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)-12x^(2)+36x-4`

Answer

To find the points of local maxima and minima for the function \( f(x) = x^3 - 12x^2 + 36x - 4 \), we will follow these steps:

### Step 1: Find the first derivative
We start by differentiating the function \( f(x) \):

\[ f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(12x^2) + \frac{d}{dx}(36x) - \frac{d}{dx}(4) \]

Calculating each term gives:

\[ f'(x) = 3x^2 - 24x + 36 \]

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

\[ 3x^2 - 24x + 36 = 0 \]

Dividing the entire equation by 3 simplifies it:

\[ x^2 - 8x + 12 = 0 \]

### Step 3: Factor the quadratic equation
Next, we factor the quadratic:

\[ (x - 6)(x - 2) = 0 \]

This gives us the critical points:

\[ x = 6 \quad \text{and} \quad x = 2 \]

### Step 4: Find the second derivative
To determine whether these critical points are local maxima or minima, we find the second derivative:

\[ f''(x) = \frac{d}{dx}(3x^2 - 24x + 36) = 6x - 24 \]

### Step 5: Evaluate the second derivative at the critical points
Now we evaluate the second derivative at the critical points \( x = 2 \) and \( x = 6 \):

1. For \( x = 2 \):

\[ f''(2) = 6(2) - 24 = 12 - 24 = -12 \]

Since \( f''(2) < 0 \), this indicates that \( x = 2 \) is a local maximum.

2. For \( x = 6 \):

\[ f''(6) = 6(6) - 24 = 36 - 24 = 12 \]

Since \( f''(6) > 0 \), this indicates that \( x = 6 \) is a local minimum.

### Step 6: Find the local maximum and minimum values
Now we calculate the function values at these critical points:

1. For \( x = 2 \):

\[ f(2) = (2)^3 - 12(2)^2 + 36(2) - 4 = 8 - 48 + 72 - 4 = 28 \]

Thus, the local maximum value is \( 28 \).

2. For \( x = 6 \):

\[ f(6) = (6)^3 - 12(6)^2 + 36(6) - 4 = 216 - 432 + 216 - 4 = -4 \]

Thus, the local minimum value is \( -4 \).

### Conclusion
The points of local maxima and minima are:
- Local maximum at \( x = 2 \) with a value of \( 28 \)
- Local minimum at \( x = 6 \) with a value of \( -4 \)

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