## ksaimouli 2 years ago find g(-1) and g''(-1)

1. ksaimouli

|dw:1356718690979:dw|

2. ksaimouli

$g(x)=\int\limits_{0}^{x}f(t) dt$

3. ksaimouli

graph above is f

4. ksaimouli

|dw:1356718836761:dw|

5. ksaimouli

find g(-1)

6. ksaimouli

@hartnn

7. hartnn

can u find the point of intersection of those 2 lines ? or, equation of those lines ??

8. mathmate

Is g(x) a constant? I don't see x in the definition of g(x). Does f(x) = 0 for x>2, or does it decrease forever?

9. ksaimouli

@hba

10. ksaimouli

i know that f'(-1)=0

11. ksaimouli

what i have no idea how to find the f(-1)

12. hartnn

|dw:1356720077842:dw|

13. ksaimouli

slope is 3

14. hba

Now you can find the equation :)

15. hartnn

and u also have point , so equation is ?

16. ksaimouli

y=3x+3

17. hartnn

x=-1 thats your f(t) for 0 to -1

18. hartnn

$$g(-1)=\int\limits_{0}^{-1}(3x+3) dt$$

19. hartnn

can u find g(-1) now

20. ksaimouli

yes

21. hartnn

good :)

22. hba

-(3x+3)

23. ksaimouli

-3/2

24. ksaimouli

25. hartnn

26. ksaimouli

yes

27. ksaimouli

|dw:1356720626985:dw|

28. hartnn

ok, $$g'(x)=(d/dx)\int_0^xf(t)dt = f(x)\\so,g''(x)=f'(x)\\g"(-1)=f(-1)$$ can u find f(-1) ?? and no

29. ksaimouli

why not

30. hartnn

why ?

31. hartnn

why u want to use that line ?? f(-1) is simple..

32. ksaimouli

ok

33. ksaimouli

0

34. hartnn

yes, correct.