anonymous
  • anonymous
Given the derivative of some inegrable function f(x) is f'(x)=e^x+(4/3)x^(-2/3). What is f(x) if f(1)=e?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
and...what is (f^-1)'(3)? Soooooo confusing!!!!
anonymous
  • anonymous
f(x) = (e^x) + (4x^1/3)
anonymous
  • anonymous
How did you find that?

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More answers

anonymous
  • anonymous
integration and derivated from e^x is e^x integration from (4/3)x^(-2/3) is (-2/3 +1)=1/3 then (4/3)/(1/3)= 4x^1/3
anonymous
  • anonymous
hmm...
anonymous
  • anonymous
So, you just simply integrate the original equation?
anonymous
  • anonymous
f'(x) means first derivative from f(x) to find f(x) you have to integrate it
anonymous
  • anonymous
When you integrate don't you get e^x+4x^(1/3)?
anonymous
  • anonymous
yup
anonymous
  • anonymous
alright and then that is just simply f(x) right?
anonymous
  • anonymous
f(1) means change x with 1
anonymous
  • anonymous
Okay. How would I exactly find the inverse of f(x)?
anonymous
  • anonymous
Follow this example. f(x) = (x+3)/5 To find its inverse, write y=f(x) y= (x+3)/5 Interchange x and y x = (y+3)/5 solve for y in terms of x 5x=y+3 y=5x-3 The inverse of f(x) is f^-1(x) = 5x-3
anonymous
  • anonymous
I got this far...x=e^y+4y^(1/3) and now I'm stuck.
anonymous
  • anonymous
log both side log x = y + 1/3 log 4y
anonymous
  • anonymous
Then you get lnx=y+(1/3)ln4+(1/3)lny?
anonymous
  • anonymous
isn't it ln x = y + 1/3 ln 4y, i dont understand this, confused too
anonymous
  • anonymous
Well, I was thinking that you would add because usually when you have a product in a ln then that means you add. right?
anonymous
  • anonymous
i made mistake, for example 2 ln 2y = ln 4y^2 so 4y^1/3 = 1/3 * 4 ln y
anonymous
  • anonymous
Okay, thank you.

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