Question

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?

Answer 1

and...what is (f^-1)'(3)? Soooooo confusing!!!!

Answer 2

f(x) = (e^x) + (4x^1/3)

Answer 3

How did you find that?

Answer 4

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

Answer 5

hmm...

Answer 6

So, you just simply integrate the original equation?

Answer 7

f'(x) means first derivative from f(x)
to find f(x) you have to integrate it

Answer 8

When you integrate don't you get e^x+4x^(1/3)?

Answer 9

yup

Answer 10

alright and then that is just simply f(x) right?

Answer 11

f(1) means change x with 1

Answer 12

Okay. How would I exactly find the inverse of f(x)?

Answer 13

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

Answer 14

I got this far...x=e^y+4y^(1/3) and now I'm stuck.

Answer 15

log both side
log x = y + 1/3 log 4y

Answer 16

Then you get lnx=y+(1/3)ln4+(1/3)lny?

Answer 17

isn't it ln x = y + 1/3 ln 4y, i dont understand this, confused too

Answer 18

Well, I was thinking that you would add because usually when you have a product in a ln then that means you add. right?

Answer 19

i made mistake, for example 2 ln 2y = ln 4y^2
so 4y^1/3 = 1/3 * 4 ln y

Answer 20

Okay, thank you.