If the tangent to a curve y `= x^(3) -x ` at x =2 is parallel to the line y = mx -19 then the value of m is

Options

A. 11
B. `(19)/(2)`
C. 6
D. `-19`

Answer

To solve the problem, we need to find the value of \( m \) such that the tangent to the curve \( y = x^3 - x \) at \( x = 2 \) is parallel to the line \( y = mx - 19 \).

### Step-by-Step Solution:

1. **Identify the curve**:
The given curve is \( y = x^3 - x \).

2. **Find the derivative**:
To find the slope of the tangent to the curve, we need to compute the derivative \( \frac{dy}{dx} \).
\[ \frac{dy}{dx} = 3x^2 - 1 \]

3. **Evaluate the derivative at \( x = 2 \)**:
Now, we will substitute \( x = 2 \) into the derivative to find the slope of the tangent at that point.
\[ \frac{dy}{dx} \bigg|_{x=2} = 3(2^2) - 1 = 3(4) - 1 = 12 - 1 = 11 \]

4. **Identify the slope of the line**:
The line given is \( y = mx - 19 \). The slope of this line is \( m \).

5. **Set the slopes equal**:
Since the tangent to the curve is parallel to the line, their slopes must be equal. Therefore, we set:
\[ m = 11 \]

6. **Conclusion**:
The value of \( m \) is \( 11 \).

### Final Answer:
The value of \( m \) is \( 11 \).

---