# Extrema: Maxima & Minima

`Extrema` are one type of Critical points, which includes `Maxima` & `Minima`. And there're two types of Max and Min, `Global Max & Local Max`, `Global Min & Local Min`. We can all them `Global Extrema` or `Local Extrema`.

And actually we can call them in different ways, e.g.:

* `Global Max` & `Local Max` or in short of `glo max` & `loc max`
* `Absolute Max` & `Relative Max` or in short of`abs max` & `rel max`

![image](https://user-images.githubusercontent.com/14041622/40529128-14cde61c-6026-11e8-8ebc-3dadc933afac.png)

### How to identify Extrema

We need two kind of conditions to identify the Max or Min. Now If we have a Non-Endpoint Minimum or Maximum point at `a`, then it must satisfies these conditions:

* **Geometric condition:** (It should be understood in a more intuitive way)
  * in the interval `[a-h, a+h]` there's no point above or below `f(a)` or
  * `f'(a-h)` & `f'(a+h)` have different sign, one negative another positive.
* **Derivative condition:**
  * `f'(a) = 0` or&#x20;
  * `f'(a) is undefined`

### How to find Extrema

[Refer to Khan academy lecture: Finding critical points](https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-analyze-functions/modal/v/finding-critical-numbers)

We just need to assume `f'(x) = 0` or `f'(x) is undefined`, and solve the equation to see what `x` value makes it then.

## `Increasing & Decreasing Intervals`

We can easily tell at a point of a function, it's at the trending of increasing or decreasing, by just looking at the `instantaneous slope` of the point, aka. the derivate. If the derivative, the slope is positive, then it's increasing. Otherwise it's decreasing.

### Finding the trending at a point

Just been said above, we assume at point `a`, it's value is `f(a)`. So the slope of it is `f'(a)`. And if `f'(a) < 0`, then it's decreasing; If `f'(a) > 0`, then it's increasing.

### `Finding a decreasing or increasing interval`

[Refer to Khan lecture.](https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-analyze-functions/modal/v/increasing-decreasing-intervals-given-the-function)

It's just doing the same thing in the opposite way. For find a decreasing interval, we assume `f'(x) < 0`, and by solving the inequality equation we will get the interval.

Strategy:

* Get Critical points.
* Separate intervals according to `critical points`, `undefined points` and `endpoints`.
* Try easy numbers in EACH intervals, to decide its TRENDING (going up/down):
  * If `f'(x) > 0` then the trending of this interval is **Increasing**.
  * If `f'(x) < 0` then the trending of this interval is **Decreasing**.

### Example

![image](https://user-images.githubusercontent.com/14041622/40609152-624738b8-62a0-11e8-8725-509e4e5a78a3.png) Solve:

* Set `f'(x) = 0 or undefined`, get `x = -2 or -1/3`
* Separate intervals to `(-∞, -2, -1/3, ∞)`

  ![image](https://user-images.githubusercontent.com/14041622/40609259-ad80e068-62a0-11e8-81cd-f4bdb5070dbe.png)
* Try some easy numbers in each interval: `-3, -1, 0`:

  ![image](https://user-images.githubusercontent.com/14041622/40609319-ddc615cc-62a0-11e8-90e5-8829f92d6e04.png)

## `How to find Relative Extrema`

> Remember that an `Absolute extreme` is also a `Relative extreme`.

[Refer to khan: Worked example: finding relative extrema](https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-analyze-functions/modal/v/finding-relative-maximum-example) [Refer to Khan Academy article: Finding relative extrema](https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-analyze-functions/modal/a/applying-the-first-derivative-test-to-find-extrema)

![image](https://user-images.githubusercontent.com/14041622/40531186-21ecfd44-602e-11e8-8018-0409609a6167.png)

Strategy:

* Get Critical points.
* Separate intervals according to `critical points`, `undefined points` and `endpoints`.
* Try easy numbers in EACH intervals, to decide its TRENDING (going up/down).
* Decide each critical point is Max, Min or Not Extreme.

[Refer to an awesome article: Using calculus to learn more about the shapes of functions](http://xaktly.com/CurveSketching.html)

![image](https://user-images.githubusercontent.com/14041622/41715223-51a450b6-7585-11e8-82bf-69c81598989f.png)

### Example

![image](https://user-images.githubusercontent.com/14041622/40608781-3c071840-629f-11e8-90b1-0225538eb324.png) Solve:

* Set `f'(x)=0 or undefined`, get `x=0 or -2 or 1`
* Separate intervals to `(-∞, -2, 0, 1, ∞)`

  ![image](https://user-images.githubusercontent.com/14041622/40608852-752ce6ea-629f-11e8-8b65-d15ac105fb52.png)
* Try easy number in each interval: `-3, -1, 0.5, 2` and get the trendings:

  ![image](https://user-images.githubusercontent.com/14041622/40608921-af3594c2-629f-11e8-9cc4-a39b476a14e4.png)
* Identify critical points' **concavity**:

  ![image](https://user-images.githubusercontent.com/14041622/40608947-c1bfb44c-629f-11e8-9e7b-b4e38185c1ec.png)

## `How to find Absolute Extrema`

[Refer to Khan academy article: Absolute minima & maxima review](https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-analyze-functions/modal/a/absolute-minima-and-maxima-review) [Refer to Khan academy lecture: Finding absolute extrema on a closed interval](https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-analyze-functions/modal/v/using-extreme-value-theorem)

Strategy:

* Find all `Relative extrema`
  * Get Critical points.
  * Separate intervals according to critical points & endpoints.
  * Try easy numbers in EACH intervals, to decide its TRENDING (going up/down).
  * Decide each critical point is Max, Min or Not Extreme.
* Input all the extreme point into original function `f(x)` and get extreme value.

### Example

![image](https://user-images.githubusercontent.com/14041622/40609657-e09e250e-62a1-11e8-9e43-f1a69c217231.png) Solve:

* Set `g'(x) = 0 or undefined` get `x=0`
* Separate intervals to `[-2, 0, 3]` according to the critical point & endpoints of the given condition.
* Try some easy numbers of each interval `-1, 2` into `g'(x)`
* Get the trending of each interval: `(+, +)`
* So the minimum must be the left endpoint of the given condition, which is `x=-2`.

### Example

![image](https://user-images.githubusercontent.com/14041622/40618317-bbc9bb7c-62c3-11e8-9723-962c727c7d4d.png) Solve:

* Differentiate to set `f'(x)=0` and got `x=0, -1, 1`
* Separate intervals to `(-∞, -1, 0, 1, +∞)`
* Try some easy numbers in each interval for `f'(x)` and got the signs: `-, +, -, +`
* According to the signs we know that `-, +` gives us a `Relative maximum`
* But at the end it's `+` again, and the interval is going up to `+∞`, means `f(x)` will go infinitely high.
* So there's NO Absolute maximum.
