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- anonymous

If you only had a derivative of a function, and you had to work out what the plot of the original curve was, and maybe the derivative too, how would you approach that problem?

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- anonymous

- katieb

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- Michele_Laino

what is:
E^2x

- anonymous

I just made it up.. how about E^x

- anonymous

\[f'[x]=(x+2)/E^x\]

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- Michele_Laino

we can integrate yor first derivative, nevertheless I don't know how, since I don't know the meaning of E^x

- anonymous

is that not E to the power of x?

- Michele_Laino

yes! Is E the base of natural logarithms

- anonymous

yes

- anonymous

So I would need to integrate.. I haven't learned that yet.. I guess that's coming next..

- Michele_Laino

ok! then we have to compute this integral:
\[\Large \int {\left( {x + 2} \right){e^{ - x}}} dx\]

- Michele_Laino

after a simple computation, we get:
\[\Large \int {\left( {x + 2} \right){e^{ - x}}} dx = - {e^{ - x}}\left( {x + 3} \right) + C\]
where C is the usually arbitrary real constant

- Michele_Laino

in other words you have a family of functions, which differ each other by an additive constant C
|dw:1435230843014:dw|

- Michele_Laino

where:
\[\Large f\left( x \right) = - {e^{ - x}}\left( {x + 3} \right)\]

- anonymous

so now its just a matter of plotting the functions out.. or working out the points.. thank you michele, I was just curious on this one.. I need to read up on the rules for integration.

- Michele_Laino

better is if you draw the graph of the function:
\[\Large f\left( x \right) = - {e^{ - x}}\left( {x + 3} \right)\]
applying the theorems of Mathematical Analysis.
Once you got that graph, you only shift it up or down by an arbitrary constant C, so you will get a family of functions, whose first derivative is given by the derivative which you provided

- anonymous

ahh right.. d/dx C would be 0 so you would have a number of potential functions

- Michele_Laino

that's right!

- anonymous

ah well, then hopefully the problem has at least one point defined.. like 0,1

- anonymous

and we restrict the family.

- Michele_Laino

yes! that's right!

- anonymous

cool, thanks michele..

- Michele_Laino

:)

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