The function `f : (-oo, 3] to (o,e ^(7)]` defined by `f (x)=e ^(x^(3)-3x^(2) -9x+2)` is

Options

A. Many one and onto
B. Many one and into
C. One to one and onto
D. One to one and into

Answer

To determine the nature of the function \( f : (-\infty, 3] \to (0, e^7] \) defined by \( f(x) = e^{x^3 - 3x^2 - 9x + 2} \), we will analyze the function step by step.

### Step 1: Understand the function
The function is given as:
\[ f(x) = e^{x^3 - 3x^2 - 9x + 2} \]
This is an exponential function where the exponent is a polynomial \( g(x) = x^3 - 3x^2 - 9x + 2 \).

### Step 2: Differentiate the function
To analyze the nature of the function, we need to find the derivative \( f'(x) \):
\[ f'(x) = e^{g(x)} \cdot g'(x) \]
where \( g'(x) \) is the derivative of the polynomial \( g(x) \).

Calculating \( g'(x) \):
\[ g'(x) = 3x^2 - 6x - 9 \]

Thus, we have:
\[ f'(x) = e^{g(x)} \cdot (3x^2 - 6x - 9) \]

### Step 3: Find critical points
To find the critical points, we set \( g'(x) = 0 \):
\[ 3x^2 - 6x - 9 = 0 \]
Dividing the entire equation by 3:
\[ x^2 - 2x - 3 = 0 \]
Factoring the quadratic:
\[ (x - 3)(x + 1) = 0 \]
Thus, the critical points are:
\[ x = 3 \quad \text{and} \quad x = -1 \]

### Step 4: Analyze intervals
We will analyze the sign of \( g'(x) \) in the intervals determined by the critical points \( x = -1 \) and \( x = 3 \):
- For \( x < -1 \): Choose \( x = -2 \):
\[ g'(-2) = 3(-2)^2 - 6(-2) - 9 = 12 + 12 - 9 = 15 \quad (\text{positive}) \]
- For \( -1 < x < 3 \): Choose \( x = 0 \):
\[ g'(0) = 3(0)^2 - 6(0) - 9 = -9 \quad (\text{negative}) \]
- For \( x > 3 \): Choose \( x = 4 \):
\[ g'(4) = 3(4)^2 - 6(4) - 9 = 48 - 24 - 9 = 15 \quad (\text{positive}) \]

### Step 5: Determine the nature of the function
From the analysis:
- \( f(x) \) is increasing on \( (-\infty, -1) \)
- \( f(x) \) is decreasing on \( (-1, 3) \)
- \( f(x) \) is increasing on \( (3, \infty) \)

Since the function decreases and then increases, it has a maximum at \( x = -1 \) and a minimum at \( x = 3 \).

### Step 6: Evaluate the function at critical points
Now we evaluate \( f(x) \) at the critical points:
- At \( x = -1 \):
\[ f(-1) = e^{-1 + 3 + 9 + 2} = e^{7} \]
- At \( x = 3 \):
\[ f(3) = e^{3^3 - 3 \cdot 3^2 - 9 \cdot 3 + 2} = e^{27 - 27 - 27 + 2} = e^{-25} \quad (\text{which approaches } 0) \]

### Step 7: Determine the range
As \( x \) approaches \( -\infty \), \( f(x) \) approaches \( 0 \). Therefore, the range of \( f(x) \) is:
\[ (0, e^7] \]

### Conclusion
The function \( f(x) \) is many-one because it decreases and then increases, meaning it takes the same value for multiple inputs. It is onto because every value in the range \( (0, e^7] \) corresponds to at least one value in the domain \( (-\infty, 3] \).

### Final Answer
The function \( f(x) \) is many-one and onto.