anonymous
  • anonymous
If f(x) = integral (3,x) sqr root (1+t^3) dt, find (f^1)'(0)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[f(x) = \int\limits_{3}^{x} \sqrt{1+t ^{3}} dt\] find \[(f ^{-1})'(0)\]
anonymous
  • anonymous
\[f(x) = \int\limits_{3}^{x} \sqrt{1+x ^{3}} dx\] \[f(x) = (1+x^3)^{1/2}\] \[f'(x) = \frac{ 1 }{ 2 }(1+x ^{3})^{^{-1/2}} (3x ^{2})\]
slaaibak
  • slaaibak
f'(x) = sqrt(1+x^3) so f'(0= 1

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anonymous
  • anonymous
is that setting up: putting the 0 from (f^-1)' \[0=(1+x^3)^{\frac{ 1 }{ 2 }}\]
slaaibak
  • slaaibak
no. so f'(0)= 1 meaning sqrt(1+0^3) = 1
anonymous
  • anonymous
you just want to sent the equation equal to 0?
anonymous
  • anonymous
oops set
slaaibak
  • slaaibak
bro, you asked me to find f'(0) and I found f'(0), what are you on about?
slaaibak
  • slaaibak
you know evaluating a function at zero and setting it equal to zero are two different things?
anonymous
  • anonymous
correct
slaaibak
  • 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?
anonymous
  • 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.
anonymous
  • anonymous
\[= \frac{ 1 }{ f'(1) }\]
slaaibak
  • slaaibak
oh! I completely overlooked the inverse over there, sorry
anonymous
  • anonymous
yeah those lovely inverses
anonymous
  • anonymous
do we need to back track some for the steps done for doing f'(0) or are we good?
slaaibak
  • slaaibak
so yeah, you solve 0 = f(x) then x=3 do you agree? Then f^-1(0) = 3 ? or alternatively, f(3) = 0
anonymous
  • anonymous
ok
slaaibak
  • slaaibak
then f'(3) = sqrt(1+3^3) = sqrt(28)
anonymous
  • anonymous
dang I didn't need to figure that ugly f'(x)?
slaaibak
  • slaaibak
you did. Well, I did. \[f'(x) ={d \over dx} \int\limits_{3}^{x} \sqrt{1+t^3}dt = \sqrt{1+x^3}\]
slaaibak
  • 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
anonymous
  • 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.
slaaibak
  • slaaibak
haha, just remember the methods/formulas, do it step by step and you'll be fine
anonymous
  • anonymous
Thanks for your help slaaibak. Have a great night.
slaaibak
  • slaaibak
No problem. Thanx, you too.
anonymous
  • 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

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