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

Question

Genesis Allen · Accepted Answer

Domain: #(-oo,-1) uu (1, +oo)#
Range: #[0, +oo)# The expression that lies below the square root will identify the function's domain. More specifically, for real numbers, the square root is defined exclusively for positive numbers, which means that #x^2-1>=0#, #(AA)x in RR#. The quadratic has two possible answers. #x^2 - 1 = 0# #x^2 = 1 => x = +-1# For numbers outside the #(-1,1)# interval, this expression will always be positive. However, for values of #x# Inside the interval, this expression will be negative, which means that the domain of the function cannot include values in the #(-1,1)# interval. The domain of the function will thus be #(-oo,-1) uu (1, +oo)#. The fact that the square root of a real number will always yield a positive value will determine the function's range. This means that the range of the will be #[0, +oo)#. sqrt(x^2-1) [-7.9, 7.904, -3.946, 3.956]} is the graph.

Ethan Adkins · Answer

undefined To find the domain of $ f(x) = \sqrt{x^2 - 1} $, we need to ensure that the expression inside the square root is non-negative:

$$ x^2 - 1 \geq 0 $$

Solving this inequality, we get:

$$ x^2 \geq 1 $$

$$ x \geq 1 $$ or $ x \leq -1 $

So, the domain of $ f(x) $ is $ x \leq -1 $ or $ x \geq 1 $.

For the range, since the square root function outputs non-negative values, the range of $ f(x) $ is all real numbers greater than or equal to 0, expressed as $ [0, +\infty) $.