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 Question:
To find the tangent equation, we need the point \( (x_1, y_1) \) on the curve and the slope \( m = dy/dx \) at that point.
Step 2: Key Formula or Approach:
1. Slope \( m = f'(x) \).
2. Tangent: \( y - y_1 = m(x - x_1) \).
Step 3: Detailed Explanation:
At \( x = 1 \), \( y = 3(1)^3 - 3(1)^2 + 1 = 1 \). Point is \( (1, 1) \).
Differentiating:
\[ \frac{dy}{dx} = 9x^2 - 6x \]
At \( x = 1 \), \( m = 9(1) - 6(1) = 3 \).
Equation:
\[ y - 1 = 3(x - 1) \Rightarrow y = 3x - 3 + 1 \Rightarrow y = 3x - 2 \]
Step 4: Final Answer:
The equation is \( y = 3x - 2 \).