Help I don't know where to start
The graph in the picture is of the derivative of the function f(t).
Given that f(0)=20, estimate of the maximum value of f(t).
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OpenStudy (anonymous):
OpenStudy (anonymous):
@satellite73 can you please help me
OpenStudy (anonymous):
you'll need to integrate this to get the function. notice that you have the two zeros of x = 0 and x = 6.
so f'(t) = -x(x-6), f'(t)= -x^2 - 6x
now you should be able to integrate and find the max value
OpenStudy (anonymous):
yeah we can do this, but first i think you need an estimate of the area under the curve
OpenStudy (anonymous):
ok so the area is 114
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OpenStudy (anonymous):
is that right
OpenStudy (anonymous):
are you still there
OpenStudy (phi):
I am wondering if this is a trick question? you get a cubic, which goes to infinity.
you do get a local maximum. Is that what they are asking for ?
OpenStudy (anonymous):
the leading coefficient is not minus one in this case
OpenStudy (phi):
there are two tails, one goes down, the other up
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OpenStudy (anonymous):
I am really confussed I wrote the problem exactly as it says
OpenStudy (anonymous):
it is asking for the max on the interval i think
OpenStudy (anonymous):
the function is not presumed to exist outside of this interval
OpenStudy (anonymous):
ok so then how do I get the estimate
OpenStudy (anonymous):
are yall still there
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OpenStudy (anonymous):
yes, i am here and i am thinking
OpenStudy (anonymous):
ok
OpenStudy (phi):
The way I did it was integrate -x(x-6) dx
and use f(0)=20 to define the constant
OpenStudy (anonymous):
ok then what
OpenStudy (anonymous):
@phi this might work, but you are making an assumption
namely that this is \(-x^2+6x\) and not \(-ax^2+6ax\)
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OpenStudy (anonymous):
so then how do you answer the question
OpenStudy (anonymous):
still thinking
OpenStudy (anonymous):
I hate to rush you but I got to get this done I am on a time limit
OpenStudy (anonymous):
got an example in the book?
OpenStudy (anonymous):
no
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OpenStudy (anonymous):
my book is worth lesss
OpenStudy (phi):
yes, it should be -15/4 * x(x-6)
to match y=30 at x=2
OpenStudy (anonymous):
ok I am following
OpenStudy (anonymous):
what next
OpenStudy (phi):
integrate -15/4 x^2 + 45/2 x
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OpenStudy (anonymous):
ok
OpenStudy (anonymous):
I got -520.86
OpenStudy (anonymous):
now what
OpenStudy (phi):
you get -15/4 * x^3/3 + 45/4 x^2
add 20 to get
y= -15/4 * x^3/3 + 45/4 x^2 +20
evaluate at x=0 to get 20
and at x=6 to get ?
OpenStudy (anonymous):
-500.86
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OpenStudy (anonymous):
just wondering where the \(-\frac{15}{4}\) came from?
OpenStudy (phi):
-15/4 came from y' graph. it is a parabola and it goes through (2,30)
OpenStudy (anonymous):
is my answer right?
OpenStudy (anonymous):
oh ok
i think you are supposed to estimate the area under the curve and add 20 to the result, but this method is probably better
OpenStudy (anonymous):
is my answer right?????
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OpenStudy (phi):
no, I think they want you to find the area under the curve.
and add the constant. It seems faster..
OpenStudy (anonymous):
what is the area then
OpenStudy (anonymous):
i get 155
OpenStudy (anonymous):
ok that is what I got 2
OpenStudy (anonymous):
so then take the -500.86 +155
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OpenStudy (anonymous):
lol then go with that one,
@phi did you get that too?
OpenStudy (anonymous):
no just 155 that is the answer
OpenStudy (anonymous):
so then all O should write is 155+20=175 and that is my answer?
OpenStudy (phi):
the integral is 155. (that includes the 20 already)
OpenStudy (anonymous):
oh ok so then is 155 my final answer
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OpenStudy (anonymous):
is it?
OpenStudy (phi):
yes. if you count the area under f' more carefully you get about 135
add to get about 155