Find F inverse prime at a=4. f (x) = 2x^3+3x^2+7x+4.

Which is derived because luckily when you plug 0 into f (x):

f (0) = 4, meaning that f^-1 (4) = 0, which means that f^-1 (a) = 0.

From there you can use the Theorem (f^-1) (a) = 1 / f (f^-1 (a))

f^-1 (a) = 0

(f^-1) (a) = 1 / f (0)

f = 6x^2 + 6x + 7

f (0) = 7

(f^-1) (a) = 1/7