Question:

Step 1: Find the point of tangency on the curve $y = g(x)$. Since $g(x)$ is the inverse function of $f(x)$, we have $f(g(x)) = x$ for all $x$ in the domain of $g$. To find $g(3)$, we need to find the value $a$ such that $f(a) = 3$. $$f(a) = 3$$ $$a^3 + a + 1 = 3$$ $$a^3 + a - 2 = 0$$ Factoring this cubic equation: $$(a-1)(a^2 + a + 2) = 0$$ Since $a^2 + a + 2$ has discriminant $1 - 8 = -7 < 0$, it has no real roots. Therefore, $a = 1$. Thus, $g(3) = 1$, and the point of tangency is $(3, 1)$. Step 2: Find the derivative of the inverse function. Using the inverse function derivative formula: $$g'(x) = \frac{1}{f'(g(x))}$$ First, compute $f'(x)$: $$f'(x) = 3x^2 + 1$$ Step 3: Calculate the slope of the tangent line at $x = 3$. Evaluate $f'$ at $g(3) = 1$: $$f'(g(3)) = f'(1) = 3(1)^2 + 1 = 4$$ Therefore, the slope of the tangent to $y = g(x)$ at $x = 3$ is: $$g'(3) = \frac{1}{f'(1)} = \frac{1}{4}$$ Step 4: Write the equation of the tangent line. Using the point-slope form with point $(3, 1)$ and slope $m = \frac{1}{4}$: $$y - 1 = \frac{1}{4}(x - 3)$$ Multiply both sides by 4: $$4y - 4 = x - 3$$ Rearrange to standard form: $$x - 4y + 1 = 0$$ **Final Answer:** The equation of the tangent to $y = g(x)$ at $x = 3$ is $x - 4y + 1 = 0$, which corresponds to **Option 1: (a)**.