Find all critical numbers of f and classify the extreme values given x ∈ [0,12] and f(x)=x^2-22x+7.

- anonymous

Find all critical numbers of f and classify the extreme values given x ∈ [0,12] and f(x)=x^2-22x+7.

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- anonymous

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

- swissgirl

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

- anonymous

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

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- anonymous

Right, which is 2(x-11)

- anonymous

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

- anonymous

So the critical numbers are found using the first derivative, right? I get x=11 as the only critical number.

- anonymous

And the second derivative test gives the local min/max of that point. The second derivative in this case is only 2. Is it better to use a slope chart in this case?

- swissgirl

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

- swissgirl

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

- anonymous

Yes, I think so. You find the first derivative, set it equal to zero, and then solve to find the critical numbers.

- swissgirl

So what I tend to do was just plug in a 11 back into the original equation and you will right away know if its the min or the max

- anonymous

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

- swissgirl

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

- swissgirl

Not exactly sure what a slope chart is but ya

- swissgirl

We must take into consideration our endpoints too

- anonymous

So does this answer choice sound right?:
C. Critical no. 11; absolute max f(0) = 7; local and absolute min f(11) = -114.

- swissgirl

Correct

- anonymous

These problems get a little more interesting, tricky, whatever, when the critical point falls outside the allowed region of values for x.

- swissgirl

For example @JakeV8 ?

- anonymous

I guess that's where I get stuck then. I get how to find absolute max, min, critical points and all of that. But I don't know how to arrive at the last bit of information.

- swissgirl

ok so let me draw a pic to illustrate

- anonymous

What am I plugging 11 into to get -114?

- anonymous

If you restrict x such that it doesn't include x = 11, then you can use the same curve, but there is no critical point in the region. So you end up with max and min points at the two ends of f(x). Wow, a picture would be much easier!!

- swissgirl

|dw:1351566181002:dw|

- swissgirl

We are only looking at the graph between 0 and 12

- anonymous

yes, I shouldn't have jumped in a minute ago... was offering a hypothetical different problem, but that only confuses things.

- anonymous

@cuzzin are you following this? It seemed like you were, then you said you were confused.

- swissgirl

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

- anonymous

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

- swissgirl

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

- swissgirl

So how do we go about this? Well we know that the max or min occur at the point where f'(x)=0 IN this case we found out that the derivative equals 0 when x=11 And then we found out if it was a max or min which it was a min in this case. Now since its a parabola we know that the vertex is the global minimum point. Then we plugged 11 back into our original equation which was f(11)=(11)^2-22(11)+7=-114 So our global minimum is at (11,-114)

- swissgirl

Now we know another thing that the end point usually are a max or min. Imagine if we didnt have a boundry this parabola would just go on for forever and we would have no maximum point. It would just continue on to infinity. When we have a boundry it stop the parabola at some point so we end up having a maximum. So look back at my picture and you can see that the maximum is at the boundry.
So what we do is, we plug in 0 and 12 back into our original equation and try to see which one is a maximum
f(0)=0^2-22(0)+7=7
f(12)=12^2-22(12)+7=-113
Now as you seee (0,7) is the maximum point

- anonymous

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

- anonymous

Thanks for being so in-depth.

- anonymous

@cuzzin... so one last thought... think if the region for x had been 0 to 2. That region would not include the critical point at x = 11. So within that region, you could calculate the points f(0) and f(2)... these would be the max and min points over the region x = 0 to x = 2, but the region would NOT include the critical point or the global minimum.
That wasn't important in this question, but you may see one like that sometime.

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