If the function f(x) = x³ + ex/2 and g(x) = f-1(x), then the value of g'(1) is

g(x) is the inverse of f(x). Hence g(f(x)) = x ⇒ g'(f(x)).f'(x) = 1 ⇒ g'(f(x)) = 1/f'(x)
We need to find out the value of x such that f(x) = 1. Which is when x = 0
So g'(1) = 1/f'(0)
f'(x) = 3x² + ex/2
f'(0) = 0 + e0/2 = 1/2
So g'(1) = 2