consider the function f(X) =`3x^(4)+4x^(3)-12x^(2)`
The range of the function y=f(x) is

Options

A. `(-oo,oo)`
B. `[-32,oo)`
C. `[0,oo)`
D. none of these

Explanation

To find the range of the function \( f(x) = 3x^4 + 4x^3 - 12x^2 \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \) with respect to \( x \).

\[ f'(x) = \frac{d}{dx}(3x^4 + 4x^3 - 12x^2) \]

Using the power rule for differentiation, we get:

\[ f'(x) = 12x^3 + 12x^2 - 24x \]

### Step 2: Factor the derivative
Next, we can factor out the common terms in \( f'(x) \):

\[ f'(x) = 12x(x^2 + x - 2) \]

Now, we can factor the quadratic expression:

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

Thus, we have:

\[ f'(x) = 12x(x - 1)(x + 2) \]

### Step 3: Find critical points
To find the critical points, we set the derivative equal to zero:

\[ 12x(x - 1)(x + 2) = 0 \]

This gives us the critical points:

\[ x = 0, \quad x = 1, \quad x = -2 \]

### Step 4: Determine the nature of critical points
We will use the first derivative test to determine whether these critical points are maxima or minima. We can analyze the sign of \( f'(x) \) around these points:

- For \( x < -2 \): \( f'(x) > 0 \) (increasing)
- For \( -2 < x < 0 \): \( f'(x) < 0 \) (decreasing)
- For \( 0 < x < 1 \): \( f'(x) > 0 \) (increasing)
- For \( x > 1 \): \( f'(x) > 0 \) (increasing)

From this analysis:
- At \( x = -2 \): Local minimum (since it changes from increasing to decreasing)
- At \( x = 0 \): Local maximum (since it changes from decreasing to increasing)
- At \( x = 1 \): Local minimum (since it changes from increasing to increasing)

### Step 5: Evaluate the function at critical points
Now we will evaluate \( f(x) \) at the critical points to find the minimum and maximum values.

1. **At \( x = -2 \)**:
\[ f(-2) = 3(-2)^4 + 4(-2)^3 - 12(-2)^2 = 3(16) + 4(-8) - 12(4) = 48 - 32 - 48 = -32 \]

2. **At \( x = 0 \)**:
\[ f(0) = 3(0)^4 + 4(0)^3 - 12(0)^2 = 0 \]

3. **At \( x = 1 \)**:
\[ f(1) = 3(1)^4 + 4(1)^3 - 12(1)^2 = 3 + 4 - 12 = -5 \]

### Step 6: Determine the range
From our evaluations:
- The local minimum value is \( -32 \) at \( x = -2 \).
- The local maximum value is \( 0 \) at \( x = 0 \).
- As \( x \) approaches \( +\infty \), \( f(x) \) approaches \( +\infty \).

Thus, the range of the function \( f(x) \) is:

\[ \text{Range} = [-32, +\infty) \]