# Extreme Values

• November 7th 2009, 03:16 PM
john-1
Extreme Values
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?
• November 7th 2009, 03:20 PM
VonNemo19
Quote:

Originally Posted by john-1
Determine the extreme values of each function on the given interval.

a) f(x)= 1 + (x+3)^2 , -2<=x<=6

what are you have trouble with here? Taking the derivative, interval notation, etc...?
• November 7th 2009, 03:32 PM
VonNemo19
Quote:

Originally Posted by john-1
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?

What you have looks good to me. Since $f'(x)>0\text{ for all }x\epsilon[-2,6]\Rightarrow\text{ abs min at }(-2,f(-2))$
Are you sure that the given interval is $[-2,6]$ and not $(-2,6]$?
• November 7th 2009, 03:34 PM
john-1
The interval is [-2,6] << that is including -2 and including 6.

Would the local max and local min be the same as abs max and abs min? If not, what's the difference between them and how do I find it?
• November 7th 2009, 03:44 PM
VonNemo19
Quote:

Originally Posted by john-1
The interval is [-2,6] << that is including -2 and including 6.

Would the local max and local min be the same as abs max and abs min? If not, what's the difference between them and how do I find it?

To be considered as a local max, there must be some interval (a,b) around some number $c$, that is $a such that $f(c)>f(x)$ for all x in $[a,b]$.

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 $'c'$ value.
• November 7th 2009, 07:22 PM
hjortur
Quote:

Originally Posted by VonNemo19
To be considered as a local max, there must be some interval (a,b) around some number $c$, that is $a such that $f(c)>f(x)$ for all x in $[a,b]$.

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 $'c'$ value.

Actually the definition of local extrema varies.
See for example

http://www.mathhelpforum.com/math-he...um-values.html

Quote:

To be considered as a local max, there must be some interval (a,b) around some number c, that is $a such that $f(c)>f(x)$ for all x in $[a,b]$
The definition I was tought:

Let S be the domain of f, for a point c to be considered a local max there must be an open
interval $I$ about c and $f(c)\geq f(x)$ for
all $x\in I\cap S$

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.
• November 7th 2009, 07:27 PM
VonNemo19
Quote:

Originally Posted by hjortur
Actually the definition of local extrema varies.
See for example

http://www.mathhelpforum.com/math-he...um-values.html

The definition I was tought:

Let S be the domain of f, for a point c to be considered a local max there must be an open
interval $I$ about c and $f(c)\geq f(x)$ for
all $x\in I\cap S$

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.

Note that - by definition - "there must be an open interval $I$"

We've said the same thing. I cannot imagine placing an open interval around an endpoint, can you?

(Wink)
• November 8th 2009, 03:27 AM
hjortur
No, the difference is the open interval doesnt have to be part of the domain.

Lets say you have a function $f:[c,d]\to\mathbb{R}$. Now lets say that
c is an absolute maximum. There is an interval $]a,b[\$ with $a, such that $c\in]a,b[$.
and $f(c)\geq f(x)$ for all $x\in]a,b[\cap[c,d]$

So c is a local maxima, and an endpoint.