# Thread: Extreme Values

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?

2. 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...?

3. 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 $\displaystyle f'(x)>0\text{ for all }x\epsilon[-2,6]\Rightarrow\text{ abs min at }(-2,f(-2))$
Are you sure that the given interval is $\displaystyle [-2,6]$ and not $\displaystyle (-2,6]$?

4. 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?

5. 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 $\displaystyle c$, that is $\displaystyle a<c<b$ such that $\displaystyle f(c)>f(x)$ for all x in $\displaystyle [a,b]$.

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 $\displaystyle 'c'$ value.

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

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 $\displaystyle 'c'$ value.
Actually the definition of local extrema varies.
See for example

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

To be considered as a local max, there must be some interval (a,b) around some number c, that is $\displaystyle a<c<b$ such that $\displaystyle f(c)>f(x)$ for all x in $\displaystyle [a,b]$
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 $\displaystyle I$ about c and $\displaystyle f(c)\geq f(x)$ for
all $\displaystyle x\in I\cap S$

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.

7. 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 $\displaystyle I$ about c and $\displaystyle f(c)\geq f(x)$ for
all $\displaystyle x\in I\cap S$

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 $\displaystyle I$"

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

8. No, the difference is the open interval doesnt have to be part of the domain.

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

So c is a local maxima, and an endpoint.