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

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

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

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.3f'(x) = sqrt(1+x^3) so f'(0= 1

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

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.3no. so f'(0)= 1 meaning sqrt(1+0^3) = 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you just want to sent the equation equal to 0?

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

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

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.3so 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
 3 years ago
Best ResponseYou've already chosen the best response.0Yes 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[= \frac{ 1 }{ f'(1) }\]

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.3oh! I completely overlooked the inverse over there, sorry

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah those lovely inverses

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

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

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.3then f'(3) = sqrt(1+3^3) = sqrt(28)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dang I didn't need to figure that ugly f'(x)?

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

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.3it'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
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.3haha, just remember the methods/formulas, do it step by step and you'll be fine

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks for your help slaaibak. Have a great night.

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.3No problem. Thanx, you too.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ok 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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