Question

For the curve \(y = 3x^3 - 3x^2 + 1\) at \(x = 1\), find the equation of the tangent.

Hint:

Always verify your tangent equation by substituting the point \((x_1, y_1)\) back into it. For \(3x - y - 2 = 0\), \(3(1) - 1 - 2 = 0\). If it satisfies the equation, your answer is likely correct!

Answer 1

Step 1: Understanding the Concept:
The equation of a tangent to a curve at a specific point \((x_1, y_1)\) is found using the point-slope form: \(y - y_1 = m(x - x_1)\).
The slope \(m\) is the value of the first derivative \(\frac{dy}{dx}\) evaluated at the given \(x\)-coordinate.

Step 2: Key Formula or Approach:
1. Find the y-coordinate: Substitute \(x=1\) into the curve equation.
2. Find the slope: Differentiate the curve and evaluate at \(x=1\).
3. Use the line equation formula: \(y - y_1 = m(x - x_1)\).

Step 3: Detailed Explanation:
1. Calculate the point of tangency \((x_1, y_1)\):
At \(x = 1\):
\[ y = 3(1)^3 - 3(1)^2 + 1 = 3 - 3 + 1 = 1 \] So, the point is \((1, 1)\).
2. Calculate the slope (\(m\)):
Differentiating \(y = 3x^3 - 3x^2 + 1\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}(3x^3) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(1) \] \[ \frac{dy}{dx} = 9x^2 - 6x \] At \(x = 1\):
\[ m = \left[ \frac{dy}{dx} \right]_{x=1} = 9(1)^2 - 6(1) = 3 \] 3. Form the equation of the tangent:
Using the point \((1, 1)\) and slope \(m = 3\):
\[ y - 1 = 3(x - 1) \] \[ y - 1 = 3x - 3 \] \[ 3x - y - 2 = 0 \]
Step 4: Final Answer:
The equation of the tangent is \(3x - y - 2 = 0\).