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Mathematics 15 Online
OpenStudy (anonymous):

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).

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

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

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

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 \(-ax^2+6ax\)

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

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

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

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?????

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

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

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

OpenStudy (phi):

**add 20

OpenStudy (anonymous):

ok yup I got it then thanks

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