## anonymous 3 years ago If f(x) = integral (3,x) sqr root (1+t^3) dt, find (f^1)'(0)

1. anonymous

$f(x) = \int\limits_{3}^{x} \sqrt{1+t ^{3}} dt$ find $(f ^{-1})'(0)$

2. anonymous

$f(x) = \int\limits_{3}^{x} \sqrt{1+x ^{3}} dx$ $f(x) = (1+x^3)^{1/2}$ $f'(x) = \frac{ 1 }{ 2 }(1+x ^{3})^{^{-1/2}} (3x ^{2})$

3. slaaibak

f'(x) = sqrt(1+x^3) so f'(0= 1

4. anonymous

is that setting up: putting the 0 from (f^-1)' $0=(1+x^3)^{\frac{ 1 }{ 2 }}$

5. slaaibak

no. so f'(0)= 1 meaning sqrt(1+0^3) = 1

6. anonymous

you just want to sent the equation equal to 0?

7. anonymous

oops set

8. slaaibak

bro, you asked me to find f'(0) and I found f'(0), what are you on about?

9. slaaibak

you know evaluating a function at zero and setting it equal to zero are two different things?

10. anonymous

correct

11. slaaibak

so you gave f(x). I found f'(x) and then I evaluated it at zero (f'(0)) That's the question, isn't it?

12. anonymous

Yes we evaluated f'(x) at zero and we ant to put that into $(f ^{-1})'(0) = \frac{ 1 }{ f'(f ^{-1}(0) }$ Correct? Or am I too tired and my brain is done for the day.

13. anonymous

$= \frac{ 1 }{ f'(1) }$

14. slaaibak

oh! I completely overlooked the inverse over there, sorry

15. anonymous

yeah those lovely inverses

16. anonymous

do we need to back track some for the steps done for doing f'(0) or are we good?

17. slaaibak

so yeah, you solve 0 = f(x) then x=3 do you agree? Then f^-1(0) = 3 ? or alternatively, f(3) = 0

18. anonymous

ok

19. slaaibak

then f'(3) = sqrt(1+3^3) = sqrt(28)

20. anonymous

dang I didn't need to figure that ugly f'(x)?

21. slaaibak

you did. Well, I did. $f'(x) ={d \over dx} \int\limits_{3}^{x} \sqrt{1+t^3}dt = \sqrt{1+x^3}$

22. slaaibak

it's kinda obvious if you think about it. you should get the same answer back, since you are integrating and then differentiating again. BUT there is a chain rule adjustment that has to be made if there are more complex bounds

23. anonymous

I wish these derivative inverses made more sense to me. I am pretty confident in my inverse solving. But these inverse derivatives are killing me.

24. slaaibak

haha, just remember the methods/formulas, do it step by step and you'll be fine

25. anonymous

Thanks for your help slaaibak. Have a great night.

26. slaaibak

No problem. Thanx, you too.

27. anonymous

ok can some one work this problem for me from 1st step to last step. I see you plug x in for t. and 3 in for x but what about the (f^-1)(0) what does that fit in thanks