1. **State the problem:** We are given the function $$h(x) = x^3 - 6x^2 + 7x + 6$$ and asked to find: (i) The derivative $$h'(x)$$. (ii) The slope of the tangent line to the curve at $$x=1$$. 2. **Recall the derivative rules:** - The derivative of $$x^n$$ is $$nx^{n-1}$$. - The derivative of a constant is 0. - The derivative of a sum is the sum of the derivatives. 3. **Find the derivative $$h'(x)$$:** $$h'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(6x^2) + \frac{d}{dx}(7x) + \frac{d}{dx}(6)$$ Using the power rule: $$h'(x) = 3x^2 - 6 \cdot 2x + 7 + 0 = 3x^2 - 12x + 7$$ 4. **Evaluate the slope at $$x=1$$:** Substitute $$x=1$$ into $$h'(x)$$: $$h'(1) = 3(1)^2 - 12(1) + 7 = 3 - 12 + 7$$ Simplify: $$h'(1) = (3 - 12) + 7 = -9 + 7 = -2$$ **Final answer:** (i) $$h'(x) = 3x^2 - 12x + 7$$ (ii) The slope of the tangent line at $$x=1$$ is $$-2$$.