Question:

Step 1: Establish the relationship between $f$ and $g$. Since $g$ is the inverse function of $f$, we have the fundamental property: $$f(g(x)) = x$$ Step 2: Find the first derivative of $g$ using implicit differentiation. Differentiating both sides of $f(g(x)) = x$ with respect to $x$: $$f'(g(x)) \cdot g'(x) = 1$$ Therefore: $$g'(x) = \frac{1}{f'(g(x))}$$ Step 3: Determine the value of $g(5)$. We need to find what value of $x$ satisfies $f(x) = 5$. Given $f(x) = x^3 + 3x + 1$, let's evaluate: $$f(1) = 1^3 + 3(1) + 1 = 1 + 3 + 1 = 5$$ Therefore: $$g(5) = 1$$ Step 4: Calculate $f'(x)$ and find $f'(1)$. Taking the derivative of $f(x) = x^3 + 3x + 1$: $$f'(x) = 3x^2 + 3$$ At $x = 1$: $$f'(1) = 3(1)^2 + 3 = 3 + 3 = 6$$ Step 5: Find $g'(5)$. Using the formula from Step 2: $$g'(5) = \frac{1}{f'(g(5))} = \frac{1}{f'(1)} = \frac{1}{6}$$ Step 6: Find the second derivative of $g$ using implicit differentiation. Differentiating $g'(x) = \frac{1}{f'(g(x))}$ with respect to $x$: $$g''(x) = -\frac{f''(g(x)) \cdot g'(x)}{[f'(g(x))]^2}$$ Step 7: Calculate $f''(x)$ and find $f''(1)$. Taking the second derivative of $f(x) = x^3 + 3x + 1$: $$f''(x) = 6x$$ At $x = 1$: $$f''(1) = 6(1) = 6$$ Step 8: Calculate $g''(5)$. Substituting the values we found into the formula from Step 6: $$g''(5) = -\frac{f''(g(5)) \cdot g'(5)}{[f'(g(5))]^2}$$ $$g''(5) = -\frac{f''(1) \cdot g'(5)}{[f'(1)]^2}$$ $$g''(5) = -\frac{6 \cdot \frac{1}{6}}{6^2}$$ $$g''(5) = -\frac{1}{36}$$ **Final Answer:** The value of $g''(5) = \dfrac{-1}{36}$ This corresponds to **Option 4**.