# 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: 

## 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>