The equation of the tangent to the curve y = | vedclass

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(C) Step $1$: Find the $y$-coordinate at $x = 1$. $y = 3(1)^3 - 3(1)^2 + 1 = 3 - 3 + 1 = 1$. The point of tangency is $(1, 1)$.Step $2$: Find the slop

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$4x - y - 3 = 0$

Vedclass · Answer

(C) Step $1$: Find the $y$-coordinate at $x = 1$. $y = 3(1)^3 - 3(1)^2 + 1 = 3 - 3 + 1 = 1$. The point of tangency is $(1, 1)$.Step $2$: Find the slope of the tangent by differentiating $y$ with respect to $x$. $\frac{dy}{dx} = 9x^2 - 6x + 1$.Step $3$: Evaluate the slope $m$ at $x = 1$. $m = 9(1)^2 - 6(1) + 1 = 9 - 6 + 1 = 4$.Step $4$: Use the point-slope form $y - y_1 = m(x - x_1)$ to find the equation. $y - 1 = 4(x - 1)$.Step $5$: Simplify the equation. $y - 1 = 4x - 4 \implies 4x - y - 3 = 0$.

The equation of the tangent to the curve $y = 3x^3 - 3x^2 + x$ at $x = 1$ is

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