`f(x)=24x^(3)+3x^(2)-3x+7`

Options

A. `I_(1)=(-(1)/(4),(1)/(6)), I_(2)=((1)/(6), infty)`
B. `I_(1)=(-infty,-(1)/(4)),uu ((1)/(6), infty), I_(2)=(-(1)/(4),(1)/(6))`
C. `I_(1)=(-(1)/(6),(1)/(4)), I_(2)=((1)/(4), (1)/(3))uu((1)/(3), infty)`
D. none of these

Answer

To determine where the function \( f(x) = 24x^3 + 3x^2 - 3x + 7 \) is increasing or decreasing, we will follow these steps:

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

\[ f'(x) = \frac{d}{dx}(24x^3 + 3x^2 - 3x + 7) \]

Using the power rule, we differentiate each term:

- The derivative of \( 24x^3 \) is \( 72x^2 \).
- The derivative of \( 3x^2 \) is \( 6x \).
- The derivative of \( -3x \) is \( -3 \).
- The derivative of the constant \( 7 \) is \( 0 \).

Thus, we have:

\[ f'(x) = 72x^2 + 6x - 3 \]

### Step 2: Set the derivative greater than zero for increasing intervals
To find where the function is increasing, we set the derivative greater than zero:

\[ 72x^2 + 6x - 3 > 0 \]

### Step 3: Simplify the inequality
We can simplify the inequality by dividing all terms by 3:

\[ 24x^2 + 2x - 1 > 0 \]

### Step 4: Factor the quadratic expression
Next, we will factor the quadratic expression \( 24x^2 + 2x - 1 \). We look for two numbers that multiply to \( 24 \times -1 = -24 \) and add to \( 2 \). The numbers \( 6 \) and \( -4 \) work. Thus, we can rewrite the quadratic as:

\[ 24x^2 + 6x - 4x - 1 > 0 \]

Now, we can factor by grouping:

\[ 6x(4x + 1) - 1(4x + 1) > 0 \]

This gives us:

\[ (6x - 1)(4x + 1) > 0 \]

### Step 5: Find the critical points
To find the critical points, we set each factor to zero:

1. \( 6x - 1 = 0 \) gives \( x = \frac{1}{6} \)
2. \( 4x + 1 = 0 \) gives \( x = -\frac{1}{4} \)

### Step 6: Test intervals on the number line
We will test the intervals determined by the critical points \( x = -\frac{1}{4} \) and \( x = \frac{1}{6} \):

- **Interval 1**: \( (-\infty, -\frac{1}{4}) \)
- **Interval 2**: \( (-\frac{1}{4}, \frac{1}{6}) \)
- **Interval 3**: \( (\frac{1}{6}, \infty) \)

We will choose test points from each interval:

1. For \( x = -1 \) (in Interval 1):
\[ (6(-1) - 1)(4(-1) + 1) = (-7)(-3) > 0 \quad \text{(positive)} \]

2. For \( x = 0 \) (in Interval 2):
\[ (6(0) - 1)(4(0) + 1) = (-1)(1) < 0 \quad \text{(negative)} \]

3. For \( x = 1 \) (in Interval 3):
\[ (6(1) - 1)(4(1) + 1) = (5)(5) > 0 \quad \text{(positive)} \]

### Step 7: Determine the intervals of increase and decrease
From our tests, we find:

- The function is increasing on \( (-\infty, -\frac{1}{4}) \) and \( (\frac{1}{6}, \infty) \).
- The function is decreasing on \( (-\frac{1}{4}, \frac{1}{6}) \).

### Final Answer
The function \( f(x) \) is increasing on the intervals \( (-\infty, -\frac{1}{4}) \) and \( (\frac{1}{6}, \infty) \), and it is decreasing on the interval \( (-\frac{1}{4}, \frac{1}{6}) \).