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But each such \((m,n)\) gives: - Decision Point

📅 May 3, 2026 👤 scraface

May 03, 2026

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Try to assume $ g(x) = x + c $. Then:
Now assume $ g $ linear: $ g(x) = mx + n $
And should equal $ 4 - 12 + 11 = 3 $
So no linear solution? But problem assumes there is.
Perhaps $ g(x) = 2 - x $? Try: $ g(2) = 0 $, $ f(0) = 5
And desired is 11, so perhaps $ g(x) = x - 1 +
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