Determine the extreme values of each function on the given interval.
a) f(x)= 1 + (x+3)^2 , -2<=x<=6
This is my work:
Critical numbers of f'(x)=2(x+3)
1) f'(x)=0 when x=-3
Test Critical numbers & end points
f(-2)=2
f(6)=82
Abs max is 82 (this answer is correct with the back of the book)
Abs min is 2 (this answer is WRONG with the back of the book)
What did I do here that is wrong?
Would the local max be 82 at x=6 and the local min be 2 at x=-2?
To be considered as a local max, there must be some interval (a,b) around some number , that is such that for all x in .
So, this implies that a local max can be an absolute max, but not all absolute maxs are local (because they can occur at endpoints where their is no open interval that can contain the value.
Actually the definition of local extrema varies.
See for example
http://www.mathhelpforum.com/math-he...um-values.html
The definition I was tought:To be considered as a local max, there must be some interval (a,b) around some number c, that is such that for all x in
Let S be the domain of f, for a point c to be considered a local max there must be an open
interval about c and for
all
The difference is subtle, in your version local extrema can't occur at endpoints, in my version they can.
I don't know which one is the right one, I just wanted to point this out.