anonymous
  • anonymous
H-E-L-P Determine the extrema of below on the given interval f(x)=5x^3-61x^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
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
first derivative and set to zero to find the local max and min other points of interest would be the end points
anonymous
  • anonymous
THE DERIVATIVE IS 15X^2-122X+16
anonymous
  • anonymous
DO WE FIND ALSO 2ND DERIVATIVE TOO?

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anonymous
  • anonymous
lol..here we are again...it's back to factoring.
anonymous
  • anonymous
You need to find factors of 240, which is the product of 15 and 16 which add to the coefficient of the middle term.
anonymous
  • anonymous
If you split the -122x into -120x and -2x, you will get the numbers I was referring to. Then you can factor by grouping.
anonymous
  • anonymous
yup here we are... where the 240 come from...
anonymous
  • anonymous
i dont believe 2nd derivative is necessary for determining extremas
anonymous
  • anonymous
http://www.sophia.org/finding-extrema-on-an-interval--2/finding-extrema-on-an-interval--8-tutorial
anonymous
  • anonymous
I explained already...240 comes from multiplying the 15(coefficient of squared term) and 16 (the constant).
anonymous
  • anonymous
ok show it.
anonymous
  • anonymous
why do you multiply it?
anonymous
  • anonymous
and the middle term is -122x
anonymous
  • anonymous
Because it is not a trinomial with leading coefficient of 1.
anonymous
  • anonymous
yes. right..
anonymous
  • anonymous
That is why I said you want numbers which add to the coeffcient of the middle term.
anonymous
  • anonymous
The two numbers which multiply to give you 240 that add to -122 are -120 and -2.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[15x ^{2}-122x+16=15x ^{2}-120x-2x+16=0\] Grouping them as I have instructed, you get \[(15x ^{2}-120x) + (-2x+16)=0\]
anonymous
  • anonymous
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?
anonymous
  • anonymous
oh i got it... you just separated them ..
anonymous
  • anonymous
can you separate them by anything? ex: -61x ?
anonymous
  • anonymous
since 122/2 = 61..
anonymous
  • anonymous
nvm i understand.. you want something you can easily factor with 15 and 2
anonymous
  • anonymous
so i get.. x= 8 and 2/15
anonymous
  • anonymous
There you go.
anonymous
  • anonymous
which one is max and min?
anonymous
  • anonymous
wait the intervals are [0,4] what do we do with it?
anonymous
  • anonymous
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.
anonymous
  • anonymous
O_o
anonymous
  • anonymous
true.. but what do you mean derivative to the left and right?
anonymous
  • anonymous
plug x=0 and x=4 into the original equation in order to find the extrema points
anonymous
  • anonymous
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.
anonymous
  • anonymous
|dw:1363299185209:dw|
anonymous
  • anonymous
Hope that makes sense.
anonymous
  • anonymous
f(0)= 3 and f(4) -589
anonymous
  • anonymous
okay those are the extrema points.. but how about the work for x= 8 and x= 2/15?
anonymous
  • anonymous
@calmat01
anonymous
  • anonymous
its wrong....
anonymous
  • anonymous
i see how you got the 16 and -232... but those aren;t the min or max of the interval.
anonymous
  • anonymous
i put 16 as min ad -232 as max. its wrong.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Because the graph at x=2/15 changes direction...it goes from increasing to dercreasing thus identifying a max.
anonymous
  • anonymous
okay so far i have f'(0)=16 f'(4)=-232 and the x=8 and the other x= 2/15
dan815
  • dan815
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)
anonymous
  • anonymous
i would appreciate if i can see some work please.
dan815
  • dan815
do u understand what i said
anonymous
  • anonymous
yeah but i'm more of a visual person..
anonymous
  • anonymous
i understand what you're saying though.
dan815
  • dan815
http://www.wolframalpha.com/input/?i=16-122+x%2B15+x%5E2&lk=1&a=ClashPrefs_*Math-
dan815
  • dan815
that is the graph of ur derivative
dan815
  • dan815
to be honest if ur visual u should realize its a 3rd degree func to 2nd degree so its like this|dw:1363301376615:dw|
dan815
  • dan815
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
dan815
  • dan815
these are inflection points say for your base function x^3 ...+...+...=y
dan815
  • dan815
|dw:1363301676110:dw|
dan815
  • dan815
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 these|dw:1363301848503:dw|
dan815
  • dan815
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
dan815
  • dan815
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
dan815
  • dan815
make sense now?
dan815
  • dan815
but lets suppose this is your interval |dw:1363302163817:dw|
dan815
  • dan815
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
anonymous
  • anonymous
oh okay thank you=]
anonymous
  • anonymous
i see what method you used.
anonymous
  • anonymous
so is the minim here: -589 and the max -1213??

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