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anonymous
 3 years ago
HELP Determine the extrema of below on the given interval
f(x)=5x^361x^2+16x+3
(a) on [0,4]
The minimum is and the maximum is .
(b) on [9,9]
The minimum is and the maximum is
anonymous
 3 years ago
HELP Determine the extrema of below on the given interval f(x)=5x^361x^2+16x+3 (a) on [0,4] The minimum is and the maximum is . (b) on [9,9] The minimum is and the maximum is

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first derivative and set to zero to find the local max and min other points of interest would be the end points

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0THE DERIVATIVE IS 15X^2122X+16

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0DO WE FIND ALSO 2ND DERIVATIVE TOO?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol..here we are again...it's back to factoring.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You need to find factors of 240, which is the product of 15 and 16 which add to the coefficient of the middle term.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you split the 122x into 120x and 2x, you will get the numbers I was referring to. Then you can factor by grouping.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yup here we are... where the 240 come from...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i dont believe 2nd derivative is necessary for determining extremas

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.sophia.org/findingextremaonaninterval2/findingextremaonaninterval8tutorial

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I explained already...240 comes from multiplying the 15(coefficient of squared term) and 16 (the constant).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why do you multiply it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and the middle term is 122x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because it is not a trinomial with leading coefficient of 1.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That is why I said you want numbers which add to the coeffcient of the middle term.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The two numbers which multiply to give you 240 that add to 122 are 120 and 2.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, basic algebra tells us that if you want to factor an equation of the form \[ax ^{2}+bx+c=0 \] we must multiply a*c and find factors of a*c which add to the coeffcient of the middle term, in this case 122.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And those factors, as I already mentioned, are 120 and 2. So, just attach the x to each of them, group the first two and last two terms in separate parenthesis and factor by grouping.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[15x ^{2}122x+16=15x ^{2}120x2x+16=0\] Grouping them as I have instructed, you get \[(15x ^{2}120x) + (2x+16)=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now just factor a 15x from the first parenthesis, and a 2 from the second parenthesis, and what do you notices when you do that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh i got it... you just separated them ..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you separate them by anything? ex: 61x ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nvm i understand.. you want something you can easily factor with 15 and 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i get.. x= 8 and 2/15

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which one is max and min?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait the intervals are [0,4] what do we do with it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Check to see what happens to the derivative to the left and right of the point x=2/15 since x=8 is not in the interval.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0true.. but what do you mean derivative to the left and right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0plug x=0 and x=4 into the original equation in order to find the extrema points

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You can tell whether you have a max or min by what happens to the derivative to the left and right of the critical point. If f'(x) > o, your function is increasing, if f'(x)<0, your function is decreasing.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363299185209:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hope that makes sense.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f(0)= 3 and f(4) 589

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay those are the extrema points.. but how about the work for x= 8 and x= 2/15?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i see how you got the 16 and 232... but those aren;t the min or max of the interval.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i put 16 as min ad 232 as max. its wrong.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, I was telling you that the max occurs at x=2/15, not that the values I used were the max or min. You still have to determine the max by substituting your value for x=2/15.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because the graph at x=2/15 changes direction...it goes from increasing to dercreasing thus identifying a max.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay so far i have f'(0)=16 f'(4)=232 and the x=8 and the other x= 2/15

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0take first dervivative and find crit points and plug them into 2nd derivative to see if those crit ppts are maxima(negative for 2nd derv) or minima(positive)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i would appreciate if i can see some work please.

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0do u understand what i said

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah but i'm more of a visual person..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i understand what you're saying though.

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=16122+x%2B15+x%5E2&lk=1&a=ClashPrefs_*Math

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0that is the graph of ur derivative

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0to be honest if ur visual u should realize its a 3rd degree func to 2nd degree so its like thisdw:1363301376615:dw

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0so say you find f''(x) = 0 then you will know anything grreater than that x value will give u minimas and anything less than 0 for 2nd derv will give you maximas because they are inflection points

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0these are inflection points say for your base function x^3 ...+...+...=y

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0so whats its basically asking you is to find the critical pts and tell you if they are concave up or down so for the example above u know the critical pts are thesedw:1363301848503:dw

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0where your slope is 0 right so those are crit pts but do those crit points lie where the graph is concave up or down

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0for the first crit pt you can see its on the part of the graph where it is concave down and the 2nd one is on the part of the graph where its concave up

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0but lets suppose this is your interval dw:1363302163817:dw

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0so for the first critical point you know thats a maxima between the interval where your interval is x between those 2 vertical lines, but your minimum in that interval wont be the 2nd critival point because that is outside of the interval rather you will have to check the ends of the interval to see which one is of less value and that is your minumum

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i see what method you used.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so is the minim here: 589 and the max 1213??
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