The slope of the tangent line to the curve `y=x^3-2x+1` at x=1 is

Options

A. 1
B. `1/2`
C. `1/3`
D. `1/4`

Answer

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

### Step 1: Differentiate the function
We start with the function:
\[ y = x^3 - 2x + 1 \]
To find the slope of the tangent line, we need to calculate the first derivative of \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(2x) + \frac{d}{dx}(1) \]

### Step 2: Apply the power rule
Using the power rule for differentiation:
- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( -2x \) is \( -2 \).
- The derivative of a constant (1) is \( 0 \).

Thus, we have:
\[ \frac{dy}{dx} = 3x^2 - 2 \]

### Step 3: Evaluate the derivative at \( x = 1 \)
Now, we need to find the slope at the specific point \( x = 1 \):
\[ \frac{dy}{dx} \bigg|_{x=1} = 3(1)^2 - 2 \]
Calculating this gives:
\[ \frac{dy}{dx} \bigg|_{x=1} = 3(1) - 2 = 3 - 2 = 1 \]

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

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