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 one year 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
 one year 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

This Question is Closed

completeidiot
 one year 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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0THE DERIVATIVE IS 15X^2122X+16

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0DO WE FIND ALSO 2ND DERIVATIVE TOO?

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0lol..here we are again...it's back to factoring.

calmat01
 one year 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.

calmat01
 one year 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.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0yup here we are... where the 240 come from...

completeidiot
 one year ago
Best ResponseYou've already chosen the best response.0i dont believe 2nd derivative is necessary for determining extremas

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0http://www.sophia.org/findingextremaonaninterval2/findingextremaonaninterval8tutorial

calmat01
 one year 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).

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0why do you multiply it?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0and the middle term is 122x

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0Because it is not a trinomial with leading coefficient of 1.

calmat01
 one year 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.

calmat01
 one year 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.

calmat01
 one year 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.

calmat01
 one year 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.

calmat01
 one year 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 \[(15x ^{2}120x) + (2x+16)=0\]

calmat01
 one year 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?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0oh i got it... you just separated them ..

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0can you separate them by anything? ex: 61x ?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0since 122/2 = 61..

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so i get.. x= 8 and 2/15

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0which one is max and min?

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

calmat01
 one year 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.

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

completeidiot
 one year 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

calmat01
 one year 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.

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363299185209:dw

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0Hope that makes sense.

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

mathcalculus
 one year 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?

mathcalculus
 one year 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.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i put 16 as min ad 232 as max. its wrong.

calmat01
 one year 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.

calmat01
 one year 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.

mathcalculus
 one year 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
 one year 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)

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i would appreciate if i can see some work please.

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0yeah but i'm more of a visual person..

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i understand what you're saying though.

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

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

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

dan815
 one year 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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0oh okay thank you=]

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i see what method you used.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so is the minim here: 589 and the max 1213??
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