The slope of the curve `y = x^(3) - 2x+1` at point x = 1 is equal to :

Options

A. 0
B. 1
C. 2
D. 3

Explanation

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

### Step 1: Write down the function
The function given is:
\[ y = x^3 - 2x + 1 \]

### Step 2: Differentiate the function
To find the slope of the curve, we need to differentiate the function with respect to \( x \). The derivative \( \frac{dy}{dx} \) is calculated as follows:
\[ \frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(2x) + \frac{d}{dx}(1) \]
Calculating each term:
- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( -2x \) is \( -2 \).
- The derivative of a constant (1) is \( 0 \).

Combining these results, we get:
\[ \frac{dy}{dx} = 3x^2 - 2 \]

### Step 3: Substitute \( x = 1 \) into the derivative
Now, we will find the slope at the point where \( x = 1 \):
\[ \frac{dy}{dx} \bigg|_{x=1} = 3(1)^2 - 2 \]
Calculating this gives:
\[ \frac{dy}{dx} \bigg|_{x=1} = 3 \cdot 1 - 2 = 3 - 2 = 1 \]

### Conclusion
The slope of the curve at the point \( x = 1 \) is:
\[ \text{slope} = 1 \]