The slope of the tangent at the point of inflection of `y = x ^(3) -3x ^(2)+ 6x +2009` is equal to :

Options

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

Answer

To find the slope of the tangent at the point of inflection of the function \( y = x^3 - 3x^2 + 6x + 2009 \), we will follow these steps:

### Step 1: Find the first derivative \( y' \)
The first derivative of the function is calculated as follows:
\[ y' = \frac{d}{dx}(x^3 - 3x^2 + 6x + 2009) \]
Using the power rule:
\[ y' = 3x^2 - 6x + 6 \]

### Step 2: Find the second derivative \( y'' \)
Next, we find the second derivative:
\[ y'' = \frac{d}{dx}(3x^2 - 6x + 6) \]
Again, using the power rule:
\[ y'' = 6x - 6 \]

### Step 3: Set the second derivative equal to zero to find the point of inflection
To find the point of inflection, we set the second derivative equal to zero:
\[ 6x - 6 = 0 \]
Solving for \( x \):
\[ 6x = 6 \implies x = 1 \]

### Step 4: Find the slope of the tangent at the point of inflection
Now, we need to find the slope of the tangent at \( x = 1 \) by evaluating the first derivative at this point:
\[ y'(1) = 3(1)^2 - 6(1) + 6 \]
Calculating this:
\[ y'(1) = 3 - 6 + 6 = 3 \]

### Conclusion
The slope of the tangent at the point of inflection is \( 3 \).

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