1. ## Absolute extrema

$\displaystyle 3x^2-6x+5$

x is a member of [0,4] find the absolute extrema

2. To find the global extrema, check the local extrema and the limit of the function approaching the bounds.

Local extrema:
$\displaystyle y=3x^2-6x + 5$

$\displaystyle y' = 6x - 6 = 0$

$\displaystyle x = 1$

We have a local extremum at $\displaystyle x = 1$. What's f(x) of the point?
$\displaystyle y=3x^2-6x + 5$
$\displaystyle y=2$

Now check the bounds.
$\displaystyle \lim_{x\to 0^{+}}3x^2-6x+5 = 5$

$\displaystyle \lim_{x\to 4^{-}}3x^2-6x+5 = 29$

We have 3 possible values to be the global extrema.
$\displaystyle y=2$ -- (the local extrema, x=1)
$\displaystyle y=5$ -- (left bound)
$\displaystyle y=29$ -- (right bound)

The greatest of these values, $\displaystyle y=29$ is the global maximum, and the least, $\displaystyle y=2$ is the global minimum.

3. The answer above is enough to solve. In addition, there are a few more things that may come in handy

$\displaystyle y= 3x^2-6x + 5$

Firstly, this is a quadratic (second degree) function.
Quadratic functions are of the form $\displaystyle y=a~x^2 + b~x + c$
The first coefficient $\displaystyle a$ indicates the side of the function (upwards or downwards) and its narrowness. If $\displaystyle a>0$, the function is upwards. If $\displaystyle a<0$, the function is downwards.
In this example, $\displaystyle a=3$, which is greater than zero. This makes the function upwards. So, the vertex is the local and global minimum, and the global maximum is at the bounds.