Existence Theorems

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

Refer to Khan academy: Existence theorems intro

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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!

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Refer to Maths if fun: Intermediate Value Theorem Refer to video: Intermediate Value Theorem Explained

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.

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Refer to Khan lecture: Extreme value theorem Refer to video: Extreme Value Theorem

Mean Value Theorem (MVT)

Refer to Khan academy article: 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

Conditions for applying MVT:

• Continuous over interval (a, b)

• Differentiable over interval [a, b]

Graph

Example

Solve:

• He's totally right.