Find the points of local maxima and local minima, if any, of the following functions. Find also the local maximum and local minimum values :
`f(x)=x^(3)-6x^(2)+9x+15. 0≤x≤6`

Answer

To find the points of local maxima and local minima for the function \( f(x) = x^3 - 6x^2 + 9x + 15 \) on the interval \( 0 \leq x \leq 6 \), we will follow these steps:

### Step 1: Find the first derivative
First, we need to find the first derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 15) = 3x^2 - 12x + 9 \]

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

\[ 3x^2 - 12x + 9 = 0 \]

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

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

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

\[ (x - 3)(x - 1) = 0 \]

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

\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Step 6: Evaluate the second derivative
To determine whether these critical points are local maxima or minima, we will evaluate the second derivative.

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

### Step 7: Test the critical points
Now we will test the critical points \( x = 1 \) and \( x = 3 \):

1. For \( x = 1 \):
\[ f''(1) = 6(1) - 12 = 6 - 12 = -6 \quad (\text{less than } 0 \Rightarrow \text{local maxima}) \]

2. For \( x = 3 \):
\[ f''(3) = 6(3) - 12 = 18 - 12 = 6 \quad (\text{greater than } 0 \Rightarrow \text{local minima}) \]

### Step 8: Evaluate the function at critical points
Now we will find the values of the function at the critical points and the endpoints of the interval \( [0, 6] \):

1. \( f(1) \):
\[ f(1) = 1^3 - 6(1^2) + 9(1) + 15 = 1 - 6 + 9 + 15 = 19 \]

2. \( f(3) \):
\[ f(3) = 3^3 - 6(3^2) + 9(3) + 15 = 27 - 54 + 27 + 15 = 15 \]

3. \( f(0) \):
\[ f(0) = 0^3 - 6(0^2) + 9(0) + 15 = 15 \]

4. \( f(6) \):
\[ f(6) = 6^3 - 6(6^2) + 9(6) + 15 = 216 - 216 + 54 + 15 = 54 \]

### Step 9: Compare values
Now we compare the values obtained:

- \( f(1) = 19 \) (local maximum)
- \( f(3) = 15 \) (local minimum)
- \( f(0) = 15 \) (endpoint)
- \( f(6) = 54 \) (endpoint)

### Conclusion
The local maximum value is \( 19 \) at \( x = 1 \), and the local minimum value is \( 15 \) at \( x = 3 \).

### Final Answer
- Local maximum: \( 19 \) at \( x = 1 \)
- Local minimum: \( 15 \) at \( x = 3 \)