How do you find the domain and range of #sqrt((13x)/((x^2)-1)#?

Question

Savannah Anderson · Accepted Answer

The domain is #x in (-1,0] uu (+1,+oo)#
The range is #y in [0,+oo)# What's uner the #sqrt# sign is #>=0# Thus, #(13x)/(x^2-1)>=0# #(13x)/((x+1)(x-1))>=0# Let #f(x)=(13x)/((x+1)(x-1))# We could create a sign chart. #color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-1##color(white)(aaaaa)##0##color(white)(aaaaaaa)##+1##color(white)(aaaa)##+oo# #color(white)(aaaa)##x+1##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aa)##+# #color(white)(aaaa)##x##color(white)(aaaaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aa)##+# #color(white)(aaaa)##x-1##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+# #color(white)(aaaa)##f(x)##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+# Consequently, #f(x)>=0# when #x in (-1,0] uu (+1,+oo)# The domain is #x in (-1,0] uu (+1,+oo)# Let's #y=sqrt((13x)/(x^2-1))# When #x=-1#, #=>#, #y=+oo# When #x=0#, #=>#, #y=0# The range is #y in [0,+oo)#

Jace Anderson · Answer

undefined To find the domain and range of $ \sqrt{\frac{13x}{x^2 - 1}} $:

1. Domain: The expression is defined for all real numbers except where the denominator is zero, $ x^2 - 1 \neq 0 $. This occurs when $ x \neq 1 $ and $ x \neq -1 $. So, the domain is $ (-\infty, -1) \cup (-1, 1) \cup (1, \infty) $.

2. Range: Since the function involves a square root, the values under the square root must be non-negative. Thus, the range is $ [0, \infty) $.

Jace Anderson · Answer

undefined To find the domain and range of the function $ f(x) = \sqrt{\frac{13x}{x^2 - 1}} $:

1. **Domain**: Determine where the function is defined. Since the function involves a square root, the expression inside the square root must be non-negative, and the denominator must not equal zero.

For $ \sqrt{\frac{13x}{x^2 - 1}} $ to be real, $ \frac{13x}{x^2 - 1} \geq 0 $.

Additionally, the denominator $ x^2 - 1 $ cannot equal zero, so $ x^2 \neq 1 $.

2. **Range**: Determine the possible output values of the function.

Since the square root of a non-negative number is always non-negative, the range of $ f(x) $ will be all non-negative real numbers.

Putting these together:

1. **Domain**:
   - The function is defined for $ x $ where $ x^2 - 1 \neq 0 $, so $ x \neq \pm 1 $.
   - Additionally, the expression under the square root must be non-negative, so $ \frac{13x}{x^2 - 1} \geq 0 $.

2. **Range**:
   - The range of $ f(x) $ is all non-negative real numbers.

Therefore, the domain of the function is all real numbers except $ x = \pm 1 $, and the range is all non-negative real numbers.