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anonymous
 5 years ago
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?
anonymous
 5 years ago
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?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and...what is (f^1)'(3)? Soooooo confusing!!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(x) = (e^x) + (4x^1/3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How did you find that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integration 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
 5 years ago
Best ResponseYou've already chosen the best response.0So, you just simply integrate the original equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f'(x) means first derivative from f(x) to find f(x) you have to integrate it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When you integrate don't you get e^x+4x^(1/3)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright and then that is just simply f(x) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(1) means change x with 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. How would I exactly find the inverse of f(x)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Follow 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=5x3 The inverse of f(x) is f^1(x) = 5x3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got this far...x=e^y+4y^(1/3) and now I'm stuck.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0log both side log x = y + 1/3 log 4y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then you get lnx=y+(1/3)ln4+(1/3)lny?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isn't it ln x = y + 1/3 ln 4y, i dont understand this, confused too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, I was thinking that you would add because usually when you have a product in a ln then that means you add. right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i made mistake, for example 2 ln 2y = ln 4y^2 so 4y^1/3 = 1/3 * 4 ln y
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