Question

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

Answer 1

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

Answer 2

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

Answer 3

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

so f'(0= 1

Answer 4

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

Answer 5

no.

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

Answer 6

you just want to sent the equation equal to 0?

Answer 7

oops set

Answer 8

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

Answer 9

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

Answer 10

correct

Answer 11

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?

Answer 12

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.

Answer 13

\[= \frac{ 1 }{ f'(1) }\]

Answer 14

oh! I completely overlooked the inverse over there, sorry

Answer 15

yeah those lovely inverses

Answer 16

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

Answer 17

so yeah, you solve 0 = f(x)

then x=3

do you agree? Then f^-1(0) = 3 ? or alternatively, f(3) = 0

Answer 18

ok

Answer 19

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

Answer 20

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

Answer 21

you did. Well, I did.

\[f'(x) ={d \over dx} \int\limits_{3}^{x} \sqrt{1+t^3}dt = \sqrt{1+x^3}\]

Answer 22

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

Answer 23

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.

Answer 24

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

Answer 25

Thanks for your help slaaibak.
Have a great night.

Answer 26

No problem. Thanx, you too.

Answer 27

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