# Thread: Absolute min and max help.

1. ## Absolute min and max help.

The instructions say sketch graph by hand and use sketch to find absolute and locx and min for values of f.

$\displaystyle f(x)= ln x, 0 < x \leq 2$
The answer is abs max is f(2) = ln 2

I can see that there is no local max or min (i cheated though and didn't draw the graph, just that f'(x) can't equal to 0)

But I can see how 2 would be the absolute max because it's the biggest number in the interval, and no absolute min because the function will always try to reach 0 but doesn't reach it.

But the Extreme value theorem in the book says that
If F is continous on a closed interval [a,b], then f attains an absolute maximum value of f(c) and an absolute min at f(d) at some number c and d in [a,b]

doesn't that mean to have absolute max or min, the function has to be on a closed interval?

2. Originally Posted by dorkymichelle

But the Extreme value theorem in the book says that
If F is continous on a closed interval [a,b], then f attains an absolute maximum value of f(c) and an absolute min at f(d) at some number c and d in [a,b]

doesn't that mean to have absolute max or min, the function has to be on a closed interval?
Note that:
$\displaystyle x \in [a,b] = a \leq x \leq b$
$\displaystyle x \in (a,b] = a < x \leq b$
$\displaystyle x \in [a,b) = a \leq x < b$
$\displaystyle x \in (a,b) = a <x<b$

So the definition in your book says the max lies within [a,b]. This includes when x = a or b too.