An example case is illustrated by the sin(x) and cos(x) functions between π/2 and 2π (see link at end of post).
"Intermediate value theorem":
"for each value between the least upper bound and greatest lower bound of the image of a continuous function there is at least one point in its domain that the function maps to that value".
Since p≠q, either p>q or q>p. Assume p>q, b>a.
Then (a), and f(b)<g(b), so if
h(x)=f(x)-g(x)
then h(a)>0, and h(b)<0.
You will need to prove, using the intermediate value theorem, that h(x0)=0 at some point a<x0<b.
Proof is similar when p<q.
http://imageshack.us/photo/my-images/821/1316321469.png/
Suppose that f and g are two functions both continuous on the
interval [a, b], and such that (b) = p and f(b) = g(a) = q
where p does not equal to q. Sketch typical graphs of two such functions . Then apply the intermediate value theorem to the function
h(x) = f(x) - g(x) to show that f(c) = g(c) at some point c of
(a, b).
1 answer
1 answer