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lgbasallote
 2 years ago
find f
\[f''(x) = x^{2}, \quad x > 0,\quad f(1) = 0, \quad f(2) = 0\]
lgbasallote
 2 years ago
find f \[f''(x) = x^{2}, \quad x > 0,\quad f(1) = 0, \quad f(2) = 0\]

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myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3To find f you need to first find f' the find f' given f'' integrate both sides

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0hmm.. \[f'(x) = \frac 1x + c\] ??

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3you are missing a certain constant multiple

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0ahh \[f'(x) = \frac 1x + c\]

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3yes :) now to find the constant hmmm....you are missing a certain initial condition to do that .... your question doesn't make sense .... you need one of those to be f'(something)=another something

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0or maybe i should do it \[f'(x) = \frac 1x + c_1\]

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0i might have typoed

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0ah yes i did. it's f'(2) not just f(2)

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1get a system of 2 eau and 2 unknowns

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0hmm \[f(x) = \ln x + c_1 x + c_2\] yes?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3ok great. you can find that constant by using f(1)=0 and then do what zarkon says to find f

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1you can do the problem with two given values of f

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3or you can wait to find the first constant whatever

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0i suppose x > 0 is just there to note that ln x exists?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3oh wait.... i guess you can do it with f(something1)=another something1 and f(something2)=another something2 oops

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0f'(2) = 0 so f'(2) = 1/2 + c_1 = 0 does this mean c_1 is 1/2?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3yes adding 1/2 to both sides solves that equation for c_1

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0then f(1) = ln (1) + 1/2 x + c_2 = 0 so c_2 is 1/2?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3x is 1 so you have ln(1)+1/2(1)+c_2=0 and yes

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0oh. yeah...forgot to sub 1 into x the second time

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0so \[f(x) = \ln x + \frac 12 x  \frac 12\] ??

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0i just noticed this was my 1000th question
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