`f(x)=2x^(3)-9x^(2)+12x+5` has maximum at the point

Options

A. `(1,2)`
B. `(2, 9)`
C. `(1, 10)`
D. `(2, 10)`

Answer

To find the maximum point of the function \( f(x) = 2x^3 - 9x^2 + 12x + 5 \), we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)
To find the critical points, we need to compute the first derivative of the function.

\[ f'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x + 5) = 6x^2 - 18x + 12 \]

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

\[ 6x^2 - 18x + 12 = 0 \]

### Step 3: Simplify the equation
We can simplify the equation by dividing all terms by 6:

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

### Step 4: Factor the quadratic equation
Now, we will factor the quadratic equation:

\[ (x - 2)(x - 1) = 0 \]

### Step 5: Solve for \( x \)
Setting each factor to zero gives us the critical points:

\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Step 6: Find the second derivative \( f''(x) \)
To determine whether these critical points are maxima or minima, we need to compute the second derivative.

\[ f''(x) = \frac{d}{dx}(6x^2 - 18x + 12) = 12x - 18 \]

### Step 7: Evaluate the second derivative at the critical points
Now we will evaluate the second derivative at \( x = 1 \) and \( x = 2 \).

For \( x = 1 \):

\[ f''(1) = 12(1) - 18 = 12 - 18 = -6 \quad (\text{less than } 0) \]

For \( x = 2 \):

\[ f''(2) = 12(2) - 18 = 24 - 18 = 6 \quad (\text{greater than } 0) \]

### Step 8: Determine the nature of the critical points
Since \( f''(1) < 0 \), the function has a local maximum at \( x = 1 \). Since \( f''(2) > 0 \), the function has a local minimum at \( x = 2 \).

### Step 9: Find the maximum value of the function
Now we will find the maximum value of the function at \( x = 1 \):

\[ f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 2 - 9 + 12 + 5 = 10 \]

### Conclusion
The function \( f(x) = 2x^3 - 9x^2 + 12x + 5 \) has a maximum at the point \( (1, 10) \).

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