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3 Responses
A quick, no-fuss way is to think in terms of “undoing” the steps of the function. For f(x) = 2x + 3, the function first doubles, then adds 3, so the inverse must first subtract 3, then divide by 2. That gives f^{-1}(x) = (x − 3)/2. To get f^{-1}(17), just do (17 − 3)/2 = 14/2 = 7, and you can sanity-check by plugging back: f(7) = 2·7 + 3 = 17. If you don’t like the “swap x and y” move, you can avoid it completely: to find f^{-1}(17), simply solve 2x + 3 = 17 directly; the solution x = 7 is the answer. A handy general memory aid: for any linear f(x) = ax + b (with a ≠ 0), the inverse is f^{-1}(x) = (x − b)/a-think “subtract b, then divide by a” to keep the signs and order straight. Hope this helps!
Quickest way: think “undo the steps in reverse.” If f(x) = ax + b with a ≠ 0, then the inverse is f⁻¹(x) = (x − b)/a. That’s just solving y = ax + b for x: subtract b, then divide by a. It’s the same idea as the “swap x and y” trick, but phrased as clean algebra so you’re less likely to trip on signs. A tiny mental check: doing f, then f⁻¹, should bring you right back to x.
Example with your function: f(x) = 2x + 3. Undo add 3 by subtracting 3, then undo multiply by 2 by dividing by 2, so f⁻¹(x) = (x − 3)/2. To find f⁻¹(17), compute (17 − 3)/2 = 14/2 = 7. Same result if you just solve 2x + 3 = 17 directly: 2x = 14 → x = 7. For a quick refresher with more examples, see Khan Academy’s guide: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:functions/x2f8bb11595b61c86:invertible-functions/a/finding-inverse-functions.
Undo the steps in reverse: from y = 2x + 3, subtract 3 then divide by 2, so f^{-1}(x) = (x − 3)/2. Plugging in 17 gives (17 − 3)/2 = 7, unless I’m missing a small detail.