How do you find the domain and range of #f(x)= sqrt( x^2+4)#?

1 Answer
May 16, 2018

Answer:

Domain: All Real Numbers ℝ
Range: [2,∞)

Explanation:

#x# is squared, so any negative x-values will result in a positive value for #x^2#. The domain is not restricted for x = a positive number or x = 0 since the expression under the radical will always be positive.
Therefore, the domain is all real numbers.

For the range, #x^2# takes its minimum at #x = 0#, so plugging in #x = 0# gives #y = sqrt(x^2+4)=sqrt(0+4)=2# therefore making the range [2,∞).