`f(x)=x^(3)-9x^(2)+15x+3` has maximum at the point

Options

A. `(1, 10)`
B. `(5, -22)`
C. `(1, 2)`
D. `(1, -22)`

Answer

To find the maximum point of the function \( f(x) = x^3 - 9x^2 + 15x + 3 \), we will follow these steps:

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

\[ f'(x) = \frac{d}{dx}(x^3 - 9x^2 + 15x + 3) \]

Using the power rule:

\[ f'(x) = 3x^2 - 18x + 15 \]

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

\[ 3x^2 - 18x + 15 = 0 \]

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

\[ x^2 - 6x + 5 = 0 \]

### Step 4: Factor the quadratic equation
Next, we factor the quadratic equation:

\[ (x - 1)(x - 5) = 0 \]

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

\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \]

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

\[ f''(x) = \frac{d}{dx}(3x^2 - 18x + 15) = 6x - 18 \]

### Step 7: Evaluate the second derivative at critical points
Now we evaluate the second derivative at the critical points \( x = 1 \) and \( x = 5 \):

1. For \( x = 1 \):

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

Since \( f''(1) < 0 \), this indicates a local maximum at \( x = 1 \).

2. For \( x = 5 \):

\[ f''(5) = 6(5) - 18 = 30 - 18 = 12 \quad (\text{greater than } 0) \]

Since \( f''(5) > 0 \), this indicates a local minimum at \( x = 5 \).

### Step 8: Find the maximum value
Now we find the maximum value of the function at \( x = 1 \):

\[ f(1) = 1^3 - 9(1^2) + 15(1) + 3 \]
\[ = 1 - 9 + 15 + 3 \]
\[ = 1 - 9 + 15 + 3 = 10 \]

### Conclusion
Thus, the function \( f(x) \) has a maximum at the point \( (1, 10) \).

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