Let `f:[-1,3]-> [-8, 72]` be defined as,`f(x) = 4x^3-12x`,then f is

Options

A. injective but not surjective
B. bijective
C. neither injective not surjective
D. surjective but not injective

Explanation

To determine the nature of the function \( f(x) = 4x^3 - 12x \) defined on the interval \([-1, 3]\) with a codomain of \([-8, 72]\), we will analyze whether the function is injective (one-to-one) and surjective (onto).

### Step 1: Check for Injectivity

To check if \( f \) is injective, we need to see if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

1. Start with the equation:
\[ f(x_1) = f(x_2) \]
This gives us:
\[ 4x_1^3 - 12x_1 = 4x_2^3 - 12x_2 \]
Rearranging this, we have:
\[ 4x_1^3 - 4x_2^3 = 12x_1 - 12x_2 \]
Factoring out common terms:
\[ 4(x_1^3 - x_2^3) = 12(x_1 - x_2) \]

2. Using the difference of cubes, we can express \( x_1^3 - x_2^3 \) as:
\[ x_1^3 - x_2^3 = (x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) \]
Substituting this back, we have:
\[ 4(x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) = 12(x_1 - x_2) \]

3. If \( x_1 \neq x_2 \), we can divide both sides by \( x_1 - x_2 \):
\[ 4(x_1^2 + x_1x_2 + x_2^2) = 12 \]
Simplifying gives:
\[ x_1^2 + x_1x_2 + x_2^2 = 3 \]

4. The equation \( x_1^2 + x_1x_2 + x_2^2 = 3 \) can have solutions other than \( x_1 = x_2 \), indicating that \( f \) is not injective.

### Step 2: Check for Surjectivity

To check if \( f \) is surjective, we need to determine if the range of \( f \) matches the codomain \([-8, 72]\).

1. First, find the derivative \( f'(x) \):
\[ f'(x) = 12x^2 - 12 = 12(x^2 - 1) = 12(x - 1)(x + 1) \]
Setting \( f'(x) = 0 \) gives critical points at \( x = -1 \) and \( x = 1 \).

2. Evaluate \( f \) at the endpoints and the critical point:
- \( f(-1) = 4(-1)^3 - 12(-1) = -4 + 12 = 8 \)
- \( f(1) = 4(1)^3 - 12(1) = 4 - 12 = -8 \)
- \( f(3) = 4(3)^3 - 12(3) = 4(27) - 36 = 108 - 36 = 72 \)

3. The values obtained are:
- \( f(-1) = 8 \)
- \( f(1) = -8 \)
- \( f(3) = 72 \)

4. The minimum value of \( f \) on the interval \([-1, 3]\) is \(-8\) and the maximum value is \(72\). Therefore, the range of \( f \) is \([-8, 8]\).

Since the range \([-8, 8]\) does not cover the entire codomain \([-8, 72]\), \( f \) is not surjective.

### Conclusion

The function \( f(x) = 4x^3 - 12x \) is **not injective** and **not surjective**.