Yahoo Answers is shutting down on 4 May 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
really simple question but i don't know how to do it.help please?
ok so the equation is y = x℮‾ ˣ², but i can't find its stationary points (max, min points and horizontal inflection points)
i've found the gradient but i got x= -1, and when i checked with the graphical calculator, it's lesser than that. so i've asked my friends and they don't know it either.
1 Answer
- JordanLv 41 decade agoFavourite answer
derivative is e^(-x^2)-2e^(-x^2)x^2 = e^(-x^2)(1-2x^2) = 0 when 1-2x^2=0 which is when x=+/-sqrt(.5)
horizontal inflection points at x=sqrt(.5) and x=-sqrt(.5)
from there, you can figure out that x<-sqrt(.5) and x>sqrt(.5) you have a -ive slope (at both infinities it approaches 0), and for -sqrt(.5)<x<sqrt(.5) you have +ive slope
there's no vertical asymptotes, so no jumps in the graph
basic plot of the graph is
....v-- x=sqrt(.5)
_../\_
..\/
..^-- x=-sqrt(.5)
so you know your horizontal tangent points are at least local max/min, solve for the limit at +/- infinity to get what the value at the ends is and so you know the global max/min (the limit as x->inf is 0, so your local max/min are your global max/min)