Find the slopes of the tangent and the normal to the curve `y=x^3-x` at `x=2`

Explanation

To find the slopes of the tangent and the normal to the curve \( y = x^3 - x \) at \( x = 2 \), we can follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y = x^3 - x \) with respect to \( x \) to find the slope of the tangent line.

\[ \frac{dy}{dx} = \frac{d}{dx}(x^3 - x) = 3x^2 - 1 \]

### Step 2: Evaluate the derivative at \( x = 2 \)
Next, we substitute \( x = 2 \) into the derivative to find the slope of the tangent line at that point.

\[ \frac{dy}{dx} \bigg|_{x=2} = 3(2^2) - 1 = 3(4) - 1 = 12 - 1 = 11 \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. If \( m_1 \) is the slope of the tangent, then the slope of the normal \( m_2 \) is given by:

\[ m_2 = -\frac{1}{m_1} = -\frac{1}{11} \]

### Summary of Results
- The slope of the tangent line at \( x = 2 \) is \( 11 \).
- The slope of the normal line at \( x = 2 \) is \( -\frac{1}{11} \).