The maximum value of the function f(x)=`x^(3)+2x^(2)-4x+6` exits at

Options

A. x=-2
B. x=1
C. x=2
D. x=-1

Explanation

To find the maximum value of the function \( f(x) = x^3 + 2x^2 - 4x + 6 \), we will follow these steps:

### Step 1: Differentiate the function
We start by finding the first derivative of the function \( f(x) \).

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

Using the power rule for differentiation, we get:

\[ f'(x) = 3x^2 + 4x - 4 \]

### Step 2: Set the first derivative to zero
To find the critical points, we set the first derivative equal to zero:

\[ 3x^2 + 4x - 4 = 0 \]

### Step 3: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 3 \), \( b = 4 \), and \( c = -4 \).

Calculating the discriminant:

\[ b^2 - 4ac = 4^2 - 4 \cdot 3 \cdot (-4) = 16 + 48 = 64 \]

Now substituting into the quadratic formula:

\[ x = \frac{-4 \pm \sqrt{64}}{2 \cdot 3} = \frac{-4 \pm 8}{6} \]

This gives us two solutions:

\[ x = \frac{4}{6} = \frac{2}{3} \quad \text{and} \quad x = \frac{-12}{6} = -2 \]

### Step 4: Determine the nature of the critical points
Next, we need to find the second derivative to determine whether these critical points are maxima or minima.

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

Now we evaluate the second derivative at the critical points:

1. For \( x = \frac{2}{3} \):

\[ f''\left(\frac{2}{3}\right) = 6\left(\frac{2}{3}\right) + 4 = 4 + 4 = 8 \quad (\text{positive, hence minimum}) \]

2. For \( x = -2 \):

\[ f''(-2) = 6(-2) + 4 = -12 + 4 = -8 \quad (\text{negative, hence maximum}) \]

### Conclusion
The maximum value of the function \( f(x) \) occurs at \( x = -2 \).