The value of g′(0) is −1/3, option (A) is correct.
To find g′(0), we start by recognizing that g(x) is the inverse function of f(x). Since f(−1) = 0, it implies that g(0) = −1 because the inverse of f(−1) = 0 is g(0) = −1. We want to find the derivative of g(x) at x = 0, which is g′(0).
Using the inverse function rule,
g′(x) = 1/f′(g(x)),
we substitute x = 0 to get
g′(0) = 1/f′(g(0))1.
To find f′(g(0)), we find the derivative of f(x) and substitute x = −1 (since g(0) = −1).
f'(x) = 3x² + 6x {derivative of f(x)}
putting x = -1 we have;
f'(-1) = 3(-1)² + 6(-1)
f'(-1) = -3 - 6
f'(-1) = -3
Therefore, g′(0) = 1/f′(−1) = 1/−3
g′(0) = -1/3. Hence, the value of g′(0) iis-1/3.