The interval in which the function `2x^3 + 15` increases less rapidly than the function `9x^2 - 12x,` is –

Options

A. `(-oo ,1)`
B. `(1,2)`
C. `(2, oo)`
D. none of these

Answer

To determine the interval in which the function \( f(x) = 2x^3 + 15 \) increases less rapidly than the function \( g(x) = 9x^2 - 12x \), we need to compare their derivatives.

### Step 1: Find the derivatives of both functions.
1. The derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3 + 15) = 6x^2 \]

2. The derivative of \( g(x) \):
\[ g'(x) = \frac{d}{dx}(9x^2 - 12x) = 18x - 12 \]

### Step 2: Set up the inequality.
We want to find the interval where \( f'(x) < g'(x) \):
\[ 6x^2 < 18x - 12 \]

### Step 3: Rearrange the inequality.
Rearranging gives:
\[ 6x^2 - 18x + 12 < 0 \]
Dividing through by 6 simplifies it to:
\[ x^2 - 3x + 2 < 0 \]

### Step 4: Factor the quadratic.
Factoring the quadratic:
\[ (x - 1)(x - 2) < 0 \]

### Step 5: Determine the critical points.
The critical points from the factors are \( x = 1 \) and \( x = 2 \).

### Step 6: Test intervals around the critical points.
We will test the intervals \( (-\infty, 1) \), \( (1, 2) \), and \( (2, \infty) \):

1. For \( x < 1 \) (e.g., \( x = 0 \)):
\[ (0 - 1)(0 - 2) = 1 \cdot 2 = 2 > 0 \]

2. For \( 1 < x < 2 \) (e.g., \( x = 1.5 \)):
\[ (1.5 - 1)(1.5 - 2) = 0.5 \cdot (-0.5) = -0.25 < 0 \]

3. For \( x > 2 \) (e.g., \( x = 3 \)):
\[ (3 - 1)(3 - 2) = 2 \cdot 1 = 2 > 0 \]

### Step 7: Conclusion.
The function \( f(x) \) increases less rapidly than \( g(x) \) in the interval \( (1, 2) \).

### Final Answer:
The interval in which the function \( 2x^3 + 15 \) increases less rapidly than the function \( 9x^2 - 12x \) is \( (1, 2) \).

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