question archive The Intermediate Value Theorem (IVT) says functions that are continuous on an interval [a,b][a,b] take on all (intermediate) values between their extremes

Subject:MathPrice: Bought3

The Intermediate Value Theorem (IVT) says functions that are continuous on an interval [a,b][a,b] take on all (intermediate) values between their extremes. The Extreme Value Theorem (EVT) says functions that are continuous on [a,b][a,b] attain their extreme values (high and low).

Here's a statement of the EVT: Let ff be continuous on [a,b][a,b]. Then there exist numbers c,d∈[a,b]c,d∈[a,b] such that f(c)≤f(x)≤f(d)f(c)≤f(x)≤f(d) for all x∈[a,b]x∈[a,b]. Stated another way, the "supremum " MM and "infimum " mm of the range {f(x):x∈[a,b]}{f(x):x∈[a,b]} exist (they're finite) and there exist numbers c,d∈[a,b]c,d∈[a,b] such that f(c)=mf(c)=m and f(d)=Mf(d)=M.

Note that the function ff must be continuous on [a,b][a,b] for the conclusion to hold. For example, if ff is a function such that f(0)=0.5f(0)=0.5, f(x)=xf(x)=x for 0<x<10<x<1, and f(1)=0.5f(1)=0.5, then ff attains no maximum or minimum value on [0,1][0,1]. (The supremum and infimum of the range exist (they're 1 and 0, respectively), but the function never attains (never equals) these values.)

Note also that the interval must be closed. The function f(x)=xf(x)=x attains no maximum or minimum value on the open interval (0,1)(0,1). (Once again, the supremum and infimum of the range exist (they're 1 and 0, respectively), but the function never attains (never equals) these values.)

The function f(x)=1xf(x)=1x also does not attain a maximum or minimum value on the open interval (0,1)(0,1). Moreover, the supremum of the range does not even exist as a finite number (it's "infinity").

Here's a statement of the IVT: Let ff be continuous on [a,b][a,b] and suppose f(a)≠f(b)f(a)≠f(b). If vv is any number between f(a)f(a) and f(b)f(b), then there exists a number c∈(a,b)c∈(a,b) such that f(c)=vf(c)=v. Moreover, if vv is a number between the supremum and infimum of the range {f(x):x∈[a,b]}{f(x):x∈[a,b]}, then there exists a number c∈[a,b]c∈[a,b] such that f(c)=vf(c)=v.

If you draw pictures of various discontinuous functions, it's pretty clear why ff needs to be continuous for the IVT to be true.