## MathDad72 Group Title If f(x) = integral (3,x) sqr root (1+t^3) dt, find (f^1)'(0) one year ago one year ago

1. MathDad72 Group Title

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

2. MathDad72 Group Title

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

3. slaaibak Group Title

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

4. MathDad72 Group Title

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

5. slaaibak Group Title

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

6. MathDad72 Group Title

you just want to sent the equation equal to 0?

7. MathDad72 Group Title

oops set

8. slaaibak Group Title

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

9. slaaibak Group Title

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

10. MathDad72 Group Title

correct

11. slaaibak Group Title

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. MathDad72 Group Title

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. MathDad72 Group Title

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

14. slaaibak Group Title

oh! I completely overlooked the inverse over there, sorry

15. MathDad72 Group Title

yeah those lovely inverses

16. MathDad72 Group Title

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

17. slaaibak Group Title

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

18. MathDad72 Group Title

ok

19. slaaibak Group Title

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

20. MathDad72 Group Title

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

21. slaaibak Group Title

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 Group Title

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. MathDad72 Group Title

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 Group Title

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

25. MathDad72 Group Title

Thanks for your help slaaibak. Have a great night.

26. slaaibak Group Title

No problem. Thanx, you too.

27. cobra661966 Group Title

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