# Existence Theorems

> Existence theorems includes 3 theorems: `Intermediate Value Theorem`, `Extreme Value Theorem`, `Mean Value Theorem`.

[Refer to Khan academy: Existence theorems intro](https://www.khanacademy.org/math/ap-calculus-ab/ab-existence-theorems/modal/v/existence-theorems)

![image](https://user-images.githubusercontent.com/14041622/40358703-b872e8f2-5df2-11e8-9230-88d5a5bb8059.png)

## `Intermediate Value Theorem (IVT)`

The IVT is saying:

> When we have 2 points **connected** by a **continuous** curve: one point **below** the line, the other point **above** the line, then there will be at least one place where the curve crosses the line!

![image](https://user-images.githubusercontent.com/14041622/40359266-b30d756a-5df4-11e8-92be-474ac99a08d4.png)

[Refer to Maths if fun: Intermediate Value Theorem](https://www.mathsisfun.com/algebra/intermediate-value-theorem.html) [Refer to video: Intermediate Value Theorem Explained](https://www.youtube.com/watch?v=9wEHwFrUyOU)

### Find roots by using IVT

`IVT` is often to find `roots` of a function, which means to find the `x value when f(x)=0`. So for finding a root, the definition will be:

> If `f(x)` is continuous and has an interval `[a, b]`, which leads the function that `f(a)<0 & f(b)>0` , then it MUST has a point `f(c)=0` between interval `[a,b]`, which makes a root `c`.

### Example

Tell whether the function `f(x) = x² - x - 12` in interval `[3,5]` has a root. Solve:

* We got that at both sides of intervals: `f(3)=-6 < 0`, and `f(5)=8 > 0`
* So according to the Intermediate Value Theorem, there IS a root between `[3,5]`.
* Calculate `f(c)=0` get the root `c=4`.

## `Extreme Value Theorem (EVT)`

The EVT is saying:

> There **MUST BE** a `Max & Min` value, if the function is continuous over the closed interval.

![image](https://user-images.githubusercontent.com/14041622/40359648-16501adc-5df6-11e8-87df-67f066ec8235.png)

[Refer to Khan lecture: Extreme value theorem](https://www.khanacademy.org/math/ap-calculus-ab/ab-existence-theorems/modal/v/extreme-value-theorem) [Refer to video: Extreme Value Theorem](https://www.youtube.com/watch?v=Sx2lPZlnWfs)

## `Mean Value Theorem (MVT)`

[Refer to Khan academy article: Establishing differentiability for MVT](https://www.khanacademy.org/math/ap-calculus-ab/ab-existence-theorems/modal/a/review-establishing-differentiability-for-mvt)

The MVT is saying:

> There **MUST BE** a `tangent line` that has the same slope with the `Secant line`, if the function is **CONTINUOUS** over `[a,b]` and **DIFFERENTIABLE** over `(a,b)`.

Which also means that, if the conditions are satisfied, then there **MUST BE** a number `c` makes the **derivative** is equal to the `Average Rate of Change` between the two end points.

![Equation](https://user-images.githubusercontent.com/14041622/40409998-308d166c-5e9f-11e8-9871-829dc7f66658.png)

Conditions for applying MVT:

* Continuous over interval `(a, b)`
* Differentiable over interval `[a, b]`

![Graph](https://user-images.githubusercontent.com/14041622/41189731-18eb9ed4-6c05-11e8-9752-33b1c48c0a8a.png)

### Example

![image](https://user-images.githubusercontent.com/14041622/42626694-8a4e235a-85fd-11e8-83e6-58ad47b33760.png) Solve:

* He's totally right.
