I’m prepping for a test and inverse functions are scrambling my brain a bit. If a function is like a recipe I can undo, why do I sometimes have to “chop” the menu before I can run it backward?
For example, take f(x) = x^2 – 4x + 3. I know the drill is to swap x and y and solve for y to get the inverse. When I try that, I end up with something like y = 2 ± something (I tried completing the square and also the quadratic formula, not sure which is more relevant). But then I’m told I have to restrict the domain so the inverse is actually a function. I get the horizontal line test in theory, but in practice I keep second-guessing which side of the vertex to keep.
How do I quickly decide the correct domain restriction (like x ≥ something or x ≤ something) without graphing every time? And when I get the ±, how do I know which branch belongs to the actual inverse? Bonus confusion: when people say the domain and range “swap,” does that mean the new domain is literally the old range, even if it includes weird values I wasn’t expecting?
I’ve tried sketching rough graphs and plugging a couple of numbers to see what’s happening, but I’m not sure that’s the right approach under time pressure. Any tips to build an intuition (or a quick checklist) so I stop mixing this up?
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3 Responses
Think of restricting the domain as picking a single, one-way hallway through the house so you don’t get turned around. A function needs to be one-to-one to have an inverse that’s also a function, and quadratics like x^2 − 4x + 3 fail the horizontal line test unless you choose one side of the vertex. The fast, no-graph way: complete the square to spot the vertex, or use h = −b/(2a). Then pick either x ≥ h or x ≤ h, so the function becomes strictly monotonic on that interval. For an upward-opening parabola (a > 0), the right side (x ≥ h) is increasing and the left side (x ≤ h) is decreasing; either is fine, just be consistent.
Worked example: f(x) = x^2 − 4x + 3 = (x − 2)^2 − 1, so the vertex is at (2, −1) and the range is [−1, ∞). Swap x and y: x = (y − 2)^2 − 1 ⇒ (y − 2)^2 = x + 1 ⇒ y = 2 ± √(x + 1). Now choose a branch. If you restrict to x ≥ 2 (the increasing side), the correct inverse is y = 2 + √(x + 1). Quick check: f(3) = 0, so f⁻¹(0) should be 3; plug in 0 to 2 + √(0 + 1) = 3-nice. The minus branch, 2 − √(x + 1), corresponds to the x ≤ 2 restriction. About the “swap”: yes, the inverse’s domain is the original function’s range (after you’ve restricted the original domain). Here that’s [−1, ∞). The inverse’s range becomes the restricted domain you chose (e.g., [2, ∞) if you picked x ≥ 2).
A tiny mental checklist: rewrite as (x − h)^2 + k, choose x ≥ h (use y = h + √(x − k)) or choose x ≤ h (use y = h − √(x − k)), and remember the inverse’s domain is x ≥ k. If you forget which sign, test a single friendly value from the original function (like we did with x = 3) to see which branch lands you back where you started.
I like your “recipe you can undo” picture. The snag is: if two different ingredients cook up the same dish, you can’t tell which one to un-mix back to. That’s exactly what happens with parabolas. For f(x) = x^2 − 4x + 3 = (x − 2)^2 − 1, the vertex is at x = 2 (you can grab that fast with −b/2a). To make it undoable, you pick one side of the vertex so it’s one-to-one: either x ≥ 2 (the right, where it’s increasing) or x ≤ 2 (the left, where it’s decreasing). No graph needed-just note the vertex and remember: for an upward-opening parabola, “right side” means increasing; “left side” means decreasing. If you want the inverse to be increasing (often the standard choice), keep x ≥ 2.
When you swap x and y, you get x = (y − 2)^2 − 1, so (y − 2)^2 = x + 1 and y = 2 ± sqrt(x + 1). The ± is the two sides you just chopped apart. If you kept x ≥ 2, you take the + branch: y = 2 + sqrt(x + 1). If you kept x ≤ 2, you take the − branch: y = 2 − sqrt(x + 1). The “domain and range swap” bit is literal after you’ve restricted: on x ≥ 2, f’s range is [−1, ∞), so the inverse’s domain is [−1, ∞). (Confession: I sometimes write x > −1 to “be safe” with the square root, even though sqrt(0) is actually fine-old habit.) The inverse’s range then becomes the side you kept: y ≥ 2 for the + branch, or y ≤ 2 for the − branch.
Quick mental checklist I use: (1) put it in vertex form to spot h and k fast; (2) pick a side of x = h-right if you want increasing; (3) solve for y and match the sign to the side you kept (right side → “+”, left side → “−”). Does that help? Which side would you pick for this one, and why-and do you want your inverse to come out increasing or decreasing?
Quick rule I use (and I’m 95% sure it’s right): complete the square f(x)=(x−2)^2−1, then choose one monotone side-keep x≥2 for the right-hand branch so the inverse is y=2+√(x+1) (domain x≥−1, range x≥2), or keep x≤2 for the left-hand branch so the inverse is y=2−√(x+1) (domain x≥−1, range x≤2); yes, the inverse’s domain is literally the original range (after the restriction).
Nice refresher: https://www.khanacademy.org/math/algebra2/manipulating-functions/inverse-functions-a2/v/restricting-domains-of-functions-to-make-their-inverses-functions