---
# System prepended metadata

title: In-Class Activity 4.5
tags: [in-class-activity, '4.5']

---

# Question 1. Find the critical values for each function.

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

<details> <summary> Example: </summary>

Find the critical values for $f(x)=5x-7$:
    
### Solution:

To find the critical values of the function, we need to take the derivative of $f(x)$ and set it equal to zero.

1. **Find the derivative of $f(x)$:**
   
   Since $f(x) = 5x - 7$, the derivative of $f(x)$ is:
   
   $f'(x) = 5$

2. **Set the derivative equal to zero and solve for $x$:**

   $f'(x) = 5 = 0$

   This equation has no solution, since 5 is a constant and cannot be equal to zero.

### Conclusion:

There are no critical values for the function $f(x) = 5x - 7$, as the derivative is a constant and never equals zero.



</details>

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

<details> <summary> Example: </summary>

Find the critical numbers for $f(x) = 4x^2 + 24x + 36$

### Solution:

To find the critical numbers, we first take the derivative of $f(x)$ and then set it equal to zero.

1. **Find the derivative of $f(x)$:**

   Since $f(x) = 4x^2 + 24x + 36$, we apply the power rule to differentiate:

   $$ f'(x) = 8x + 24 $$

2. **Set the derivative equal to zero and solve for $x$:**

   $$ 8x + 24 = 0 $$

   Solving for $x$:

   $$ 8x = -24 $$

   $$ x = -3 $$

### Conclusion:

The critical number for the function $f(x) = 4x^2 + 24x + 36$ is $x = -3$.

</details>

## c. $h(x)=-x^3-3x^2+24x+5$

<details> <summary> Example: </summary>

Find the critical values for the function $h(x) = 2x^3 + 3x^2 - 36x + 5$

### Solution:

To find the critical values, we first take the derivative of $h(x)$ and then set it equal to zero.

1. **Find the derivative of $h(x)$:**

   Since $h(x) = 2x^3 + 3x^2 - 36x + 5$, we differentiate term by term:

   $$ h'(x) = 6x^2 + 6x - 36 $$

2. **Set the derivative equal to zero and solve for $x$:**

   $$ 6x^2 + 6x - 36 = 0 $$

   Simplifying by dividing through by 6:

   $$ x^2 + x - 6 = 0 $$

   Factor the quadratic equation:

   $$ (x + 3)(x - 2) = 0 $$

   Solving for $x$:

   $$ x = -3 \quad \text{or} \quad x = 2 $$

### Conclusion:

The critical values for the function $h(x) = 2x^3 + 3x^2 - 36x + 5$ are $x = -3$ and $x = 2$.


</details>

## d. $i(x)=2\ln x-x^2$

<details> <summary> Example: </summary>

Find the critical values for the function $f(x) = 8 \ln(x) - x^2$

### Solution:

To find the critical values, we first take the derivative of $f(x)$ and then set it equal to zero.

1. **Find the derivative of $f(x)$:**

   The derivative of $f(x) = 8 \ln(x) - x^2$ is:

   $$ f'(x) = \frac{8}{x} - 2x $$

2. **Set the derivative equal to zero and solve for $x$:**

   $$ \frac{8}{x} - 2x = 0 $$

   Multiply through by $x$ to eliminate the fraction:

   $$ 8 - 2x^2 = 0 $$

   Solving for $x$:

   $$ 2x^2 = 8 $$

   $$ x^2 = 4 $$

   $$ x = 2 \quad \text{(since $x > 0$ for $\ln(x)$ to be defined)} $$

### Conclusion:

The critical value for the function $f(x) = 8 \ln(x) - x^2$ is $x = 2$.



</details>

## e. $j(x)=e^{3x^2+2x}$

<details> <summary> Example: </summary>

 Find the critical values for the function $j(x) = e^{4x^2 + 5x}$

### Solution:

To find the critical values, we first take the derivative of $j(x)$ and then set it equal to zero.

1. **Find the derivative of $j(x)$:**

   Using the chain rule, the derivative of $j(x) = e^{4x^2 + 5x}$ is:

   $$ j'(x) = e^{4x^2 + 5x} \cdot (8x + 5) $$

2. **Set the derivative equal to zero and solve for $x$:**

   Since $e^{4x^2 + 5x}$ is never zero, we set the factor $(8x + 5)$ equal to zero:

   $$ 8x + 5 = 0 $$

   Solving for $x$:

   $$ x = -\frac{5}{8} $$

### Conclusion:

The critical value for the function $j(x) = e^{4x^2 + 5x}$ is $x = -\frac{5}{8}$.

</details>

# Question 2. Find the absolute minimum and maximum over the indicated intervals: ![image](https://hackmd.io/_uploads/S1KUBKYC0.png)


## a. $[1,9]$

<details> <summary> Example: </summary>

Find the absolute minimum and maximum on the interval $[1,4]$.
    
The graph has an absolute maximum at the point $(3,9)$.
    
The graph has an absolute minimum at the point $(1,5)$.
    

    
    
 

</details>

## b. $[5,8]$

<details> <summary> Example: </summary>

Find the absolute minimum and maximum of the graph on the interval $[3,10]$.
    
The graph has an absolute minimum at the point $(7,5)$.
    
The graph has an absolute maximum at the point $(10,15)$.

</details>


# Question 3. Find the absolute minimum value on $(0,\infty)$ for $$f(x)=x+\dfrac{1}{x}+\dfrac{30}{x^3}$$

<details> <summary> Example: </summary>

### Question: Find the absolute minimum value on $(0, \infty)$ for 

$$ g(x) = x + \frac{2}{x} + \frac{40}{x^3} $$

### Solution:

To find the absolute minimum value, we first determine the critical points by finding the derivative of $g(x)$, setting it equal to zero, and solving for $x$. Then, we verify if it’s a minimum using the second derivative test and evaluate the function at the critical points to find the minimum value.

1. **Find the derivative of $g(x)$:**

   Using the power rule and the chain rule, we differentiate each term of $g(x)$:

   $$ g'(x) = 1 - \frac{2}{x^2} - \frac{120}{x^4} $$

2. **Set the derivative equal to zero and solve for $x$:**

   Set $g'(x) = 0$:

   $$ 1 - \frac{2}{x^2} - \frac{120}{x^4} = 0 $$

   Multiply through by $x^4$ to eliminate the fractions:

   $$ x^4 - 2x^2 - 120 = 0 $$

   Let $z = x^2$, so the equation becomes a quadratic:

   $$ z^2 - 2z - 120 = 0 $$

   Solve this quadratic using the quadratic formula:

   $$ z = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-120)}}{2(1)} = \frac{2 \pm \sqrt{4 + 480}}{2} = \frac{2 \pm \sqrt{484}}{2} = \frac{2 \pm 22}{2} $$

   This gives:

   $$ z = 12 \quad \text{or} \quad z = -10 $$

   Since $z = x^2$, and $x^2 \geq 0$, we discard $z = -10$. Thus, $z = 12$.

   Therefore, $x^2 = 12$, so:

   $$ x = \sqrt{12} $$

3. **Use the second derivative test to verify it's a minimum:**

   We now calculate the second derivative of $g(x)$:

   $$ g''(x) = \frac{4}{x^3} + \frac{480}{x^5} $$

   Substitute $x = \sqrt{12}$ into the second derivative:

   $$ g''(\sqrt{12}) = \frac{4}{(\sqrt{12})^3} + \frac{480}{(\sqrt{12})^5} $$

   Since $\sqrt{12} = 2\sqrt{3}$, we simplify the terms:

   $$ g''(\sqrt{12}) = \frac{4}{12\sqrt{12}} + \frac{480}{144\sqrt{12}} $$

   Both terms are positive, meaning $g''(\sqrt{12}) > 0$.

   Since $g''(\sqrt{12}) > 0$, the function is concave up at $x = \sqrt{12}$, confirming that this critical point is a minimum.

4. **Evaluate $g(x)$ at $x = \sqrt{12}$:**

   Now, substitute $x = \sqrt{12}$ into the original function $g(x)$ to find the corresponding value of $g(x)$:

   $$ g(\sqrt{12}) = \sqrt{12} + \frac{2}{\sqrt{12}} + \frac{40}{(\sqrt{12})^3} $$

   Simplify the expression:

   $$ g(\sqrt{12}) = \sqrt{12} + \frac{2}{\sqrt{12}} + \frac{40}{12 \sqrt{12}} $$

   \begin{align*}
   g(\sqrt{12}) &= \sqrt{12} + \frac{2}{\sqrt{12}} + \frac{40}{12 \sqrt{12}} \\
                &= 2\sqrt{3} + \frac{2}{2\sqrt{3}} + \frac{40}{12 \cdot 2\sqrt{3}} \\
                &= 2\sqrt{3} + \frac{1}{\sqrt{3}} + \frac{40}{24\sqrt{3}} \\
                &= 2\sqrt{3} + \frac{1}{\sqrt{3}} + \frac{5}{3\sqrt{3}} \\
                &= 2\sqrt{3} + \left( \frac{1}{\sqrt{3}} + \frac{5}{3\sqrt{3}} \right) \\
                &= 2\sqrt{3} + \frac{1 \cdot 3 + 5}{3\sqrt{3}} \\
                &= 2\sqrt{3} + \frac{8}{3\sqrt{3}} \\
                &= 2\sqrt{3} + \frac{8}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} \\
                &= 2\sqrt{3} + \frac{8\sqrt{3}}{9} \\
                &= \sqrt{3} \left( 2 + \frac{8}{9} \right) \\
                &= \sqrt{3} \cdot \left(\frac{18}{9} + \frac{8}{9}\right) \\
                &= \sqrt{3} \cdot \frac{26}{9} \\
   g(\sqrt{12}) &= \frac{26\sqrt{3}}{9}
   \end{align*}

   Therefore, the absolute minimum value of $g(x)$ occurs at $x = \sqrt{12}$.

### Conclusion:

Since the second derivative at $x = \sqrt{12}$ is positive, this confirms that $x = \sqrt{12}$ is a minimum. The absolute minimum value of $g(x) = x + \frac{2}{x} + \frac{40}{x^3}$ on $(0, \infty)$ is achieved at $x = \sqrt{12}$, and the value is:

$$ g(\sqrt{12}) = \frac{26\sqrt{3}}{9} $$



</details>

# Question 4. Find the absolute minimum and maximum value of the function on the interval $[0,10]$. $$f(x)=x^2-6x+7$$

<details> <summary> Example: </summary>

Find the absolute minimum and maximum value of the function on the interval $[0,9]$. $$f(x)=x^2-8x+9$$
    
### Solution:

To find the absolute minimum and maximum values on the interval $[0, 9]$, follow these steps:

1. **Find the critical points of the function:**

   First, take the derivative of $f(x)$:

   $$ f'(x) = 2x - 8 $$

   Set the derivative equal to zero to find the critical points:

   $$ 2x - 8 = 0 $$

   Solving for $x$:

   $$ x = \frac{8}{2} = 4 $$

2. **Evaluate the function at the critical points and the endpoints:**

   We now evaluate $f(x)$ at the critical point $x = 4$, and at the endpoints of the interval, $x = 0$ and $x = 9$.

   - At $x = 0$:

     $$ f(0) = 0^2 - 8(0) + 9 = 9 $$

   - At $x = 9$:

     $$ f(9) = 9^2 - 8(9) + 9 = 81 - 72 + 9 = 18 $$

   - At $x = 4$:

     $$ f(4) = 4^2 - 8(4) + 9 = 16 - 32 + 9 = -7 $$

3. **Determine the absolute minimum and maximum values:**

   From the calculations, the function values are:

   - $f(0) = 9$
   - $f(9) = 18$
   - $f(4) = -7$

   Therefore, the **absolute minimum value** is $-7$ at $x = 4$, and the **absolute maximum value** is $18$ at $x = 9$.

### Conclusion:

- The **absolute minimum value** of $f(x)$ on the interval $[0, 9]$ is $-7$ at $x = 4$.
- The **absolute maximum value** of $f(x)$ on the interval $[0, 9]$ is $18$ at $x = 9$.

</details>