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

Calculus1
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To find f you need to first find f' the find f' given f'' integrate both sides
hmm.. \[f'(x) = \frac 1x + c\] ??
almost

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Other answers:

you are missing a certain constant multiple
ahh \[f'(x) = -\frac 1x + c\]
yes :) 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
or maybe i should do it \[f'(x) = -\frac 1x + c_1\]
i might have typoed
integrate again
ah yes i did. it's f'(2) not just f(2)
get a system of 2 eau and 2 unknowns
hmm \[f(x) = -\ln x + c_1 x + c_2\] yes?
ok great. you can find that constant by using f(1)=0 and then do what zarkon says to find f
you can do the problem with two given values of f
or you can wait to find the first constant whatever
i suppose x > 0 is just there to note that -ln x exists?
oh wait.... i guess you can do it with f(something1)=another something1 and f(something2)=another something2 oops
f'(2) = 0 so f'(2) = -1/2 + c_1 = 0 does this mean c_1 is 1/2?
yes adding 1/2 to both sides solves that equation for c_1
then f(1) = -ln (1) + 1/2 x + c_2 = 0 so c_2 is -1/2?
x is 1 so you have -ln(1)+1/2(1)+c_2=0 and yes
oh. yeah...forgot to sub 1 into x the second time
so \[f(x) = -\ln x + \frac 12 x - \frac 12\] ??
tep
nice. thanks
i just noticed this was my 1000th question

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