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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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  1. MathDad72 Group Title
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    \[f(x) = \int\limits_{3}^{x} \sqrt{1+t ^{3}} dt\] find \[(f ^{-1})'(0)\]

    • 2 years ago
  2. MathDad72 Group Title
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    \[f(x) = \int\limits_{3}^{x} \sqrt{1+x ^{3}} dx\] \[f(x) = (1+x^3)^{1/2}\] \[f'(x) = \frac{ 1 }{ 2 }(1+x ^{3})^{^{-1/2}} (3x ^{2})\]

    • 2 years ago
  3. slaaibak Group Title
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    f'(x) = sqrt(1+x^3) so f'(0= 1

    • 2 years ago
  4. MathDad72 Group Title
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    is that setting up: putting the 0 from (f^-1)' \[0=(1+x^3)^{\frac{ 1 }{ 2 }}\]

    • 2 years ago
  5. slaaibak Group Title
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    no. so f'(0)= 1 meaning sqrt(1+0^3) = 1

    • 2 years ago
  6. MathDad72 Group Title
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    you just want to sent the equation equal to 0?

    • 2 years ago
  7. MathDad72 Group Title
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    oops set

    • 2 years ago
  8. slaaibak Group Title
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    bro, you asked me to find f'(0) and I found f'(0), what are you on about?

    • 2 years ago
  9. slaaibak Group Title
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    you know evaluating a function at zero and setting it equal to zero are two different things?

    • 2 years ago
  10. MathDad72 Group Title
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    correct

    • 2 years ago
  11. slaaibak Group Title
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    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
  12. MathDad72 Group Title
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    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
  13. MathDad72 Group Title
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    \[= \frac{ 1 }{ f'(1) }\]

    • 2 years ago
  14. slaaibak Group Title
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    oh! I completely overlooked the inverse over there, sorry

    • 2 years ago
  15. MathDad72 Group Title
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    yeah those lovely inverses

    • 2 years ago
  16. MathDad72 Group Title
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    do we need to back track some for the steps done for doing f'(0) or are we good?

    • 2 years ago
  17. slaaibak Group Title
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    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
  18. MathDad72 Group Title
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    ok

    • 2 years ago
  19. slaaibak Group Title
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    then f'(3) = sqrt(1+3^3) = sqrt(28)

    • 2 years ago
  20. MathDad72 Group Title
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    dang I didn't need to figure that ugly f'(x)?

    • 2 years ago
  21. slaaibak Group Title
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    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
  22. slaaibak Group Title
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    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
  23. MathDad72 Group Title
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    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
  24. slaaibak Group Title
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    haha, just remember the methods/formulas, do it step by step and you'll be fine

    • 2 years ago
  25. MathDad72 Group Title
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    Thanks for your help slaaibak. Have a great night.

    • 2 years ago
  26. slaaibak Group Title
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    No problem. Thanx, you too.

    • 2 years ago
  27. cobra661966 Group Title
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    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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