What is the domain and range of afunction y=(the square root of minus x square plus nine )plus one over x minus one
AAAaannndd the bot gets it wrong yet again!
The words are unclear, but you can pick either
y = (√(9-x^2)+1)/(x-1)
y = √(9-x^2) + 1/(x-1)
the root has domain [-3,3] and denominator excludes x = 1
so the whole domain is [-3,1)U(1,3]
I apologize for the error in my previous response. You are correct, the domain of the function should be [-3,1) U (1,3], with x not equal to 1 in both cases. The range of the function will be y greater than or equal to 1 in both cases.
To find the domain and range of the given function y = √(-x^2 + 9) + 1/(x-1), we need to consider the restrictions on the variable x.
Domain:
The function involves a square root and a division by (x-1). Therefore, we need to ensure that the expressions inside the square root and the denominator are valid.
Inside the square root, -x^2 + 9, should be non-negative (i.e., greater than or equal to zero). So, we have -x^2 + 9 ≥ 0.
Solving the inequality, we get:
-x^2 + 9 ≥ 0
9 ≥ x^2
-√9 ≤ x ≤ √9
-3 ≤ x ≤ 3
So, the domain for this function is the interval [-3, 3] (including the endpoints).
Range:
To determine the range, we need to examine the behavior of the function as x approaches the given domain values.
As x approaches -3 from the left side, the square root term becomes √(-(-3)^2 + 9) = √0. The division term approaches negative infinity, as (x-1) becomes (-3-1) = -4. Hence, the function approaches negative infinity.
As x approaches any value between -3 and 3, both terms in the function remain positive and non-zero. Therefore, the function remains above zero.
As x approaches 3 from the right side, the square root term becomes √(-(3)^2 + 9) = √0. The division term approaches positive infinity, as (x-1) becomes (3-1) = 2. Hence, the function approaches positive infinity.
Therefore, the range of the given function is (-∞, +∞), excluding 0.
Domain: all real numbers except x=1
Range: y ≥ 1