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What is the range of this function?

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1 Answer

“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

Because y = 4 – x^2 is a parabola that opens down. It's x-intercepts are ±2 so it's maximum is halfway between at x = 0. It's minimum on the domain –2 ≤ x ≤ 2 is when it is zero at x = ±2.
But doesn't this square root function go continuously to the left and starts at -2? So why would there be 2 zeros and a vertex?

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