Question

The function g is continuous on the interval [a, b] and is differentiable on (a, b).

Suppose that g(x) = 0 for 5 distinct values of x in (a, b).
What is the minimum number, k, of z in (a, b) such that g'(z) = 0? k = __

Answer 1

k = 4

Explanation:

You want the minimum number of values z in (a, b) such that g'(z) = 0 if g(x) = 0 for 5 distinct values of x in (a, b).

Direction

Each distinct value of x such that g(x)=0 will correspond to a change in direction from g(x) approaching 0 to g(x) moving away from 0. Then, if there is to be another such value of x, there must be a point x=z where g(x) changes from "moving away" from the x-axis to "moving toward" the x-axis.

That turning point will correspond to a change in sign of g'(x). Since g(x) is differentiable, g'(x) is continuous and the intermediate value theorem guarantees there is a point associated with that sign change such that g'(z) = 0.

Turning points

The 5 distinct x-intercepts mean there must be at least 4 turning points:

k = 4