Cubic S Curve E7Ec4C

1. The problem is to find an equation for a cubic-like S-curve passing through points $(1,0)$ and $(2,1)$. 2. A general cubic function has the form $$y = ax^3 + bx^2 + cx + d$$ where $a$, $b$, $c$, and $d$ are constants. 3. Since the curve passes through $(1,0)$, substitute $x=1$, $y=0$: $$0 = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d$$ 4. Since the curve passes through $(2,1)$, substitute $x=2$, $y=1$: $$1 = a(2)^3 + b(2)^2 + c(2) + d = 8a + 4b + 2c + d$$ 5. To find a unique cubic, we need two more conditions. For an S-curve shape, the inflection point is important. The inflection point occurs where the second derivative is zero. 6. The second derivative of $y$ is: $$y'' = 6ax + 2b$$ 7. Assume the inflection point is at $x=1.5$ (midpoint between 1 and 2), so: $$0 = 6a(1.5) + 2b = 9a + 2b$$ 8. We have three equations: $$a + b + c + d = 0$$ $$8a + 4b + 2c + d = 1$$ $$9a + 2b = 0$$ 9. Solve for $b$ from the third equation: $$b = -\frac{9}{2}a$$ 10. Substitute $b$ into the first two equations: $$a - \frac{9}{2}a + c + d = 0 \Rightarrow -\frac{7}{2}a + c + d = 0$$ $$8a + 4\left(-\frac{9}{2}a\right) + 2c + d = 1 \Rightarrow 8a - 18a + 2c + d = 1 \Rightarrow -10a + 2c + d = 1$$ 11. From the first substituted equation: $$c + d = \frac{7}{2}a$$ 12. From the second substituted equation: $$2c + d = 1 + 10a$$ 13. Subtract the first from the second: $$2c + d - (c + d) = 1 + 10a - \frac{7}{2}a \Rightarrow c = 1 + 10a - \frac{7}{2}a = 1 + \frac{20}{2}a - \frac{7}{2}a = 1 + \frac{13}{2}a$$ 14. Substitute $c$ back into $c + d = \frac{7}{2}a$: $$1 + \frac{13}{2}a + d = \frac{7}{2}a \Rightarrow d = \frac{7}{2}a - 1 - \frac{13}{2}a = -1 - 3a$$ 15. Choose $a=\frac{2}{13}$ to simplify coefficients: $$a = \frac{2}{13}, b = -\frac{9}{2} \times \frac{2}{13} = -\frac{9}{13}, c = 1 + \frac{13}{2} \times \frac{2}{13} = 1 + 1 = 2, d = -1 - 3 \times \frac{2}{13} = -1 - \frac{6}{13} = -\frac{19}{13}$$ 16. The cubic equation is: $$y = \frac{2}{13}x^3 - \frac{9}{13}x^2 + 2x - \frac{19}{13}$$ 17. This equation passes through $(1,0)$ and $(2,1)$ and has an inflection point near $x=1.5$, matching the S-curve shape. Final answer: $$y = \frac{2}{13}x^3 - \frac{9}{13}x^2 + 2x - \frac{19}{13}$$