1. **Problem statement:** Given the function $f(x) = x^3 - 3x^2 + 3$, (a) Find the equation of the tangent line at the inflection point. (b) Calculate the area of the triangular region enclosed by this tangent line and the coordinate axes. 2. **Find the inflection point:** The inflection point occurs where the second derivative $f''(x)$ changes sign. Calculate first derivative: $$f'(x) = 3x^2 - 6x$$ Calculate second derivative: $$f''(x) = 6x - 6$$ Set $f''(x) = 0$ to find inflection point: $$6x - 6 = 0$$ $$6x = 6$$ $$x = 1$$ Find $y$ coordinate at $x=1$: $$f(1) = 1^3 - 3(1)^2 + 3 = 1 - 3 + 3 = 1$$ So the inflection point is $W(1,1)$. 3. **Find the slope of the tangent at $W$:** Evaluate $f'(1)$: $$f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3$$ 4. **Equation of the tangent line at $W$:** Use point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 1 = -3(x - 1)$$ $$y - 1 = -3x + 3$$ $$y = -3x + 4$$ 5. **Find intercepts of the tangent line:** - $y$-intercept: set $x=0$ $$y = -3(0) + 4 = 4$$ - $x$-intercept: set $y=0$ $$0 = -3x + 4$$ $$3x = 4$$ $$x = \frac{4}{3}$$ 6. **Calculate the area of the triangle formed by the tangent and axes:** The triangle has base on the $x$-axis from $0$ to $\frac{4}{3}$ and height on the $y$-axis from $0$ to $4$. Area formula: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ Substitute values: $$\text{Area} = \frac{1}{2} \times \frac{4}{3} \times 4 = \frac{1}{2} \times \frac{16}{3} = \frac{16}{6} = \frac{8}{3}$$ **Final answers:** (a) Tangent line equation at inflection point: $$y = -3x + 4$$ (b) Area of the triangle enclosed by the tangent and coordinate axes: $$\frac{8}{3}$$