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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
So the critical numbers are found using the first derivative, right? I get x=11 as the only critical number.
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?
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?
Yes, I think so. You find the first derivative, set it equal to zero, and then solve to find the critical numbers.
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
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
So does this answer choice sound right?: C. Critical no. 11; absolute max f(0) = 7; local and absolute min f(11) = -114.
These problems get a little more interesting, tricky, whatever, when the critical point falls outside the allowed region of values for x.
For example @JakeV8 ?
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.
ok so let me draw a pic to illustrate
What am I plugging 11 into to get -114?
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!!
We are only looking at the graph between 0 and 12
yes, I shouldn't have jumped in a minute ago... was offering a hypothetical different problem, but that only confuses things.
@cuzzin are you following this? It seemed like you were, then you said you were confused.
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
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)
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
Ah, ok then. That makes a lot of sense. Thanks for the help, I get it now.
Thanks for being so in-depth.
@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.