Thank you all! This approach, using the 'polynom' library, did the trick. > library(polynom) # -6 + 11*x - 6*x^2 + x^3 > p0 <- polynomial(c(-6, 11, -6, 1)) # has zeros at 1, 2, and 3 > p1 <- deriv(p0); p2 <- deriv(p1) # first and second derivative > xm <- solve(p1) # maxima and minima of p0 > xmax = xm[predict(p2, xm) < 0] # select the maxima > xmax # [1] 1.42265
With these tweaks: In fitting the model to the data I had to use raw=TRUE: model <- lm( y ~ poly(x, 3, raw=TRUE) ) Then I could generate p0 directly from my lm: p0 <- polynomial( coef(model) ) And I answered my second question by using the obvious: predict(p2,xmax) I don't know what I would have done if the optima weren't between x=0 and x=1, which was my constraint. In that case the "maximum" would have been one of the endpoints rather than a zero of p1. I suppose I could just have checked for it with some if/then code. Fortunately it didn't turn out to be an issue with my data. And no, this wasn't a homework problem. I didn't do the math by hand because I needed to automate this process for several subsets of my data and several fitted models. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
