recall two properties
1. for y = abc
y' = abc' + acb' + bca'
etc for multiple factors
**** However, if any of the a, b, or c are powers such as (c-4)^4 , then that base of x-4 will appear in each of the terms of y'' ***** very important !
2. if y'' > 0 for some x, then that x will produce a minimu, the curve is opening upwards
if y '' < 0 for some x, then that x will produce a maximum, the curve opens downwards
if however y''=0, then there is neither a max or min, the curve "flattens out".
So for your first derivative, we have only one linear factor, the (x-1).
It will appear in all the terms of the second derivative except in the term when it was differentiated.
So when x = 1, you will get zeros for all except
(1)(x-2)^2 (x-3)^3 (x-4)^4 (x-5)^5
finding the value for x = 1 is still easy ...
1(1)(-2)^3(-3)^4 (-4)^5
= (+)(+)(-)(+)(-) = + , notice I don't really care about the actual number value
so x=1 will produce a minimum
all the other factors will appear in each of the terms, so when we sub in x = 2,3, .... we will get 0 for each and thus no max or min for those values.
The slope of a curve Y = f(x) is given by dy/dx=(X-1)(x-2)^2(x-3)^3(x-4)^4(x-5)^5.
(a) For what value or values of x does y have a local maximum/minimum? Justify your answer.
I know the maximum/min occurs when dy/dx = 0, which in this case is 1,2,3,4,5 but how do you know which one is max/min?
2 answers
AHHH ok ok thank you so much!!
1 answer