Find the slope of the tangent to the curve `y=x^(3)-x+1` at the point whose x - coordinate is 3.

Answer

To find the slope of the tangent to the curve \( y = x^3 - x + 1 \) at the point where the x-coordinate is 3, we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y = x^3 - x + 1 \) with respect to \( x \) 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 each term:
- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( x \) is \( 1 \).
- The derivative of a constant (1) is \( 0 \).

So, we have:

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

### Step 2: Substitute \( x = 3 \)
Next, we substitute \( x = 3 \) into the derivative to find the slope at that point.

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

Calculating \( 3^2 \):

\[ 3^2 = 9 \]

Now substituting back:

\[ \frac{dy}{dx} \bigg|_{x=3} = 3(9) - 1 = 27 - 1 = 26 \]

### Step 3: Conclusion
Thus, the slope of the tangent to the curve at the point where \( x = 3 \) is:

\[ \text{slope} = 26 \]

### Final Answer
The slope of the tangent to the curve \( y = x^3 - x + 1 \) at the point where \( x = 3 \) is \( 26 \).

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