Suppose that g" is continuous on [a,b] and that g has three zeros in the interval. Show that g" has at least one zero in (a,b). Generalize your result.

Answer 1

Andrea

Suppose g(x) has zeros at z1, z2, z3 where a<=z1 < z2 < z3<=b

Consider the interval (z1,z2). g(x) must be >0 or <0 throughout this interval (if not there would be another zero between z1 and z2). That means there must be at least one turning point of g(x) in the interval (z1,z2), in other words at least one point in (z1,z2) where g'(x) = 0. [Draw a picture to convince yourself of this. mathematically linked to intermediate value theorem].

So let z4 be one of the points in (z1,z2) where g' (x) = 0

Similar argument for (z2,z3). There is a point z5 in (z2,z3) where g'(x)=0.

So there are at least two zeroes of g'(x) in (z1,z3).

Consider the interval (z4,z5) now. g'(z4) and g'(z5) are both zero. There must be at least one turning point of g'(x) between z4 and z5. In other words there must be at least one zero of g''(x) between z4 and z5.

To generalise : if we have 3 zeroes of g(x), we know g'' (x) has at least one zero.

If we had 4 zeroes, 5 zeroes ... what can we say about g''' (x), g'''' (x) etc.

Mike