Determine the intervals in which the following functions are strictly increasing or strictly decreasing :
`f(x)=(1)/(4)x^(4)+(2)/(3)x^(3)-(5)/(2)x^(2)-6x+7`

Answer

To determine the intervals in which the function \( f(x) = \frac{1}{4}x^4 + \frac{2}{3}x^3 - \frac{5}{2}x^2 - 6x + 7 \) is strictly increasing or strictly decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
The first step is to differentiate the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}\left(\frac{1}{4}x^4 + \frac{2}{3}x^3 - \frac{5}{2}x^2 - 6x + 7\right) \]

Using the power rule for differentiation, we get:

\[ f'(x) = x^3 + 2x^2 - 5x - 6 \]

### Step 2: Set the derivative equal to zero
To find the critical points where the function changes from increasing to decreasing or vice versa, we need to solve:

\[ f'(x) = 0 \]

This gives us the equation:

\[ x^3 + 2x^2 - 5x - 6 = 0 \]

### Step 3: Factor the derivative
To solve the cubic equation, we can try to factor it. We can use the Rational Root Theorem to find possible rational roots. Testing \( x = -3 \):

\[ (-3)^3 + 2(-3)^2 - 5(-3) - 6 = -27 + 18 + 15 - 6 = 0 \]

So, \( x = -3 \) is a root. We can factor \( f'(x) \) as follows:

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

Next, we can factor \( x^2 - x - 2 \):

\[ x^2 - x - 2 = (x - 2)(x + 1) \]

Thus, we have:

\[ f'(x) = (x + 3)(x - 2)(x + 1) \]

### Step 4: Find the critical points
The critical points are found by setting each factor to zero:

1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x - 2 = 0 \) → \( x = 2 \)
3. \( x + 1 = 0 \) → \( x = -1 \)

So the critical points are \( x = -3, -1, 2 \).

### Step 5: Determine the sign of the derivative
We will test the intervals determined by the critical points: \( (-\infty, -3) \), \( (-3, -1) \), \( (-1, 2) \), and \( (2, \infty) \).

- **Interval \( (-\infty, -3) \)**: Choose \( x = -4 \)
\[ f'(-4) = (-4 + 3)(-4 - 2)(-4 + 1) = (-1)(-6)(-3) < 0 \quad \text{(decreasing)} \]

- **Interval \( (-3, -1) \)**: Choose \( x = -2 \)
\[ f'(-2) = (-2 + 3)(-2 - 2)(-2 + 1) = (1)(-4)(-1) > 0 \quad \text{(increasing)} \]

- **Interval \( (-1, 2) \)**: Choose \( x = 0 \)
\[ f'(0) = (0 + 3)(0 - 2)(0 + 1) = (3)(-2)(1) < 0 \quad \text{(decreasing)} \]

- **Interval \( (2, \infty) \)**: Choose \( x = 3 \)
\[ f'(3) = (3 + 3)(3 - 2)(3 + 1) = (6)(1)(4) > 0 \quad \text{(increasing)} \]

### Step 6: Summarize the intervals
From our analysis, we find:

- \( f(x) \) is **strictly decreasing** on the intervals \( (-\infty, -3) \) and \( (-1, 2) \).
- \( f(x) \) is **strictly increasing** on the intervals \( (-3, -1) \) and \( (2, \infty) \).

### Final Answer
- **Strictly Increasing**: \( (-3, -1) \) and \( (2, \infty) \)
- **Strictly Decreasing**: \( (-\infty, -3) \) and \( (-1, 2) \)