The tangent to the curve `y= x^(3) -6x^(2) + 9x + 4, 0 le x le 5` has maximum slope at x which is equal to

Options

A. 2
B. 3
C. 4
D. none of these

Answer

To find the value of \( x \) at which the tangent to the curve \( y = x^3 - 6x^2 + 9x + 4 \) has a maximum slope, we will follow these steps:

### Step 1: Find the first derivative
The slope of the tangent line to the curve is given by the first derivative \( \frac{dy}{dx} \). We will differentiate the function with respect to \( x \).

\[ \frac{dy}{dx} = \frac{d}{dx}(x^3 - 6x^2 + 9x + 4) \]

Using the power rule:

\[ \frac{dy}{dx} = 3x^2 - 12x + 9 \]

### Step 2: Find the second derivative
To find where the slope is maximized, we need to find the critical points of the first derivative. We do this by taking the derivative of \( \frac{dy}{dx} \) to get the second derivative \( \frac{d^2y}{dx^2} \).

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 - 12x + 9) \]

Differentiating again:

\[ \frac{d^2y}{dx^2} = 6x - 12 \]

### Step 3: Set the second derivative to zero
To find the critical points, we set the second derivative equal to zero:

\[ 6x - 12 = 0 \]

Solving for \( x \):

\[ 6x = 12 \implies x = 2 \]

### Step 4: Determine if it is a maximum
To confirm whether this critical point corresponds to a maximum slope, we can evaluate the second derivative at \( x = 2 \):

\[ \frac{d^2y}{dx^2} = 6(2) - 12 = 12 - 12 = 0 \]

Since the second derivative is zero, we need to check the values of the first derivative around \( x = 2 \) to determine the nature of the critical point.

### Step 5: Evaluate the first derivative at the endpoints
Now we will evaluate the first derivative at the endpoints of the interval \( [0, 5] \) and at \( x = 2 \):

1. **At \( x = 0 \)**:
\[ \frac{dy}{dx} = 3(0)^2 - 12(0) + 9 = 9 \]

2. **At \( x = 2 \)**:
\[ \frac{dy}{dx} = 3(2)^2 - 12(2) + 9 = 3(4) - 24 + 9 = 12 - 24 + 9 = -3 \]

3. **At \( x = 5 \)**:
\[ \frac{dy}{dx} = 3(5)^2 - 12(5) + 9 = 3(25) - 60 + 9 = 75 - 60 + 9 = 24 \]

### Step 6: Conclusion
From the evaluations:
- At \( x = 0 \), the slope is \( 9 \).
- At \( x = 2 \), the slope is \( -3 \).
- At \( x = 5 \), the slope is \( 24 \).

The maximum slope occurs at \( x = 5 \).

Thus, the value of \( x \) at which the tangent to the curve has the maximum slope is:

\[ \boxed{5} \]