“f(x) = 2 sqrt(-x^2+4) - 2
the range is R:{y| -2 ≤ y ≤ 2, yER)
How do i get this answer?”
f(x) = 2 √(–x^2 + 4) - 2
Domain:
We assume f(x) is a real function, meaning the values of f(x) are real, not imaginary.
So the square root must evaluate to a real number.
So the expression under the square root must NOT evaluate to a negative number.
4 – x^2 ≥ 0
x^2 ≤ 4
|x| ≤ 2
–2 ≤ x ≤ 2 <== f(x) is only defined for these x’s; the inequality is the Domain of f(x).
Range of f(x) = 2 √(–x^2 + 4) - 2, –2 ≤ x ≤ 2:
f(±2) = 2 √(–(±2)^2 + 4) - 2 = 2 √(–4 + 4) - 2 = –2
f(0) = 2 √(–0^2 + 4) - 2 = 2 √(4) - 2 = 2
Any other value of x in the domain of f(x) will result in a value between –2 and 2 because the vertex of the downward opening parabola y = –x^2 + 4 is (0,4).
So the Range of f(x) is:
–2 ≤ f(x) ≤ 2
In "R:{y| -2 ≤ y ≤ 2, y ∈ R)", "R:{...}" means "Range is the set ...", y = f(x), "|" means "such that", "y ∈ R" means y is a member of the set of Real Numbers.
To visualize this problem, here are GeoGebra graphs:
http://www.wyzant.com/resources/files/267507/range_of_sqrt_function
Note that GeoGebra does not graph imaginary values, that’s why the graphs stop abruptly.
Comments