[toc]

## Critical Value

$x=c$ is a critical value of $f(x)$ when both are true:
1. $f'(c)=0$ or $f'(c)$ is undefined.
2. $f(c)$ exists.

### Example

$f(x)=x^2-6x+3$

$f'(x)=2x-6$

Set $f'(x)=0$ and solve:

$2x-6=0$
$2x=6$
$x=3$

Plug in $x=3$ into the original function to check that the $y$-value exists:

$f(3)=3^2-6(3)+3=9-18+3=-6$

Since $f'(3)=0$ and $f(3)=-6$ which exists, $x=3$ is a critical number of $f$.

## Absolute Maximum

An Absolute Maximum is sometimes called a Global Maximum. It is where the function's $y$ value achieves its largest value. It is the tallest mountain in the given interval $[a,b]$.

## Absolute Minimum

An Absolute Minimum is sometimes called a Global Minimum. It is where the function's $y$ value achieves its lowest value. It is the deepest valley in the given interval $[a,b]$.

- Where do these absolute minimums and maximums happen?

## Examples
| Graph | Left Endpoint | Critical Value in between | Right Endpoint |
| -------- | -------- | -------- |-------- |
|  | $x=-4$ and $y=1$ | **Absolute Maximum** $x=-1$ and $y=4$ | **Absolute Minimum** $x=3$ and $y=-1$ |
| Graph | Left Endpoint | Critical Value in between | Right Endpoint |
|  | **Absolute Minimum** $x=-3$ and $y=-2$ | | **Absolute Maximum** $x=5$ and $y=4$ |
| Graph | Left Endpoint | Critical Value in between | Right Endpoint |
|  | **Absolute Minimum** $x=-1$ and $y=0$ | **Absolute Maximum** $x=2$ and $y=3$ | $x=3$ and $y=2$ |
| Graph | Left Endpoint | Critical Value in between | Right Endpoint |
| | **Absolute Maximum** $x=-4$ and $y=2$ | $x=-2$ and $y=-1$ | **Absolute Minimum** $x=1$ and $y=-4$ |
| Graph | Left Endpoint | Critical Value in between | Right Endpoint |
|  | $x=-3$ and $y=1$ | Absolute Minimum $x=-1$ and $y=-1$ | Absolute Maximum $x=4$ and $y=4$ |

$f(x)$ only has absolute minimums or absolute maximums either at the <details> endpoints </details>
or somewhere in between when <details> $f'(c)=0$ </details>

---

## To find where $f(x)$ has an absolute maximum and minimum on interval $[a,b]$.

### Step 1. Find $f'(x)$.
### Step 2. Find $x$ when $f'(x)=0$ or is undefined. Throw out any $x$ values that are outside the interval $[a,b]$. Say there's only one critical value $c$ such that $f'(c)=0$.
### Step 3. Plug $c$ into the original function to get the $y$-value.
### Step 4. Plug the endpoints $a,b$ into the original function to get the $y$-value.
### Step 5. The largest $y$-value is the Absolute Maximum of $f(x)$ on $[a,b]$.
### Step 6. The smallest $y$-value is the Absolute Minimum of $f(x)$ on $[a,b]$.

---

## Example 1: Find the Absolute Minimum and Maximum of $f(x)=x^2-6x+3$ on the interval $[1,4]$

### Step 1. Find $f'(x)$.

<details> <summary> Solution: </summary>

$f'(x)=2x-6$

</details>

### Step 2. Find $x$ when $f'(x)=0$ or is undefined. Throw out any $x$ values that are outside the interval $[a,b]$. Say there's only one critical value $c$ such that $f'(c)=0$.

<details> <summary> Solution: </summary>

$2x-6=0$
$2x=6$
$x=3$

Note that $3$ is in the interval $[1,4]$ since $1<3<4$.

</details>

### Step 3. Plug $c$ into the original function to get the $y$-value.

<details> <summary> Solution: </summary>

Critical Value in between: $y=f(3)=3^2-6(3)+3=9-18+3=-6$

</details>

### Step 4. Plug the endpoints $a,b$ into the original function to get the $y$-value.

<details> <summary> Solution: </summary>

Note that our interval is $[1,4]$, so $a=1$ and $b=4$.

Left Endpoint $y=f(1)=1^2-6(3)+3=1-6+3=-2$
Right Endpoint $y=f(4)=4^2-6(4)+3=16-24+3=-5$

</details>

### Step 5. The largest $y$-value is the Absolute Maximum of $f(x)$ on $[a,b]$.

<details> <summary> Solution: </summary>

| $x$ | $y=f(x)$ | Min or Max |
| -------- | -------- | -------- |
| $1$ | $-2$ | Absolute Maximum |
| $3$ | $-6$ | Absolute Minimum |
| $4$ | $-5$ | |

$f(x)$ has an absolute maximum value at $x=1$ and $y=-2$.

</details>

### Step 6. The smallest $y$-value is the Absolute Minimum of $f(x)$ on $[a,b]$.

<details> <summary> Solution: </summary>

$f(x)$ has an absolute minimum value at $x=3$ and $y=-6$.


</details>

## Example 2: Find the Absolute Minimum and Maximum of $f(x)=1-x^{2/3}$ on the interval $[-1,8]$

### Step 1. Find $f'(x)$.

<details> <summary> Solution: </summary>

$f'(x)=\dfrac{2}{3}x^{2/3-1}=\dfrac{2}{3}x^{-1/3}=\dfrac{2}{3x^{1/3}}$

$f'(x)=\dfrac{2}{3x^{1/3}}$

</details>

### Step 2. Find $x$ when $f'(x)=0$ or is undefined. Throw out any $x$ values that are outside the interval $[a,b]$. Say there's only one critical value $c$ such that $f'(c)=0$.

<details> <summary> Solution: </summary>

$f'(x)=\dfrac{2}{3x^{1/3}}$ is never zero as the numerator never equals zero.

However, $f'(x)=\dfrac{2}{3x^{1/3}}$ is undefined when the denominator is zero, which is when $x=0$.

Note that $f(0)=1-0^{2/3}=1-0=1$ which exists. So $c=0$ is a critical value of $f$ in the interval $[-1,8]$.

</details>

### Step 3. Plug $c$ into the original function to get the $y$-value.

<details> <summary> Solution: </summary>

Critical Value in between: $y=f(0)=1-0^{2/3}=1-0=1$

</details>

### Step 4. Plug the endpoints $a,b$ into the original function to get the $y$-value.

<details> <summary> Solution: </summary>

Note that our interval is $[-1,8]$, so $a=-1$ and $b=8$.

Left Endpoint $y=f(-1)=1-(-1)^{2/3}=1-1=0$
Right Endpoint $y=f(8)=1-8^{2/3}=1-4=-3$

</details>

### Step 5. The largest $y$-value is the Absolute Maximum of $f(x)$ on $[a,b]$.

<details> <summary> Solution: </summary>

| $x$ | $y=f(x)$ | Min or Max |
| -------- | -------- | -------- |
| $-1$ | $0$ | |
| $0$ | $1$ | Absolute Maximum |
| $8$ | $-3$ | Absolute Minimum |

$f(x)$ has an absolute maximum value at $x=0$ and $y=1$.

</details>

### Step 6. The smallest $y$-value is the Absolute Minimum of $f(x)$ on $[a,b]$.

<details> <summary> Solution: </summary>

$f(x)$ has an absolute minimum value at $x=8$ and $y=-3$.


</details>