Slopes and roots without a limit
Enough machinery. Let us put it to work on one honest polynomial and get the things calculus is actually for — a slope, a turning point, the roots — using nothing but grade bookkeeping and, at the very end, a single square root. No limits, no solver, no approximation creeping in anywhere.
The function, the same one we differentiated two chapters back:
We already have this one. Offset the input by the graded zero, sort the expansion onto grades, and the coefficient on grade 1 is the derivative — recovered by lifting with ω and projecting, exactly as in the structural differential:
That is the additive reading — comparing f to its neighbour by subtraction. The interesting work starts when we change the chart.
Take the multiplicative reading from the chart — compare by dividing rather than subtracting. Instead of f(x+0) − f(x), form the ratio f(x+0) / f(x). Using the same offset expansion, the value in front factors straight out:
The ratio is 1 plus a grade-1 term, and that term's coefficient is the logarithmic derivative — lift by ω and project just as before:
This one object holds both things we are hunting. Its zeros are where the top vanishes — the slope is flat — so they are the critical points of f. Its poles are where the bottom vanishes — the division blows up — so they are the roots of f. Extrema upstairs, roots downstairs, in a single expression.
Set the numerator to zero. That is the whole calculation:
At x = −3/2 the slope is flat — the turning point of the parabola. Its height is f(−3/2) = −25/4, and that 25 is worth remembering: it is about to reappear on the other side of the ledger.
Now the roots. At a root, f = 0, so the logarithmic derivative diverges — the denominator hits the pole and the whole thing runs off to ω. Rather than clear the fraction and factor, concentrate the unknown into the numerator's own variable. Let z = 2x + 3, so x = (z − 3)/2, and rewrite the logarithmic derivative in z:
All of the difficulty has collected into one place. This diverges exactly when the denominator vanishes — and that is a single square root:
Walk each value back through x = (z − 3)/2:
So the roots are x = 1 and x = −4 — and indeed (x − 1)(x + 4) = x² + 3x − 4. Factoring would have found the same pair faster here, but that is the point in miniature: the method did not need to guess a factorization. It pushed every bit of the difficulty into one square root, √25 — which is exactly √(b² − 4ac), the discriminant, the same 25 that set the depth of the turning point. The extremum and the root-spacing were the one number, read once downstairs and once up.
Slope, turning point, and roots — the entire first-pass analysis of a function — came out of graded arithmetic and a single surd, with no limit taken and no decimal approached at any step. The logarithmic derivative did the organizing: it filed the extrema and the roots as the zeros and poles of one object, and the lone irrationality the problem truly contains showed up as one honest square root rather than smeared across a numerical search.1 This is what "calculus without limits" looks like when it stops being a slogan and starts doing homework.