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THE DERIVATIVE IS 15X^2-122X+16

DO WE FIND ALSO 2ND DERIVATIVE TOO?

lol..here we are again...it's back to factoring.

yup here we are... where the 240 come from...

i dont believe 2nd derivative is necessary for determining extremas

http://www.sophia.org/finding-extrema-on-an-interval--2/finding-extrema-on-an-interval--8-tutorial

ok show it.

why do you multiply it?

and the middle term is -122x

Because it is not a trinomial with leading coefficient of 1.

yes. right..

That is why I said you want numbers which add to the coeffcient of the middle term.

The two numbers which multiply to give you 240 that add to -122 are -120 and -2.

oh i got it... you just separated them ..

can you separate them by anything? ex: -61x ?

since 122/2 = 61..

nvm i understand.. you want something you can easily factor with 15 and 2

so i get.. x= 8 and 2/15

There you go.

which one is max and min?

wait the intervals are [0,4] what do we do with it?

O_o

true.. but what do you mean derivative to the left and right?

plug x=0 and x=4 into the original equation in order to find the extrema points

|dw:1363299185209:dw|

Hope that makes sense.

f(0)= 3 and f(4) -589

okay those are the extrema points.. but how about the work for x= 8 and x= 2/15?

its wrong....

i see how you got the 16 and -232... but those aren;t the min or max of the interval.

i put 16 as min ad -232 as max. its wrong.

okay so far i have f'(0)=16 f'(4)=-232 and the x=8 and the other x= 2/15

i would appreciate if i can see some work please.

do u understand what i said

yeah but i'm more of a visual person..

i understand what you're saying though.

http://www.wolframalpha.com/input/?i=16-122+x%2B15+x%5E2&lk=1&a=ClashPrefs_*Math-

that is the graph of ur derivative

these are inflection points say for your base function x^3 ...+...+...=y

|dw:1363301676110:dw|

make sense now?

but lets suppose this is your interval |dw:1363302163817:dw|

oh okay thank you=]

i see what method you used.

so is the minim here: -589 and the max -1213??