There are two basic facts about continuous functions that we should spell out here. They are not really astounding, but to prove them precisely is not as trivial as it may seem.
Theorem 1 If f is continuous at b and lim x® a g(x) = b, then lim x® a f(g(x)) = f(b) = f æè lim x® a g(x) öø .
Theorem 2 If g is continuous at a and f is continuous at g(a) , then f°g(x) = f(g(x)) is continuous at a .
Theorem 3 If f is continuous on a closed interval [a,b] , then there is a point x0 Î [a,b] where f is largest ( f(x) £ f(x0) for all x Î [a,b] ), and a point x1 Î [a,b] where f is smallest ( f(x) ³ f(x0) for all x Î [a,b] ).
The hard part of this fact is that it doesn't tell you where the maximum or minimum might be, just that they do exist.
If f is continuous on a closed interval [a,b] . Let m be a number between f(a) and f(b) . Then, there is a c Î [a,b] so that f(c) = m .
So, for example, if the function is continuous, and at some point its value is positive, while at another point the value is negative, then there must have been some point in between where the value was 0.
Theorem 4 If f(x) is continuous at x0 , and f(x0) > 0 , then there is a d > 0 so that f(x) > 0 for |x-x0| < d.
The strength of these theorems is not really in these examples. They will be used in the rest of this course, from time to time, as theoretical tools to help justify other facts. Certainly, when we get to the chapter on max/min problems we will rely on the ``max/min'' theorem above to let us know that a maximum or a minimum actually does exist.
Classification of discontinuities
Exercise 1 If f(x) = cos(x) , find the maximum and minimum of f on [-p/2,p] .
Exercise 2 If f(x) = x2 , on the interval [-1,2) , then does f have a maximum? How does this relate to the theorem?
Example 1 Show that f(x) = x7+11x3-15x+1 has at least one root in [0,1] .
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Copyright (c) 2000 by David L. Johnson. File translated from TEX by TTH, version 2.61.On 18 Oct 2000, 00:55.
Copyright (c) 2000 by David L. Johnson.