Find f. f '(t) = 8 / 1 + t² , f(1) = 0

Solution:

Inverse function is represented by f^-1 with regards to the original function f and the domain of the original function becomes the range of inverse function and the range of the given function becomes the domain of the inverse function. 

Given f '(t) = 8 / 1 + t² and f(1) = 0

We know that by definition of integral of tan⁻¹(x) = 1/1 + t²

f '(t) = 8 / 1 + t² = 8(1/ 1 + t²)

Integrate on both sides, we get

f(t) = 8 tan⁻¹(t) + c

f(1) = 0 ;

f(1) = 8 tan⁻¹(1) + c = 0

c = -8 (π/4) = -2π.

Therefore, f(x) = 8 tan⁻¹(t) - 2π.

Find f. f '(t) = 8 / 1 + t² , f(1) = 0

Summary:

f '(t) = 8 / 1 + t² , f(1) = 0. The function f(x) is 8tan⁻¹(t) - 2π