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

Answer

To find the slope of the tangent to the curve \( y = 2x^3 - x^2 + 2 \) at the point \( \left( \frac{1}{2}, 2 \right) \), we will follow these steps:

### Step 1: Differentiate the function
We need to find the derivative of the function \( y \) with respect to \( x \). The derivative \( \frac{dy}{dx} \) gives us the slope of the tangent line at any point on the curve.

Given:
\[ y = 2x^3 - x^2 + 2 \]

Differentiating term by term:
- The derivative of \( 2x^3 \) is \( 6x^2 \) (using the power rule).
- The derivative of \( -x^2 \) is \( -2x \).
- The derivative of the constant \( 2 \) is \( 0 \).

So, we have:
\[ \frac{dy}{dx} = 6x^2 - 2x \]

### Step 2: Evaluate the derivative at the given point
Now, we need to evaluate the derivative at the point \( x = \frac{1}{2} \).

Substituting \( x = \frac{1}{2} \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=\frac{1}{2}} = 6\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) \]

Calculating \( \left(\frac{1}{2}\right)^2 \):
\[ \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

Now substituting this back:
\[ = 6 \cdot \frac{1}{4} - 2 \cdot \frac{1}{2} \]
\[ = \frac{6}{4} - 1 \]
\[ = \frac{3}{2} - 1 \]
\[ = \frac{3}{2} - \frac{2}{2} = \frac{1}{2} \]

### Conclusion
The slope of the tangent to the curve \( y = 2x^3 - x^2 + 2 \) at the point \( \left( \frac{1}{2}, 2 \right) \) is \( \frac{1}{2} \).

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