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f'(x)=2x-22
f'(x)=2(x-11)
critical numbers: x=11

Well firstly you gotta find the derivative of f(x)

What's my next move here? I'm stuck.

Right, which is 2(x-11)

Oh, and problem correction. It is x^2-22x+7

wait just trying to remember how I use to solve this give me a sec

Well when f'(x)=0 its either a max or a min correct?

I found, using a slope chart, that 11 is a local min.

in this case its a parabola so its simple to know the shape|dw:1351565881044:dw|

Not exactly sure what a slope chart is but ya

We must take into consideration our endpoints too

Correct

ok so let me draw a pic to illustrate

What am I plugging 11 into to get -114?

|dw:1351566181002:dw|

We are only looking at the graph between 0 and 12

No what he is missing is that we are only looking at the area in within the boundry

I follow everything except how we arrived at the (0,7) and (11,-114) points.

We would like to find the max and the min within that boundry

Ah, ok then. That makes a lot of sense. Thanks for the help, I get it now.

Thanks for being so in-depth.