How do you find domain and range for # f(x) = (x^2 - 16) / (x^2 - 25)#?

Question

Genesis Allen · Accepted Answer

Domain = #x in RR: x!=-5, 5#
Range = #x in RR: x!=1# First solve for domain which for inverse functions is equal to all real numbers except for when the denominator is equal to zero (so solve #x^2-25=0 rarr x=-5,5# so the domain is equal to all real numbers except -5 and 5 To solve for range I had to graph it using (https://tutor.hix.ai), there is a mechanical way to find it but graphing is much easier and most of the time you should have access to a graphing calculator

Savannah Anderson · Answer

undefined To find the domain of $ f(x) = \frac{x^2 - 16}{x^2 - 25} $, identify values that make the denominator equal to zero. The domain is all real numbers except $ x = -5 $ and $ x = 5 $.

To determine the range, analyze the behavior of the function. As $ x $ approaches $ \pm \infty $, $ f(x) $ approaches 1. Therefore, the range is $ (-\infty, 1) $ excluding any values of $ f(x) $ when $ x = -5 $ or $ x = 5 $.

Jameson Allen · Answer

undefined To find the domain and range of $ f(x) = \frac{x^2 - 16}{x^2 - 25} $:

1. **Domain**: Identify any values of $ x $ that would make the denominator equal to zero. The domain consists of all real numbers except those that would result in division by zero. In this case, the denominator $ x^2 - 25 $ cannot equal zero, so we exclude values of $ x $ that satisfy $ x^2 - 25 = 0 $. Solve $ x^2 - 25 = 0 $ to find these values.

$$ x^2 - 25 = 0 $$
$$ (x + 5)(x - 5) = 0 $$

The solutions are $ x = -5 $ and $ x = 5 $. Therefore, the domain is all real numbers except $ x = -5 $ and $ x = 5 $.

2. **Range**: To find the range, observe the behavior of the function as $ x $ approaches positive and negative infinity. Notice that as $ x $ approaches positive or negative infinity, the function approaches 1. Thus, the range is all real numbers except for the value 1.