---
# System prepended metadata

title: In-Class Activity 4.2
tags: [in-class-activity, '4.2']

---

# Question 1 ![{F2B9541B-EF52-47EA-9F77-BCE55FA2CF29}](https://hackmd.io/_uploads/HkMZ6lvAR.png)

a. Identify all intervals on which $f(x)$ is decreasing.

b. Identify all intervals on which $f(x)$ is increasing.

c. Identify all intervals on which $f(x)$ is concave up.

d. Identify all intervals on which $f(x)$ is concave down.


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# Question 2. Find the second derivative for each function:

a. $f(x)=6x-4$

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### Problem: Find the second derivative of $f(x) = 5x - 7$

#### Step 1: Find the first derivative

We start by differentiating $f(x)$ to find the first derivative:

$$
f'(x) = \frac{d}{dx}(5x - 7) = 5
$$

#### Step 2: Find the second derivative

Now, we differentiate $f'(x)$ to find the second derivative:

$$
f''(x) = \frac{d}{dx}(5) = 0
$$

### Conclusion:

The second derivative of $f(x) = 5x - 7$ is $f''(x) = 0$.

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b. $g(x)=3x^2+5x+12$

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<summary> Example: </summary>

### Problem: Find the second derivative of $f(x) = 5x^2 + 6x + 7$

#### Step 1: Find the first derivative

We start by differentiating $f(x)$ to find the first derivative:

$$
f'(x) = \frac{d}{dx}(5x^2 + 6x + 7) = 10x + 6
$$

#### Step 2: Find the second derivative

Now, we differentiate $f'(x)$ to find the second derivative:

$$
f''(x) = \frac{d}{dx}(10x + 6) = 10
$$

### Conclusion:

The second derivative of $f(x) = 5x^2 + 6x + 7$ is $f''(x) = 10$.


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c. $h(x)=-x^3+2x^2-5$

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### Problem: Find the second derivative of $f(x) = -x^3 + 4x^2 - 6$

#### Step 1: Find the first derivative

We start by differentiating $f(x)$ to find the first derivative:

$$
f'(x) = \frac{d}{dx}(-x^3 + 4x^2 - 6) = -3x^2 + 8x
$$

#### Step 2: Find the second derivative

Now, we differentiate $f'(x)$ to find the second derivative:

$$
f''(x) = \frac{d}{dx}(-3x^2 + 8x) = -6x + 8
$$

### Conclusion:

The second derivative of $f(x) = -x^3 + 4x^2 - 6$ is $f''(x) = -6x + 8$.


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d. $i(x)=\ln x$

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### Problem: Find the second derivative of $f(x) = 3\ln(4x)$

#### Step 1: Find the first derivative

We start by differentiating $f(x) = 3\ln(4x)$. Using the chain rule:

$$
f'(x) = 3 \cdot \frac{d}{dx}(\ln(4x)) = 3 \cdot \frac{1}{4x} \cdot 4 = \frac{3}{x}
$$

Now, rewrite $\frac{3}{x}$ as $3x^{-1}$:

$$
f'(x) = 3x^{-1}
$$

#### Step 2: Find the second derivative

Now, differentiate $f'(x) = 3x^{-1}$ using the power rule.

$$
f''(x) = \frac{d}{dx}(3x^{-1}) = 3 \cdot (-1) \cdot x^{-2}
$$

Simplify the result:

$$
f''(x) = -\frac{3}{x^2}
$$

### Conclusion:

The second derivative of $f(x) = 3\ln(4x)$ is $f''(x) = -\frac{3}{x^2}$.



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e. $j(x)=e^{3x}$

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<summary> Example: </summary>

### Problem: Find the second derivative of $f(x) = e^{4x}$

#### Step 1: Find the first derivative

We start by differentiating $f(x) = e^{4x}$. Using the chain rule:

$$
f'(x) = \frac{d}{dx}(e^{4x}) = e^{4x} \cdot \frac{d}{dx}(4x) = 4e^{4x}
$$

#### Step 2: Find the second derivative

Now, we differentiate $f'(x) = 4e^{4x}$ to find the second derivative. Again, applying the chain rule:

$$
f''(x) = \frac{d}{dx}(4e^{4x}) = 4 \cdot e^{4x} \cdot \frac{d}{dx}(4x) = 16e^{4x}
$$

### Conclusion:

The second derivative of $f(x) = e^{4x}$ is $f''(x) = 16e^{4x}$.


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# Question 3 (Inflation) On commonly used measure of inflation is the annual rate of change of the Consumer Price Index (CPI). A newspaper headline proclaims that the rate of change of inflation for consumer prices is increasing. What does this say about the shape of the graph of the CPI?

# Question 4: (Inflation) Another commonly used measure of inflation is the annual rate of change of the Producer Price Index (PPI). A government report states that the rate of change of inflation for producer prices is decreasing. What does this say about the shape of the graph of the PPI?

# Question 5: Match the indicated conditions with one of the graphs (A)-(D) shown in the figure.

a. $f'(x) > 0$ and $f''(x) > 0$ on $(a, b)$.

b. $f'(x) > 0$ and $f''(x) < 0$ on $(a, b)$.

c. $f'(x) < 0$ and $f''(x) > 0$ on $(a, b)$.

d. $f'(x) < 0$ and $f''(x) < 0$ on $(a, b)$.


![{E71C6A8B-F18C-41DD-86F5-BE239754640A}](https://hackmd.io/_uploads/Bk9PClD0C.png)


# Question 6. Find the intervals on which the graph of f(x) is concave upward, the intervals on which the graph of f(x) is concave downward, and the inflection points. $$f(x)=x^4-2x^3-5x+3$$

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<summary> Example: </summary>


Given $f(x)=\dfrac{5}{12}x^4+\dfrac{5}{3}x^3+4x+7$
### Problem: Analyze the concavity and inflection points of $f(x)$

We are given the function:

$$
f(x) = \frac{5}{12}x^4 + \frac{5}{3}x^3 + 4x + 7
$$

### Step 1: Find the second derivative

First, we need to compute the first and second derivatives of $f(x)$.

#### First derivative:

The first derivative is:

$$
f'(x) = \frac{d}{dx} \left( \frac{5}{12}x^4 + \frac{5}{3}x^3 + 4x + 7 \right) = \frac{5}{3}x^3 + 5x^2 + 4
$$

#### Second derivative:

The second derivative is:

$$
f''(x) = \frac{d}{dx} \left( \frac{5}{3}x^3 + 5x^2 + 4 \right) = 5x^2 + 10x
$$

### Step 2: Solve $f''(x) = 0$

To find potential inflection points, we solve $f''(x) = 0$:

$$
5x^2 + 10x = 0
$$

Factor the equation:

$$
5x(x + 2) = 0
$$

The solutions are:

$$
x = 0 \quad \text{or} \quad x = -2
$$

These are the critical points where the concavity might change, so we will analyze the sign of $f''(x)$ around these points.

### Step 3: Sign chart for the second derivative

We now analyze the sign of $f''(x) = 5x(x + 2)$ in the intervals determined by the critical points $x = 0$ and $x = -2$. These points divide the real line into three intervals: $(-\infty, -2)$, $(-2, 0)$, and $(0, \infty)$. We will test the sign of $f''(x)$ in each interval.

| Interval          | Test $x$ value | $f''(x) = 5x(x + 2)$             | Sign     |
|-------------------|----------------|----------------------------------|----------|
| $(-\infty, -2)$   | $x = -3$       | $f''(-3) = 5(-3)((-3) + 2) = 15$ | Positive |
| $(-2, 0)$         | $x = -1$       | $f''(-1) = 5(-1)((-1) + 2) = -5$ | Negative |
| $(0, \infty)$     | $x = 1$        | $f''(1) = 5(1)(1 + 2) = 15$      | Positive |

### Step 5: Find the $y$ values of the inflection points

To find the $y$ values of the inflection points, we substitute the $x$-coordinates of the inflection points ($x = -2$ and $x = 0$) into the original function $f(x)$.

- For $x = -2$:

$$
\begin{align*}
f(-2) &= \frac{5}{12}(-2)^4 + \frac{5}{3}(-2)^3 + 4(-2) + 7 \\
      &= \frac{5}{12}(16) + \frac{5}{3}(-8) + (-8) + 7 \\
      &= \frac{80}{12} - \frac{40}{3} - 8 + 7 \\
      &= \frac{80}{12} - \frac{160}{12} - \frac{96}{12} + \frac{84}{12} \\
      &= \frac{80 - 160 - 96 + 84}{12} \\
      &= \frac{-92}{12} = -\frac{23}{3}
\end{align*}
$$

So, the inflection point at $x = -2$ is $(-2, -\frac{23}{3})$.

- For $x = 0$:

$$
\begin{align*}
f(0) &= \frac{5}{12}(0)^4 + \frac{5}{3}(0)^3 + 4(0) + 7 \\
     &= 7
\end{align*}
$$

So, the inflection point at $x = 0$ is $(0, 7)$.

### Step 6: Conclusion

- The graph of $f(x)$ is **concave upward** on the intervals $(-\infty, -2) \cup (0, \infty)$.
- The graph of $f(x)$ is **concave downward** on the interval $(-2, 0)$.
- The **inflection points** are at $(-2, -\frac{23}{3})$ and $(0, 7)$.





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