anonymous one year ago 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?

1. Michele_Laino

what is: E^2x

2. anonymous

3. anonymous

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

4. Michele_Laino

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

5. anonymous

is that not E to the power of x?

6. Michele_Laino

yes! Is E the base of natural logarithms

7. anonymous

yes

8. anonymous

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

9. Michele_Laino

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

10. 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

11. Michele_Laino

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

12. Michele_Laino

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

13. 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.

14. 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

15. anonymous

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

16. Michele_Laino

that's right!

17. anonymous

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

18. anonymous

and we restrict the family.

19. Michele_Laino

yes! that's right!

20. anonymous

cool, thanks michele..

21. Michele_Laino

:)