1. **State the problem:** We are given two functions: $$g(x) = x^3 - 4x + 1$$ and $$L(x) = 8(x - 2) + 1$$ We want to verify that the line $L(x)$ is tangent to the curve $g(x)$ at the point $(2,1)$. 2. **Recall the tangent line conditions:** - The tangent line touches the curve at exactly one point. - At the point of tangency, the function values are equal: $$g(a) = L(a)$$. - The slopes are equal at that point: $$g'(a) = L'(a)$$. 3. **Check the point of tangency:** Evaluate $g(2)$: $$g(2) = 2^3 - 4(2) + 1 = 8 - 8 + 1 = 1$$ Evaluate $L(2)$: $$L(2) = 8(2 - 2) + 1 = 8(0) + 1 = 1$$ Since $g(2) = L(2) = 1$, the line passes through the point $(2,1)$ on the curve. 4. **Find the derivative of $g(x)$:** $$g'(x) = \frac{d}{dx}(x^3 - 4x + 1) = 3x^2 - 4$$ Evaluate $g'(2)$: $$g'(2) = 3(2)^2 - 4 = 3(4) - 4 = 12 - 4 = 8$$ 5. **Find the slope of $L(x)$:** Rewrite $L(x)$: $$L(x) = 8(x - 2) + 1 = 8x - 16 + 1 = 8x - 15$$ The slope of $L(x)$ is the coefficient of $x$, which is 8. 6. **Compare slopes:** Since $g'(2) = 8$ and the slope of $L(x)$ is also 8, the slopes are equal at $x=2$. 7. **Conclusion:** The line $L(x)$ passes through the point $(2,1)$ on the curve $g(x)$ and has the same slope as $g(x)$ at that point. Therefore, $L(x)$ is tangent to $g(x)$ at $(2,1)$. **Final answer:** The line $L(x) = 8(x - 2) + 1$ is tangent to the curve $g(x) = x^3 - 4x + 1$ at the point $(2,1)$.