Summary
Highlights
A function f is continuous at a number c if three conditions are satisfied: f(c) is defined, the limit of f(x) as x approaches c exists, and the limit of f(x) as x approaches c equals f(c).
Discontinuities are categorized into three types: removable discontinuity, where the function has a hole or a point moved from its expected place; jump discontinuity, where the function 'jumps' from one value to another; and infinite discontinuity, where the function approaches infinity at a certain point.
The lecture provides various examples with graphs to distinguish between continuous functions and different types of discontinuities: a continuous straight line, a removable discontinuity with a hole, another removable discontinuity where a point is displaced, an infinite discontinuity in 1/x, and a jump discontinuity in the absolute value function.
Polynomial functions are always continuous at any real number. Rational functions are continuous at all points except where the denominator is zero. To find points of discontinuity for a rational function, set the denominator to zero and solve for x.
For the function f(x) = (x^2 - 4) / (x^3 + x^2 - 2x), the discontinuities are found by setting the denominator to zero, yielding x = 0, x = -2, and x = 1 as points of discontinuity.
To prove a function is continuous on a closed interval [a, b], it must first be shown as continuous on the open interval (a, b). Additionally, specific one-sided limits must be checked at the endpoints: the limit as x approaches 'a' from the right must equal f(a), and the limit as x approaches 'b' from the left must equal f(b).
To prove f(x) = sqrt(9 - x^2) is continuous on [-3, 3], one must first show it's continuous on the open interval (-3, 3) by checking the three continuity conditions for an arbitrary 'c'. Then, for the endpoints, verify that lim (x->-3+) f(x) = f(-3) and lim (x->3-) f(x) = f(3).
If functions f and g are continuous at point c, then their sum, difference, and product are also continuous at c. Their quotient, f/g, is continuous at c, provided g(c) is not zero.
If g is continuous at c, and f is continuous at g(c), then the composite function (f ∘ g) is continuous at c. This means that the limit of f(g(x)) as x approaches c is F(g(c)).
The Intermediate Value Theorem states that if a function f is continuous on a closed interval [a, b], and if 'w' is any number strictly between f(a) and f(b), then there exists at least one number 'c' in the open interval (a, b) such that f(c) = w.
The IVT can prove the existence of a zero for a function within an interval. For f(x) = x^5 + 2x^4 - 6x^3 + 2x - 3, we show there's a zero between 1 and 2. Since f(1) is negative and f(2) is positive, and the function is continuous (as it's a polynomial), by IVT, there must be a point 'c' between 1 and 2 where f(c) = 0.