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We are given the function:

\[
f(x) = 2x^3 - 27x^2 + 48x + 8
\]

We need to find the local minimum and local maximum. Let's start by finding the critical points where the derivative \( f'(x) = 0 \).

### Step 1: Find the first derivative of \( f(x) \)
\[
f'(x) = \frac{d}{dx} \left( 2x^3 - 27x^2 + 48x + 8 \right) = 6x^2 - 54x + 48
\]

### Step 2: Solve \( f'(x) = 0 \) to find critical points
\[
6x^2 - 54x + 48 = 0
\]
Dividing the equation by 6:
\[
x^2 - 9x + 8 = 0
\]
Now, factor the quadratic equation:
\[
(x - 8)(x - 1) = 0
\]
So, the critical points are \( x = 8 \) and \( x = 1 \).

### Step 3: Determine whether these points are minima or maxima
We use the second derivative test.

Find the second derivative:
\[
f''(x) = \frac{d}{dx}(6x^2 - 54x + 48) = 12x - 54
\]

Now evaluate \( f''(x) \) at the critical points:

- For \( x = 1 \):
  \[
  f''(1) = 12(1) - 54 = 12 - 54 = -42
  \]
  Since \( f''(1) < 0 \), there is a local maximum at \( x = 1 \).

- For \( x = 8 \):
  \[
  f''(8) = 12(8) - 54 = 96 - 54 = 42
  \]
  Since \( f''(8) > 0 \), there is a local minimum at \( x = 8 \).

### Step 4: Find the values of the function at these points
To find the actual values at the local minimum and maximum, substitute \( x = 1 \) and \( x = 8 \) into the original function \( f(x) \).

- For \( x = 1 \):
  \[
  f(1) = 2(1)^3 - 27(1)^2 + 48(1) + 8 = 2 - 27 + 48 + 8 = 31
  \]
  So, the local maximum value is \( f(1) = 31 \).

- For \( x = 8 \):
  \[
  f(8) = 2(8)^3 - 27(8)^2 + 48(8) + 8 = 2(512) - 27(64) + 48(8) + 8 = 1024 - 1728 + 384 + 8 = -312
  \]
  So, the local minimum value is \( f(8) = -312 \).

### Final Answer:
- The local minimum is at \( x = 8 \) with value \( -312 \).
- The local maximum is at \( x = 1 \) with value \( 31 \).

Would you like a more detailed explanation or further clarification on any steps? 

Here are 5 related questions you could explore:
1. What is the graphical interpretation of local maxima and minima?
2. How do the critical points relate to the shape of the graph of the function?
3. What happens if both the first and second derivatives at a critical point are zero?
4. How does one determine global maxima and minima on a closed interval?
5. How can we apply this method to functions with more complex terms?

**Tip:** Always remember that a second derivative greater than zero indicates concavity upwards (local minimum), while less than zero indicates concavity downwards (local maximum).

The function f(x) = 2x^3 - 27x^2 + 48x + 8 has one local minimum and one local maximum. Find the local minimum and local maximum along with their values.

This solution guides you through finding the local minima and maxima of the cubic function f(x) = 2x^3 - 27x^2 + 48x + 8 using calculus techniques. Learn how to find critical points, apply the second derivative test, and calculate the function values at these points. Suitable for high school students in Grades 10-12 or early college-level calculus learners.

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