anonymous
  • anonymous
Find f if f''(x)=2e^t+3sin(t), f(0)=7, f(pi)=-8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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FireKat97
  • FireKat97
because you have been given f"(x) you need to integrate twice to get to f'(x) and then f(x)
anonymous
  • anonymous
I have been doing that but have not gotten the correct response
FireKat97
  • FireKat97
Can you post your working so far?

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anonymous
  • anonymous
I have quite a few different problems But I only have one more chance to get it right.
anonymous
  • anonymous
Like I have worked them a lot
FireKat97
  • FireKat97
but can you show me how you went about trying to solve the problem? So your working out so I can see where you're going wrong..
anonymous
  • anonymous
I am not exactly sure how to take a photo and attach a file since I am on my computer.
FireKat97
  • FireKat97
do you think you could type up your solution?
anonymous
  • anonymous
its a kinda complicated solution thats why i am on here trying to get an answer for it so i an compare the correct answer to my answer to see where i went wrong
FireKat97
  • FireKat97
okay so when you integrate 2e^x what do you get?
anonymous
  • anonymous
2e^x anything e^x is still e^x
FireKat97
  • FireKat97
yup thats correct
FireKat97
  • FireKat97
sorry what do you get when you integrate 3sint
anonymous
  • anonymous
-3cost
FireKat97
  • FireKat97
Yup, thats correct too, so we know that the integral of f"(x) is 2e^t - 3cost + c
FireKat97
  • FireKat97
so f'(x) = 2e^t - 3cost so now we have to integrate again
anonymous
  • anonymous
I am running out of time its an online homework due at a specific time Can you tell me the answer? I only have one last attempt and 2/23 problems left that I have been struggling with for the past hour and a half
FireKat97
  • FireKat97
you're not going to learn if i just tell you the answer..
FireKat97
  • FireKat97
just try integrating f'(x) = 2e^t - 3cost + c in the same way and tell me what you get
anonymous
  • anonymous
The tricky part with this question is that you have two constants since you have to integrate twice I understand the mechanics of how to do the problem and with you not telling me your answer I can't know if it is correct so if the answer you have is wrong then I am learning the wrong thing I have tried 4 different answers so if you tell me your answer I can check to see if that was one of the answers I got
anonymous
  • anonymous
2e^t+3sint+cx+c
freckles
  • freckles
he and you are both correct and saying f'(x)=2e^t-3cos(t)+c you can integrate first... or you can use your constraint to find c...wait isn't one of your constraints written incorrectly...one of them should be f'(a)=b where a and b are some numbers ...whichever one it is use the constraint then integrate note: if you integrate first make sure you don't use the same letter for the other constant you get after the next integration
anonymous
  • anonymous
But what is the final answer that you get ?
freckles
  • freckles
No one can know the problem is incorrectly stated
anonymous
  • anonymous
What do you mean it is incorrectly stated ?
freckles
  • freckles
As I was saying earlier isn't one of your constraints is suppose to be in the form f'(a)=b where and b are some numbers?
anonymous
  • anonymous
yes it is supposed to be but the two questions I was stuck on do not give it in the form of f(x) and f'(x) which is why it is difficult But it can be done I know the steps but keep getting the wrong answer
freckles
  • freckles
where a and b are some numbers*
FireKat97
  • FireKat97
No, not necessarily.
freckles
  • freckles
err... do you not have a condition on f'?
FireKat97
  • FireKat97
In this case, because f(0) is given, cx can be eliminated giving the constant d
freckles
  • freckles
you have f(0)=7 and f(pi)=-8 one of these is really suppose to be about f'
FireKat97
  • FireKat97
what I mean is f(x) = 2e^t - 3sint + cx + d we can find d and hence when we sub in f(pi) = -8, the value of d is already known so we can find c as well
anonymous
  • anonymous
no you are supposed to integrate twice fine the constant of the first integrated equation integrate again and then solve for the second constant and plug everything back into the integrated problem that gave the equation for f(x)
FireKat97
  • FireKat97
because since f(0) = 7 is known, everything except d is equal to 0, so d = 7
anonymous
  • anonymous
FireKat97 that is what I have been doing and keep getting the wrong answer Can you tell me your answer so that I can see if it is correct and then send me a picture of your work ?
FireKat97
  • FireKat97
then we know that f(pi) = -8 = 2e^π - 3sinπ + cπ + 7 and c can now be found too
FireKat97
  • FireKat97
I got f(x) = 2e^t - 3sin(t) + t(-15 - 2e^t)/π + 7 hoping I didn't make any silly mistakes along the way...
freckles
  • freckles
I guess you guys were right about f(0)=7 and f(pi)=-8 I was just thinking they normally give a condition for each integration step but I got a different d and c above using the conditions
FireKat97
  • FireKat97
Unless f(0) is given, something about f'(x) is required
freckles
  • freckles
right I see that now
anonymous
  • anonymous
the answer you gave was incorrect FireKat97
FireKat97
  • FireKat97
do they need it in decimals if its an online quiz thing
freckles
  • freckles
but yeah f(0)=7 doesn't give d=7 you should get 2-3(0)+c(0)+d=7 which this will give us a different c in the end also
FireKat97
  • FireKat97
OH wait whoops! yeah
anonymous
  • anonymous
no, no decimals
anonymous
  • anonymous
so wait what should the answer be ?
FireKat97
  • FireKat97
because e^0 = 1 not 0
FireKat97
  • FireKat97
d = 5 my badddd
anonymous
  • anonymous
so instead of 7 it should be 5 ?
FireKat97
  • FireKat97
yeah and so when you sub d in again for f(pi) = -8, use d as 5, so c should be different too
anonymous
  • anonymous
okay i give up I just won't get points for this problem lol
FireKat97
  • FireKat97
don't give up, you're almost there :p
zepdrix
  • zepdrix
broski, ask for help on a question sooner than 30 minutes before the due date! XD that's crazy sauce!
anonymous
  • anonymous
I still don't have an answer lol so I do not feel almost there
anonymous
  • anonymous
I have more than 30 minutes..
zepdrix
  • zepdrix
For future reference, when you get into a jam like this, https://www.wolframalpha.com/input/?i=f%27%27%28t%29%3D2e%5Et%2B3sin%28t%29%2C+f%280%29%3D7%2C+f%28pi%29%3D-8 Wolfram is really helpful with problems such as this one. Unless your teacher considers that cheating of course :p I find it to be a useful tool to check work though.
zepdrix
  • zepdrix
yay \c:/
anonymous
  • anonymous
I have already tried wolfram and it does not work either not for this problem
FireKat97
  • FireKat97
i copied and pasted and forgot to get fix the 5 :p haha
zepdrix
  • zepdrix
inputting into wolfram is a little tricky :O I think I did it correctly though, didn't i?
anonymous
  • anonymous
so that is not the correct answer FIREKAT?
freckles
  • freckles
@Asiah321 we said d=5 and you had f(t)= 2e^t - 3sint + cx + 5 so now use this other condition f(pi) = -8 plug in pi for t and replace f(pi) wth -8 and solve for c
freckles
  • freckles
oops that x should be t but whatever
freckles
  • freckles
solve for c: -8=2e^pi-3sin(pi)+c(pi)+5
anonymous
  • anonymous
Where is the typo ?
Jhannybean
  • Jhannybean
\[f''(x)=2e^t+3\sin(t)\]\[f'(x)=\int (2e^t+3\sin(t))dt = 2e^t-3\cos(t)+c\]\[f(x) = \int (2e^t-3\cos(t)+C)dt = 2e^t-3\sin(t) +Cx+D\]\[7=2e^{0}-3\sin(0)+C(0)+D \implies D = 5\]\[-8 = 2e^{\pi} -3\sin(\pi)+\pi C+5\]\[\pi C = -2e^{\pi}-13 \qquad \implies C=\frac{-2e^{\pi}-13}{\pi}\]\[f(x) = ....\]
Jhannybean
  • Jhannybean
Therefore, what im getting for \(f(x)\) is.. \[f(x) = 2e^{t}-3\sin(t)+\left(\frac{-2e^{\pi}-13}{\pi}\right)t+5\]
Jhannybean
  • Jhannybean
some variable errors. But you can figure out that much
freckles
  • freckles
lol is it f(x) or f(t)
Jhannybean
  • Jhannybean
Haha, it's f(t).
anonymous
  • anonymous
Jhannybean it was correct !!!! thank you so muchhhhh !!!!
Jhannybean
  • Jhannybean
Whatever, it's f(o)
freckles
  • freckles
I only wanted to point that out because there has been a bunch of mixing of variables in this post
anonymous
  • anonymous
jhannybean can you check your messages please
Jhannybean
  • Jhannybean
Yeah, that's why i just went with what I thought it was.
Jhannybean
  • Jhannybean
You shouldn't be thanking me, honestly. I was working it out while @freckles and @FireKat97 we're helping you solve it:P Sometimes you have to work with people instead of asking fr the answer.
anonymous
  • anonymous
I really appreciate you all ! I have been struggling with this problem and another one for a while
Jhannybean
  • Jhannybean
post your new question in a new thread :)

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