Pre-requisite Know how to take the limit of a function Basic algebra

Criteria for continuity at a point:

Criteria for continuity at a point If f is continuous at c then: f(c) is defined exists = f(c) *All must be true

Continuity on an open interval:

Continuity on an open interval A function is continuous on an open interval ( a,b ) if it’s continuous at each point.

Is the function f(x) continuous when x =0?:

Is the function f(x) continuous when x =0? f(c) is defined exists = f(c) c = 0 f(0) =-1 does not exist The function f(x ) is discontinuous when x=0.

Is the function f(x) continuous when x =2?:

Is the function f(x) continuous when x =2? f(c) is defined exists = f(c) c = 2 f(2) = 8 = 8 = f(2) The function f(x ) is continuous when x=2.

Common mistake:

Common mistake Not following all three rules of continuity Is the function f(x) continuous when x =-2 ? f(-2) = 2 =6 f (-2) Since rule number three is not true, f(x ) is not continuous at -2.

Helpful hint:

Helpful hint f(x) is not continuous at a point if you have to pick up your pencil while graphing the point Is the function f(x) continuous when x =-2? Since you have to pick up your pencil at -2, f(x) is not continuous at -2.

Intermediate Value Theorem(IVT):

Intermediate Value Theorem(IVT) If a function is continuous on the closed interval [ a,b ] and w is any number between f(a) and f(b) then there is at least one number w in [ a,b ] such that f(c) = w.

Use the IVT to show that there is a root of the given equation in the specified interval:

Use the IVT to show that there is a root of the given equation in the specified interval (-2,2) f(x) is continuous between -2 and 2 because it’s a polynomial so it’s continuous everywhere Plug in the values -5 = 3 Since f(-2)= -5 and f(x) is continuous on the interval, there exists a number c in [-2,2] such that f(c) = 0 by the IVT.

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