Determine the intervals in which the following functions are strictly increasing or strictly decreasing :
`f(x)=(3)/(10)x^(4)-(4)/(5)x^(3)-3x^(2)+(36)/(5)x+11`

Answer

To determine the intervals in which the function \( f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11 \) 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 \( f(x) \).

\[ f'(x) = \frac{d}{dx} \left( \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11 \right) \]

Calculating the derivative term by term:

- The derivative of \( \frac{3}{10}x^4 \) is \( \frac{12}{10}x^3 = \frac{6}{5}x^3 \)
- The derivative of \( -\frac{4}{5}x^3 \) is \( -\frac{12}{5}x^2 \)
- The derivative of \( -3x^2 \) is \( -6x \)
- The derivative of \( \frac{36}{5}x \) is \( \frac{36}{5} \)
- The derivative of a constant \( 11 \) is \( 0 \)

Combining these, we have:

\[ f'(x) = \frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5} \]

### Step 2: Set the derivative equal to zero to find critical points
Next, we need to find the critical points by setting \( f'(x) = 0 \):

\[ \frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5} = 0 \]

To simplify, multiply the entire equation by \( 5 \) to eliminate the fraction:

\[ 6x^3 - 12x^2 - 30x + 36 = 0 \]

### Step 3: Factor the polynomial
Now we will factor the polynomial. We can try to find rational roots using the Rational Root Theorem. Testing \( x = 2 \):

\[ 6(2)^3 - 12(2)^2 - 30(2) + 36 = 48 - 48 - 60 + 36 = -24 \quad (\text{not a root}) \]

Testing \( x = 3 \):

\[ 6(3)^3 - 12(3)^2 - 30(3) + 36 = 162 - 108 - 90 + 36 = 0 \quad (\text{is a root}) \]

Now we can factor \( (x - 3) \) out of \( 6x^3 - 12x^2 - 30x + 36 \). Using synthetic division:

\[ 6x^3 - 12x^2 - 30x + 36 = (x - 3)(6x^2 + 6x - 12) \]

Now we can factor \( 6x^2 + 6x - 12 \):

\[ 6(x^2 + x - 2) = 6(x - 1)(x + 2) \]

Thus, we have:

\[ f'(x) = 6(x - 3)(x - 1)(x + 2) \]

### Step 4: Determine the sign of the derivative
The critical points are \( x = -2, 1, 3 \). We will test the intervals determined by these points: \( (-\infty, -2) \), \( (-2, 1) \), \( (1, 3) \), and \( (3, \infty) \).

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

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

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

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

### Step 5: Summarize the results
- \( f(x) \) is **strictly decreasing** on the interval \( (-\infty, -2) \) and \( (1, 3) \).
- \( f(x) \) is **strictly increasing** on the interval \( (-2, 1) \) and \( (3, \infty) \).

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