The interval in which the function `x^3` increases less rapidly than `6x^2+15x+5`

Options

A. `(-oo,-1)`
B. (-5,1)
C. (-1,5)
D. `(5,oo)`

Answer

To find the interval in which the function \( x^3 \) increases less rapidly than \( 6x^2 + 15x + 5 \), we need to compare their rates of increase, which can be determined by taking the derivatives of both functions.

### Step-by-Step Solution:

1. **Differentiate both functions**:
- The derivative of \( x^3 \) is:
\[ \frac{d}{dx}(x^3) = 3x^2 \]
- The derivative of \( 6x^2 + 15x + 5 \) is:
\[ \frac{d}{dx}(6x^2 + 15x + 5) = 12x + 15 \]

2. **Set up the inequality**:
We want to find the interval where the rate of increase of \( x^3 \) is less than that of \( 6x^2 + 15x + 5 \):
\[ 3x^2 < 12x + 15 \]

3. **Rearrange the inequality**:
Move all terms to one side:
\[ 3x^2 - 12x - 15 < 0 \]

4. **Factor the quadratic expression**:
First, factor out the common term (3):
\[ 3(x^2 - 4x - 5) < 0 \]
Now, we can focus on the quadratic:
\[ x^2 - 4x - 5 < 0 \]

5. **Factor the quadratic**:
To factor \( x^2 - 4x - 5 \), we look for two numbers that multiply to \(-5\) and add to \(-4\). These numbers are \(-5\) and \(1\):
\[ (x - 5)(x + 1) < 0 \]

6. **Find the critical points**:
The critical points are found by setting each factor to zero:
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

7. **Test intervals**:
We need to test the intervals determined by the critical points \( x = -1 \) and \( x = 5 \):
- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2 - 5)(-2 + 1) = (-7)(-1) > 0 \]
- For \( -1 < x < 5 \) (e.g., \( x = 0 \)):
\[ (0 - 5)(0 + 1) = (-5)(1) < 0 \]
- For \( x > 5 \) (e.g., \( x = 6 \)):
\[ (6 - 5)(6 + 1) = (1)(7) > 0 \]

8. **Conclusion**:
The inequality \( (x - 5)(x + 1) < 0 \) holds true in the interval:
\[ (-1, 5) \]

### Final Answer:
The interval in which the function \( x^3 \) increases less rapidly than \( 6x^2 + 15x + 5 \) is:
\[ (-1, 5) \]