# Beginning Numerical Analysis question. Still calculus

1. Oct 1, 2013

### jaqueh

1. The problem statement, all variables and given/known data
Show f(x)=(x−2)sinxln(x+2) has f'(x)=0 somewhere on [-1,3]

3. The attempt at a solution
I tried using Rolle's theorem, but f(-1)≠f(3). Then I tried the mean value theorem, but didn't get 0 either.

2. Oct 1, 2013

### Dick

Check some obvious values in between -1 and 3. Maybe x=0 and x=2? You are giving up too easily.

3. Oct 1, 2013

### pasmith

You can find f'(x) analytically, so maybe you should do that and then apply the intermediate value theorem.

4. Oct 1, 2013

### jackmell

Well if it's numerical analysis, why are you using those theorems? Suppose it was some function that you didn't even know what the form was, how could you show the derivative is zero somewhere in that interval? If it was mine, I'd generate "numerically" since this is numerical analysis, a set of equally spaced points in that interval and then compute the value of the function and then I'd inspect the list for the "monotonicity" change, that is, when the numbers are increasing then decreasing or vice-a-versa. However if there is an inflection point, the derivative could still be zero without this change in monotonicity. But we could at least rule out the former case this way. Anyway, lots of tough problems won't give you the luxury of using nice theorems and you'll have to muscle-through the data in this way or another.

I got an idea, why don't you write a short Mathematica program to generate these numbers then pick out the monotonicity change?

5. Oct 1, 2013

### jaqueh

I got it, make [0,1] a subset of [-1,3] then i can use Rolle's theorem and then determine that there must be a c in [0,1] so there's gotta be a c in [-1,3]

6. Oct 1, 2013

### LCKurtz

What does the interval [0,1] do for you?

7. Oct 2, 2013

### jaqueh

sorry i meant for the interval to be [0,2]
then i get f(0)=f(2)=0 => rolle's

Last edited: Oct 2, 2013
8. Oct 2, 2013

### LCKurtz

That's better.