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MathDad72Best ResponseYou've already chosen the best response.0
\[f(x) = \int\limits_{3}^{x} \sqrt{1+t ^{3}} dt\] find \[(f ^{1})'(0)\]
 one year ago

MathDad72Best ResponseYou've already chosen the best response.0
\[f(x) = \int\limits_{3}^{x} \sqrt{1+x ^{3}} dx\] \[f(x) = (1+x^3)^{1/2}\] \[f'(x) = \frac{ 1 }{ 2 }(1+x ^{3})^{^{1/2}} (3x ^{2})\]
 one year ago

slaaibakBest ResponseYou've already chosen the best response.3
f'(x) = sqrt(1+x^3) so f'(0= 1
 one year ago

MathDad72Best ResponseYou've already chosen the best response.0
is that setting up: putting the 0 from (f^1)' \[0=(1+x^3)^{\frac{ 1 }{ 2 }}\]
 one year ago

slaaibakBest ResponseYou've already chosen the best response.3
no. so f'(0)= 1 meaning sqrt(1+0^3) = 1
 one year ago

MathDad72Best ResponseYou've already chosen the best response.0
you just want to sent the equation equal to 0?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.3
bro, you asked me to find f'(0) and I found f'(0), what are you on about?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.3
you know evaluating a function at zero and setting it equal to zero are two different things?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.3
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?
 one year ago

MathDad72Best ResponseYou've already chosen the best response.0
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.
 one year ago

MathDad72Best ResponseYou've already chosen the best response.0
\[= \frac{ 1 }{ f'(1) }\]
 one year ago

slaaibakBest ResponseYou've already chosen the best response.3
oh! I completely overlooked the inverse over there, sorry
 one year ago

MathDad72Best ResponseYou've already chosen the best response.0
yeah those lovely inverses
 one year ago

MathDad72Best ResponseYou've already chosen the best response.0
do we need to back track some for the steps done for doing f'(0) or are we good?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.3
so yeah, you solve 0 = f(x) then x=3 do you agree? Then f^1(0) = 3 ? or alternatively, f(3) = 0
 one year ago

slaaibakBest ResponseYou've already chosen the best response.3
then f'(3) = sqrt(1+3^3) = sqrt(28)
 one year ago

MathDad72Best ResponseYou've already chosen the best response.0
dang I didn't need to figure that ugly f'(x)?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.3
you did. Well, I did. \[f'(x) ={d \over dx} \int\limits_{3}^{x} \sqrt{1+t^3}dt = \sqrt{1+x^3}\]
 one year ago

slaaibakBest ResponseYou've already chosen the best response.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
 one year ago

MathDad72Best ResponseYou've already chosen the best response.0
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.
 one year ago

slaaibakBest ResponseYou've already chosen the best response.3
haha, just remember the methods/formulas, do it step by step and you'll be fine
 one year ago

MathDad72Best ResponseYou've already chosen the best response.0
Thanks for your help slaaibak. Have a great night.
 one year ago

slaaibakBest ResponseYou've already chosen the best response.3
No problem. Thanx, you too.
 one year ago

cobra661966Best ResponseYou've already chosen the best response.0
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
 7 months ago
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