12th Calculus
posted by elley .
use the derivative of the function
y=f(x)to find the points at which f has a
local maximum
local minimum
point of inflection
y'=(x1)^2(x2)(x4)

if f(x)=uvw
then
f'(x)=uvw'+ uwv'+ wvu' 
I will assume you really meant to write the derivative y' and not the function y although I would have thought the book would ask you to find that derivative.
Wherever that derivative, y', is zero, the curve is horizontal
This is twice at x = 1 and at x =2 and at x = 4
Look at the second derivative to see what happens there.
y'=(x1)^2 (x^26x+8)
so
y" = (x1)^2(2x6)+2(x1)(x^26x+8)
now what is y"at x = 1?
y"(1) = Zero, so that is an inflection point
what is y" at x = 2?
y"(2) = 2+0 = 2
That is a maximum because y" is negative
what is y"(4)?
y"(4) = 9(2)+2*3(1624+8)
y"(4) = 18+0 = 18
so it is Minimum at x = 4
y"