Find the slope of the normal to the curve :
`y=x^(3)-x+1" at "x=2`

Answer

To find the slope of the normal to the curve \( y = x^3 - x + 1 \) at \( x = 2 \), we will follow these steps:

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

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

Calculating the derivatives, we get:

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

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

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

### 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 \( m_t \) is the slope of the tangent, then the slope of the normal \( m_n \) is given by:

\[ m_n = -\frac{1}{m_t} \]

Substituting the value of \( m_t \):

\[ m_n = -\frac{1}{11} \]

### Final Answer
Thus, the slope of the normal to the curve at \( x = 2 \) is:

\[ \boxed{-\frac{1}{11}} \]

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