I’m practicing domains and ranges and I’m stuck on this function: f(x) = sqrt(4 – x^2) / (x + 2).
My domain attempt: 4 – x^2 ≥ 0 gives -2 ≤ x ≤ 2, and the denominator says x ≠ -2. So I wrote the domain as (-2, 2]. That seems right to me, but I’m open to correction if I’ve overlooked something.
Range is where I’m getting confused. Since sqrt(·) ≥ 0 and for x in (-2, 2] we have x + 2 > 0, I figured f(x) ≥ 0. Then I set y = sqrt(4 – x^2)/(x + 2) and squared to remove the root: y^2(x + 2)^2 = 4 – x^2. I rewrote this as a quadratic in x and tried using the discriminant to get conditions on y. I somehow ended up with 0 ≤ y ≤ 1, but that doesn’t seem right because near x = -2^+ the expression looks like it should get very large. So I think I’m mishandling a restriction when squaring, or not tracking the domain constraint properly after solving for x.
What’s the clean way to find the range here without introducing fake y-values when squaring? Do I need to keep x ∈ (-2, 2] explicitly in the final inequality, or is there a better trick?
Any help appreciated!
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3 Responses
You’ve nailed the domain: from 4 − x^2 ≥ 0 we get −2 ≤ x ≤ 2, and the denominator rules out x = −2, so the domain is (−2, 2]. For the range, a clean trick is to factor the inside of the root: 4 − x^2 = (2 − x)(2 + x). Since x + 2 > 0 on the domain, you can write f(x) = sqrt((2 − x)(2 + x))/(x + 2) = sqrt(2 − x) sqrt(2 + x)/(x + 2) = sqrt(2 − x)/sqrt(2 + x) = sqrt((2 − x)/(2 + x)). That makes the behavior transparent: as x → −2^+, the denominator sqrt(2 + x) → 0^+ while the numerator stays near sqrt(4), so f(x) → ∞; at x = 2, f(2) = 0; and at x = 0, f(0) = 1. In fact (2 − x)/(2 + x) decreases continuously from +∞ to 0 on (−2, 2], so the range is [0, ∞). I used to make the same “square-it” detour and got 0 ≤ y ≤ 1 too, because I treated y^2(x + 2)^2 = 4 − x^2 without keeping track that x + 2 is small near −2; it’s like forgetting you’re dividing by a tiny number. The first time a teacher showed me the substitution x = 2 cos θ (then f = tan(θ/2)), it finally clicked why the values go all the way up to infinity.
Your domain (-2, 2] is spot on; a clean range trick is to set x = 2 cosθ (think “sliding around a radius-2 semicircle”), so √(4 − x²) = 2 sinθ and f(x) = 2 sinθ / (2(1 + cosθ)) = tan(θ/2) with θ ∈ [0, π), hence the range is [0, ∞).
Example: x = 0 (θ = π/2) gives f(0) = √4/2 = 1, and as x → −2⁺ (θ → π⁻), tan(θ/2) → ∞.
Domain (-2, 2] is correct. For the range, set x = 2 cos θ with θ in [0, π], then f(x) = 2 sin θ / (2(1 + cos θ)) = tan(θ/2), so y ≥ 0 and the range is [0, ∞) (the ≤ 1 came from squaring without keeping the domain in view).