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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
 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
 one year ago

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

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

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

calmat01Best ResponseYou've already chosen the best response.0
lol..here we are again...it's back to factoring.
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
You need to find factors of 240, which is the product of 15 and 16 which add to the coefficient of the middle term.
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
If you split the 122x into 120x and 2x, you will get the numbers I was referring to. Then you can factor by grouping.
 one year ago

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

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

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

calmat01Best ResponseYou've already chosen the best response.0
I explained already...240 comes from multiplying the 15(coefficient of squared term) and 16 (the constant).
 one year ago

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

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

calmat01Best ResponseYou've already chosen the best response.0
Because it is not a trinomial with leading coefficient of 1.
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
That is why I said you want numbers which add to the coeffcient of the middle term.
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
The two numbers which multiply to give you 240 that add to 122 are 120 and 2.
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
So, 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.
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
And 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.
 one year ago

calmat01Best 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 \[(15x ^{2}120x) + (2x+16)=0\]
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
Now just factor a 15x from the first parenthesis, and a 2 from the second parenthesis, and what do you notices when you do that?
 one year ago

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

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

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

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

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

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

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

calmat01Best ResponseYou've already chosen the best response.0
Check 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.
 one year ago

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

completeidiotBest ResponseYou've already chosen the best response.0
plug x=0 and x=4 into the original equation in order to find the extrema points
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
You 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.
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
dw:1363299185209:dw
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
Hope that makes sense.
 one year ago

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

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

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

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

calmat01Best ResponseYou've already chosen the best response.0
No, 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.
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
Because the graph at x=2/15 changes direction...it goes from increasing to dercreasing thus identifying a max.
 one year ago

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

dan815Best ResponseYou've already chosen the best response.0
take 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)
 one year ago

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

dan815Best ResponseYou've already chosen the best response.0
do u understand what i said
 one year ago

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

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

dan815Best ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=16122+x%2B15+x%5E2&lk=1&a=ClashPrefs_*Math
 one year ago

dan815Best ResponseYou've already chosen the best response.0
that is the graph of ur derivative
 one year ago

dan815Best ResponseYou've already chosen the best response.0
to be honest if ur visual u should realize its a 3rd degree func to 2nd degree so its like thisdw:1363301376615:dw
 one year ago

dan815Best ResponseYou've already chosen the best response.0
so 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
 one year ago

dan815Best ResponseYou've already chosen the best response.0
these are inflection points say for your base function x^3 ...+...+...=y
 one year ago

dan815Best ResponseYou've already chosen the best response.0
so 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
 one year ago

dan815Best ResponseYou've already chosen the best response.0
where your slope is 0 right so those are crit pts but do those crit points lie where the graph is concave up or down
 one year ago

dan815Best ResponseYou've already chosen the best response.0
for 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
 one year ago

dan815Best ResponseYou've already chosen the best response.0
but lets suppose this is your interval dw:1363302163817:dw
 one year ago

dan815Best ResponseYou've already chosen the best response.0
so 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
 one year ago

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

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

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