# Find f. f'(x)= 3/(1-x^2)^(1/2)) , f(1/2)=9 Find f. f'(x)= 3/(1-x^2)^(1/2)) , f(1/2)=9   f(x)=_______________   I found; sin^-1 (x) +9 -pi/6 but it is wrong ????

You need to perform the reverse operation to differentiation, hence you need to integrate the function f'(x) such that:

`f(x) = int f'(x) dx`

`f(x) = int 3/(sqrt(1-x^2))dx`

`f(x) = 3int (dx)/(sqrt(1-x^2))`

`f(x) = 3 sin^(-1) x + c `

You need to find the constant term c, hence you need to substitute `1/2`  for x in equation of function such that:

`f(1/2) = 9 =gt 3 sin^(-1)(1/2) + c = 9`

`c = 9 - 3sin^(-1)(1/2)`

`c = 9 - 3pi/6 =gt c = 9 - pi/2`

`f(x) = 3 sin^(-1) x + 9 - pi/2`

Hence, evaluating the equation of function f(x) under given conditions yields `f(x) = 3 sin^(-1) x + 9 - pi/2` .

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