Use the second derivative to determine each of the following for the function given below. $$ f(x)=\frac{x^{4}}{6}-\frac{1}{3} x^{3}-12 x^{2}+9 x+5 $$ Determine intervals on which the function is concave up. (Enter your answer using interval notation.) $$ (-\infty,-3) \cup(4, \infty) $$ Determine intervals on which the function is concave down. (Enter your answer using interval notation.) $$ (-3,4) $$ $$ (x, y)=\left(-3,-\frac{215}{2}\right. $$ $$ (x, y)=\left(4,-\frac{389}{3}\right. $$ Locate the points of inflection. (If an answer does not exist, enter DNE.) (smaller $$ x $$-value) (larger $$ x $$-value)

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To determine the concavity of the function $$ f(x) = \frac{x^{4}}{6} - \frac{1}{3} x^{3} - 12 x^{2} + 9 x + 5 $$, we will follow these steps:

1. **Find the first derivative $$ f'(x) $$**.
2. **Find the second derivative $$ f''(x) $$**.
3. **Determine the critical points of $$ f''(x) $$**.
4. **Analyze the sign of $$ f''(x) $$ to find intervals of concavity**.
5. **Locate points of inflection where $$ f''(x) = 0 $$**.

### Step 1: Find the first derivative $$ f'(x) $$

The first derivative is calculated as follows:

$$
f'(x) = \frac{d}{dx}\left(\frac{x^{4}}{6} - \frac{1}{3} x^{3} - 12 x^{2} + 9 x + 5\right)
$$

### Step 2: Find the second derivative $$ f''(x) $$

The second derivative is calculated as follows:

$$
f''(x) = \frac{d}{dx}(f'(x))
$$

### Step 3: Determine the critical points of $$ f''(x) $$

We will set $$ f''(x) = 0 $$ and solve for $$ x $$.

### Step 4: Analyze the sign of $$ f''(x) $$

We will test intervals around the critical points to determine where $$ f''(x) $$ is positive (concave up) and where it is negative (concave down).

### Step 5: Locate points of inflection

The points of inflection occur where $$ f''(x) = 0 $$ and the concavity changes.

Let's perform the calculations for the derivatives and find the necessary information.

First, I will calculate the first and second derivatives.

### Calculating the first and second derivatives

1. **First Derivative**:
   $$
   f'(x) = \frac{4}{6}x^{3} - \frac{3}{3}x^{2} - 24x + 9 = \frac{2}{3}x^{3} - x^{2} - 24x + 9
   $$

2. **Second Derivative**:
   $$
   f''(x) = \frac{d}{dx}\left(\frac{2}{3}x^{3} - x^{2} - 24x + 9\right) = 2x^{2} - 2x - 24
   $$

Now, let's set $$ f''(x) = 0 $$ and solve for $$ x $$:

$$
2x^{2} - 2x - 24 = 0
$$

Dividing the entire equation by 2:

$$
x^{2} - x - 12 = 0
$$

Now, we will solve this quadratic equation.

### Solving the quadratic equation

Using the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$:

Here, $$ a = 1, b = -1, c = -12 $$.

Let's calculate the roots.

$$
b^2 - 4ac = (-1)^2 - 4(1)(-12) = 1 + 48 = 49
$$

Now substituting into the quadratic formula:

$$
x = \frac{-(-1) \pm \sqrt{49}}{2(1)} = \frac{1 \pm 7}{2}
$$

Calculating the two roots:

1. $$ x = \frac{8}{2} = 4 $$
2. $$ x = \frac{-6}{2} = -3 $$

Now we have the critical points $$ x = -3 $$ and $$ x = 4 $$.

### Step 4: Analyze the sign of $$ f''(x) $$

We will test the intervals $$ (-\infty, -3) $$, $$ (-3, 4) $$, and $$ (4, \infty) $$:

1. **For $$ x < -3 $$** (e.g., $$ x = -4 $$):
   $$
   f''(-4) = 2(-4)^{2} - 2(-4) - 24 = 32 + 8 - 24 = 16 \quad (\text{positive, concave up})
   $$

2. **For $$ -3 < x < 4 $$** (e.g., $$ x = 0 $$):
   $$
   f''(0) = 2(0)^{2} - 2(0) - 24 = -24 \quad (\text{negative, concave down})
   $$

3. **For $$ x > 4 $$** (e.g., $$ x = 5 $$):
   $$
   f''(5) = 2(5)^{2} - 2(5) - 24 = 50 - 10 - 24 = 16 \quad (\text{positive, concave up})
   $$

### Conclusion on concavity

- **Concave Up**: $$ (-\infty, -3) \cup (4, \infty) $$
- **Concave Down**: $$ (-3, 4) $$

### Step 5: Locate points of inflection

The points of inflection occur at $$ x = -3 $$ and $$ x = 4 $$.

Now, we will find the corresponding $$ y $$-values for these points:

1. For $$ x = -3 $$:
   $$
   f(-3) = \frac{(-3)^{4}}{6} - \frac{1}{3}(-3)^{3} - 12(-3)^{2} + 9(-3) + 5
   $$

2. For $$ x = 4 $$:
   $$
   f(4) = \frac{(4)^{4}}{6} - \frac{1}{3}(4)^{3} - 12(4)^{2} + 9(4) + 5
   $$

Let's calculate these values.

### Calculating $$ f(-3) $$ and $$ f(4) $$

1. **Calculating $$ f(-3) $$**:
   $$
   f(-3) = \frac{81}{6} + 9 - 108 - 27 + 5 = \frac{81}{6} + 9 - 108 - 27 + 5
   $$
   \[
   = \frac{81}{6} + \frac{54}{6} - \frac{648}{6} - \frac{162}{6} + \frac{30}{6} = \
**Concave Up Intervals:** $$ (-\infty, -3) $$ and $$ (4, \infty) $$ **Concave Down Interval:** $$ (-3, 4) $$ **Points of Inflection:** - $$ x = -3 $$ with $$ y = -\frac{215}{2} $$ - $$ x = 4 $$ with $$ y = -\frac{389}{3} $$

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