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MathDad72
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If f(x) = integral (3,x) sqr root (1+t^3) dt, find (f^1)'(0)
 2 years ago
 2 years ago
MathDad72 Group Title
If f(x) = integral (3,x) sqr root (1+t^3) dt, find (f^1)'(0)
 2 years ago
 2 years ago

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MathDad72 Group TitleBest ResponseYou've already chosen the best response.0
\[f(x) = \int\limits_{3}^{x} \sqrt{1+t ^{3}} dt\] find \[(f ^{1})'(0)\]
 2 years ago

MathDad72 Group TitleBest ResponseYou've already chosen the best response.0
\[f(x) = \int\limits_{3}^{x} \sqrt{1+x ^{3}} dx\] \[f(x) = (1+x^3)^{1/2}\] \[f'(x) = \frac{ 1 }{ 2 }(1+x ^{3})^{^{1/2}} (3x ^{2})\]
 2 years ago

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

MathDad72 Group TitleBest ResponseYou've already chosen the best response.0
is that setting up: putting the 0 from (f^1)' \[0=(1+x^3)^{\frac{ 1 }{ 2 }}\]
 2 years ago

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

MathDad72 Group TitleBest ResponseYou've already chosen the best response.0
you just want to sent the equation equal to 0?
 2 years ago

slaaibak Group TitleBest 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?
 2 years ago

slaaibak Group TitleBest 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?
 2 years ago

slaaibak Group TitleBest 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?
 2 years ago

MathDad72 Group TitleBest 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.
 2 years ago

MathDad72 Group TitleBest ResponseYou've already chosen the best response.0
\[= \frac{ 1 }{ f'(1) }\]
 2 years ago

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

MathDad72 Group TitleBest ResponseYou've already chosen the best response.0
yeah those lovely inverses
 2 years ago

MathDad72 Group TitleBest 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?
 2 years ago

slaaibak Group TitleBest 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
 2 years ago

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

MathDad72 Group TitleBest ResponseYou've already chosen the best response.0
dang I didn't need to figure that ugly f'(x)?
 2 years ago

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

slaaibak Group TitleBest 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
 2 years ago

MathDad72 Group TitleBest 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.
 2 years ago

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

MathDad72 Group TitleBest ResponseYou've already chosen the best response.0
Thanks for your help slaaibak. Have a great night.
 2 years ago

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

cobra661966 Group TitleBest 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
 one year ago
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