f(x)=x^3+4.5x^2-12x-1. find the relative maxima of f'(x)

To find the relative maxima of \( f(x) = x^3 + 4.5x^2 - 12x - 1 \), we follow these steps:

### Step 1: Find the first derivative \( f'(x) \)
The derivative of the given function \( f(x) \) is:

\[
f'(x) = \frac{d}{dx}(x^3 + 4.5x^2 - 12x - 1)
\]

Using standard differentiation rules:

\[
f'(x) = 3x^2 + 9x - 12
\]

### Step 2: Find critical points by setting \( f'(x) = 0 \)
To find the critical points, we solve the equation \( f'(x) = 0 \):

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

We can simplify by dividing through by 3:

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

This quadratic equation can be factored:

\[
(x + 4)(x - 1) = 0
\]

So, the critical points are:

\[
x = -4 \quad \text{and} \quad x = 1
\]

### Step 3: Determine the nature of the critical points (Maxima/Minima)
To find whether these critical points correspond to maxima or minima, we use the second derivative test.

#### Step 3.1: Find the second derivative \( f''(x) \)

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

#### Step 3.2: Evaluate \( f''(x) \) at \( x = -4 \) and \( x = 1 \)

- For \( x = -4 \):

\[
f''(-4) = 6(-4) + 9 = -24 + 9 = -15
\]

Since \( f''(-4) < 0 \), this means \( x = -4 \) is a **local maximum**.

- For \( x = 1 \):

\[
f''(1) = 6(1) + 9 = 6 + 9 = 15
\]

Since \( f''(1) > 0 \), this means \( x = 1 \) is a **local minimum**.

### Conclusion:
The relative maximum of \( f'(x) \) occurs at \( x = -4 \).

Would you like to know more details or have any questions? 

Here are 5 related questions for further exploration:

1. How do we interpret the significance of the second derivative test?
2. Can a function have more than one relative maximum?
3. How would the relative extrema change if we modified the coefficients of \( f(x) \)?
4. What is the difference between a global and relative maximum?
5. How can we use this information to sketch the graph of \( f(x) \)?

**Tip**: The second derivative test is a quick way to determine concavity and identify maxima or minima at critical points.

f(x)=x^3+4.5x^2-12x-1. Find the relative maxima of f'(x).

Explore how to find the relative maxima of f'(x) for the function f(x) = x^3 + 4.5x^2 - 12x - 1. This step-by-step solution covers critical points, the second derivative test, and solving quadratic equations. Suitable for advanced high school calculus students (Grades 11-12).

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