The function `f(x)=e^(x+1)(4x^2-16x+11)` is :

Options

A. decreasing in `(-oo,-5//2) uu (1//2, oo)`
B. decreasing in `(-oo,1//2) uu (5//2 , oo)`
C. increasing in `(-oo,-2) uu (2,oo)`
D. increasing in `(-oo,-1//2)uu (5//2, oo)`

Answer

To analyze the function \( f(x) = e^{(x+1)}(4x^2 - 16x + 11) \) for intervals of increase and decrease, we will follow these steps:

### Step 1: Differentiate the function
We need to find the derivative \( f'(x) \) using the product rule. The product rule states that if \( f(x) = u(x)v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \).

Let:
- \( u(x) = e^{(x+1)} \)
- \( v(x) = 4x^2 - 16x + 11 \)

Now, we differentiate both \( u(x) \) and \( v(x) \):
- \( u'(x) = e^{(x+1)} \) (since the derivative of \( e^x \) is \( e^x \))
- \( v'(x) = 8x - 16 \) (using the power rule)

Now, applying the product rule:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) = e^{(x+1)}(4x^2 - 16x + 11) + e^{(x+1)}(8x - 16) \]

### Step 2: Set the derivative to zero
To find the critical points, we set \( f'(x) = 0 \):
\[ e^{(x+1)}(4x^2 - 16x + 11 + 8x - 16) = 0 \]
Since \( e^{(x+1)} \) is never zero, we focus on the quadratic:
\[ 4x^2 - 8x - 5 = 0 \]

### Step 3: Solve the quadratic equation
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4, b = -8, c = -5 \):
\[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 4 \cdot (-5)}}{2 \cdot 4} \]
Calculating the discriminant:
\[ 64 + 80 = 144 \]
Thus,
\[ x = \frac{8 \pm 12}{8} \]
Calculating the two roots:
1. \( x = \frac{20}{8} = \frac{5}{2} \)
2. \( x = \frac{-4}{8} = -\frac{1}{2} \)

### Step 4: Analyze the sign of \( f'(x) \)
We have critical points at \( x = -\frac{1}{2} \) and \( x = \frac{5}{2} \). We will test intervals around these points to determine where \( f'(x) \) is positive or negative.

- **Interval 1**: \( (-\infty, -\frac{1}{2}) \)
- **Interval 2**: \( (-\frac{1}{2}, \frac{5}{2}) \)
- **Interval 3**: \( (\frac{5}{2}, \infty) \)

Choose test points:
1. For \( x = -1 \) in Interval 1:
\[ f'(-1) = e^{0}(4(-1)^2 - 8(-1) - 5) = 4 + 8 - 5 = 7 > 0 \]
So, \( f'(x) > 0 \) in this interval.

2. For \( x = 0 \) in Interval 2:
\[ f'(0) = e^{1}(4(0)^2 - 8(0) - 5) = e^{1}(-5) < 0 \]
So, \( f'(x) < 0 \) in this interval.

3. For \( x = 3 \) in Interval 3:
\[ f'(3) = e^{4}(4(3)^2 - 8(3) - 5) = e^{4}(36 - 24 - 5) = e^{4}(7) > 0 \]
So, \( f'(x) > 0 \) in this interval.

### Conclusion
- The function \( f(x) \) is **increasing** on the intervals \( (-\infty, -\frac{1}{2}) \) and \( (\frac{5}{2}, \infty) \).
- The function \( f(x) \) is **decreasing** on the interval \( (-\frac{1}{2}, \frac{5}{2}) \).