1. We are given the function $$f(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5}$$ and its inverse function $$g(x) = f^{-1}(x)$$. We want to find the value of $$g''(0)$$, the second derivative of $$g$$ at $$0$$. 2. Recall that if $$y = f(x)$$ and $$x = g(y)$$ are inverse functions, they satisfy the relation: $$ f(g(x)) = x $$. 3. Differentiating both sides with respect to $$x$$, we get: $$ f'(g(x)) \cdot g'(x) = 1 $$. Evaluating at $$x=0$$: $$ f'(g(0)) \cdot g'(0) = 1 $$. Since $$g(0)$$ is such that $$f(g(0))=0$$, and $$f(0) = 0$$, so $$g(0) = 0$$. 4. Compute $$f'(x)$$: $$ f'(x) = 1 + x + x^2 + x^3 + x^4 $$. Therefore, $$f'(0) = 1$$. 5. From step 3: $$ 1 \cdot g'(0) = 1 \Rightarrow g'(0) = 1 $$. 6. Differentiate $$ f'(g(x)) \cdot g'(x) = 1 $$ again using product and chain rules: $$ f''(g(x)) \cdot (g'(x))^2 + f'(g(x)) \cdot g''(x) = 0 $$. Plug in $$x=0$$: $$ f''(0) \cdot (g'(0))^2 + f'(0) \cdot g''(0) = 0 $$. 7. Compute $$f''(x)$$: $$ f''(x) = 0 + 1 + 2x + 3x^2 + 4x^3 $$, thus $$f''(0) = 1$$. 8. Substitute values: $$ 1 \cdot 1^2 + 1 \cdot g''(0) = 0 \Rightarrow 1 + g''(0) = 0 $$, which gives $$ g''(0) = -1 $$. Final answer: $$\boxed{-1}$$