}{x+|x|}=\frac{1}{x+(-x)}=\frac{1}{x-x}=\frac{1}{0}$

And 1/0 = ∞

⇒ The given function also does not hold for x = 0 .

It holds only for |x| = x   for   x > 0

So

Domain = ( 0 , + ∞ )

Answer 2

Answer:

dom (f)=(0,∞)

Step-by-step explanation:

Find the domain of the real valued function:

$f(x) = \frac{1}{x + |x|}$

now

$|x|=\left \{ {{x,when\:x\geq 0} \atop {-x,when x<0}} \right. \\\\so\\\\x+|x|=\left \{ {{x+x,when\:x\geq 0} \atop {x-x,when x<0}} \right. \\\\\\x+|x|=\left \{ {{2x,when\:x\geq 0} \atop {0,when x<0}} \right.$

so for

$f(x)=\frac{1}{x+|x|}$

assue those values for which x+|x|≠ 0,

thus

dom (f)=(0,∞)