Here's the question you clicked on:
MathDad72
If f(x) = integral (3,x) sqr root (1+t^3) dt, find (f^1)'(0)
\[f(x) = \int\limits_{3}^{x} \sqrt{1+t ^{3}} dt\] find \[(f ^{-1})'(0)\]
\[f(x) = \int\limits_{3}^{x} \sqrt{1+x ^{3}} dx\] \[f(x) = (1+x^3)^{1/2}\] \[f'(x) = \frac{ 1 }{ 2 }(1+x ^{3})^{^{-1/2}} (3x ^{2})\]
f'(x) = sqrt(1+x^3) so f'(0= 1
is that setting up: putting the 0 from (f^-1)' \[0=(1+x^3)^{\frac{ 1 }{ 2 }}\]
no. so f'(0)= 1 meaning sqrt(1+0^3) = 1
you just want to sent the equation equal to 0?
bro, you asked me to find f'(0) and I found f'(0), what are you on about?
you know evaluating a function at zero and setting it equal to zero are two different things?
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?
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.
\[= \frac{ 1 }{ f'(1) }\]
oh! I completely overlooked the inverse over there, sorry
yeah those lovely inverses
do we need to back track some for the steps done for doing f'(0) or are we good?
so yeah, you solve 0 = f(x) then x=3 do you agree? Then f^-1(0) = 3 ? or alternatively, f(3) = 0
then f'(3) = sqrt(1+3^3) = sqrt(28)
dang I didn't need to figure that ugly f'(x)?
you did. Well, I did. \[f'(x) ={d \over dx} \int\limits_{3}^{x} \sqrt{1+t^3}dt = \sqrt{1+x^3}\]
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
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.
haha, just remember the methods/formulas, do it step by step and you'll be fine
Thanks for your help slaaibak. Have a great night.
No problem. Thanx, you too.
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