To find 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 \( f'(x) \)
The first step is to differentiate the function \( f(x) \) with respect to \( x \).
\[
f'(x) = \frac{d}{dx}\left(\frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11\right)
\]
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 the 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
To find the critical points where the function changes from increasing to decreasing or vice versa, we set \( f'(x) = 0 \):
\[
\frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5} = 0
\]
Multiplying through by \( 5 \) to eliminate the fraction gives:
\[
6x^3 - 12x^2 - 30x + 36 = 0
\]
### Step 3: Factor the cubic equation
To factor this cubic equation, we can use the Rational Root Theorem or synthetic division. Testing possible rational roots, we find that \( x = 1 \) is a root.
Using synthetic division to divide \( 6x^3 - 12x^2 - 30x + 36 \) by \( x - 1 \):
\[
6x^3 - 12x^2 - 30x + 36 = (x - 1)(6x^2 - 6x - 36)
\]
Next, we simplify \( 6x^2 - 6x - 36 \) to find its roots:
\[
6(x^2 - x - 6) = 0
\]
Factoring gives:
\[
6(x - 3)(x + 2) = 0
\]
Thus, the critical points are:
\[
x = 1, \quad x = 3, \quad x = -2
\]
### Step 4: Determine the sign of \( f'(x) \) in each interval
We will test the sign of \( f'(x) \) in the intervals determined by the critical points: \( (-\infty, -2) \), \( (-2, 1) \), \( (1, 3) \), and \( (3, \infty) \).
1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \)
\[
f'(-3) = 6(-3)^3 - 12(-3)^2 - 30(-3) + 36 = -162 - 108 + 90 + 36 < 0 \quad \text{(decreasing)}
\]
2. **Interval \( (-2, 1) \)**: Choose \( x = 0 \)
\[
f'(0) = 6(0)^3 - 12(0)^2 - 30(0) + 36 = 36 > 0 \quad \text{(increasing)}
\]
3. **Interval \( (1, 3) \)**: Choose \( x = 2 \)
\[
f'(2) = 6(2)^3 - 12(2)^2 - 30(2) + 36 = 48 - 48 - 60 + 36 < 0 \quad \text{(decreasing)}
\]
4. **Interval \( (3, \infty) \)**: Choose \( x = 4 \)
\[
f'(4) = 6(4)^3 - 12(4)^2 - 30(4) + 36 = 384 - 192 - 120 + 36 > 0 \quad \text{(increasing)}
\]
### Step 5: Summarize the intervals
From our analysis, we conclude:
- \( f(x) \) is **strictly increasing** on the intervals \( (-2, 1) \) and \( (3, \infty) \).
- \( f(x) \) is **strictly decreasing** on the intervals \( (-\infty, -2) \) and \( (1, 3) \).
### Final Answer
(a) \( f(x) \) is strictly increasing on \( (-2, 1) \) and \( (3, \infty) \).
(b) \( f(x) \) is strictly decreasing on \( (-\infty, -2) \) and \( (1, 3) \).
