# Derivatives and Zeros

1. Apr 10, 2004

### Caldus

Got some questions:

Say I have the function f(x) = x*sin(3x/4).

1. What is the relationship between the sign of f'(x) and the graph of f(x)?
2. What can be observed about the relative locations of the roots of f(x) and f'(x)?
3. If f has zeros at r1 and r2 and if f is differentiable on the interval [r1, r2], then f' must...?

I cannot for the life of me figure out any of these. For the first question, I'm guessing that f increases whenever f' is positive and f decreases when f' is negative?

Thanks for any assistance.

2. Apr 11, 2004

### Chen

Do you know what the meaning of the derivative is? That will answer the first question. For the second you will have to find the derivative and see for yourself. As for the third one... what does the function look like between r1 and r2? What must it have in that interval?

3. Apr 11, 2004

### HallsofIvy

Your "guess" is correct for the first one since positive derivative means the tangent line has positive slope and negative derivative means the tangent line has negative slope.

For the second one, have a look at the Mean Value Theorem (or, more specifically, Rolle's theorem).

4. Apr 11, 2004

### Caldus

I think I know the last one now. If f(x) contains 2 zeros, then its derivative must contain 2 zeros as well, correct?

I still cannot figure out the second one though.

5. Apr 11, 2004

### Caldus

No...I think I know now. For the second and last one, it is based on the fact that the derivative function between two zeros of the original function must pass through the x-axis at least once if the original function is differentiable between those two zeros.

6. Apr 11, 2004

### Chen

That is a rather awkward way of putting it, but yes. What this means is that the derivative gets a value of 0 at least once, so that the original function has a minimum or maximum point in that interval.

7. Apr 12, 2004

### HallsofIvy

Basically, that is Rolle's theorem: if f is continous on [a,b], differentiable on (a,b), f(a)= 0, and f(b)= 0, then for some c between a and b, f'(c)= 0.

That's a special case of the "mean value theorem": as long as f is continuous on [a,b] and differentiable on (a,b), then there exist a number c between a and b such that f'(c)= (f(b)-f(a))/(b-a).

8. Apr 12, 2004

### kioria

This MVT is a very obvious one... if there are two roots, at x = a, b, and the function is c.t.s on [a,b], then there must exist a point x = c, where f'(c) = 0.