---
# System prepended metadata

title: 4.5 - Absolute Minimum and Maximum
tags: ['4.5', lecture]

---

[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 |
| -------- | -------- | -------- |-------- |
| ![image](https://hackmd.io/_uploads/rkbGrINxkx.png)  |   $x=-4$ 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 |
| ![image](https://hackmd.io/_uploads/HyXhB8Vxkg.png)  |   **Absolute Minimum** $x=-3$ and $y=-2$   |      |  **Absolute Maximum** $x=5$ and $y=4$     |
| Graph | Left Endpoint | Critical Value in between | Right Endpoint |
| ![image](https://hackmd.io/_uploads/rJ4TUUVeyg.png)  |   **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 |
| ![image](https://hackmd.io/_uploads/HyrtOUVx1l.png)|   **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 |
| ![image](https://hackmd.io/_uploads/r1hQOUVxJg.png)  |  $x=-3$ 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 $y=-6$.

![image](https://hackmd.io/_uploads/r1z7C8Exyg.png)

</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 $y=-3$.

![image](https://hackmd.io/_uploads/H1vyzwVlyl.png)


</details>
