Question

Assume f is differentiable on [-1,1] and f'(-1)=f'(1)=1 and assume that f(-1)=f(1)=0.

a) Explain why f has a zero in it (-1,1)
b) Explain why -f' must have at least two zeros in (-1,1)

I need to know this for a test, but i have no idea how to do it. Can you someone guide me through and show me work too?

Answer 1

you can see it from the picture, lets see if we can put in in math

first off, since \(f\) in increasing on some interval \((-1,c)\) and increasing one some interval \((d,1)\) and since \(f(-1)=f(1)\) then on \((-1,c)\) \(f>0\) and on \((d,1),f<0\)
since \(f\) is continuous, (because it is differentiable) and takes on positive and negative values, it must be zero somewhere

Answer 2

now you need to show \(f'\) has two zeros, mvt shows that is has one, since \(f(-1)=f(1)\)
we know \(f\) is increasing on some open interval to the right of \(-1\) and to the left of \(1\) and since \(f(-1)=f(1)\) it must be decreasing somewhere in \((-1,1)\)
since it is continuous, increases, decreases, increases, (at the very least, it could go up and down and up and down etc) it must have a relative max and a relative min in the interval. since it is differentiable, the derivative must be zero at those two points, and so \(f'\) has two zeros, not just one
you can probably say this more eloquently

Answer 3

By part (1) there is a root between -1 and 1. call it a

by the mvt

\[0=\frac{f(a)-f(-1)}{a-(-1)}=f'(c_1)\text { with }-1<c_1<a\]

and

\[0=\frac{f(1)-f(a)}{1-a}=f'(c_2)\text { with }a<c_2<1\]

Answer 4

The function has a positive slope at both endpoints of the interval. Taking into account that the function is differentiable, that indicates f(-1 + epsilon) is positive, and f(1-epsilon) is negative. Apply the mean value theorem to conclude the function has a zero on the interval.