# Homework Help: Math question

1. Dec 24, 2004

### UrbanXrisis

I'm not quite sure what the question is asking:
"If $$f(x)=e^xsin(x)$$, then the number of zeros of $$f$$ on the interval [o,2pi] is?"

I took the derivative of this and found where it was equal to zero:
$$f(x)=e^xsin(x)$$
$$f'(x)=e^xsin(x)+e^xcos(x)$$
$$0=sin(x)+cos(x)$$

I got zero. However, tha answer is 3, any suggestions?

2. Dec 24, 2004

### Nylex

No, it wants where the function is zero, not where the gradient of the curve is zero. Where is sin x zero in that interval?

3. Dec 24, 2004

### UrbanXrisis

you mean when $$0=e^xsin(x)$$?

4. Dec 24, 2004

### dextercioby

$$e^{x}\sin x=0\Rightarrow \sin x=0$$
Solve the last equation on the interval $[0,2\pi]$

Daniel.

5. Dec 24, 2004

### Nylex

Yes, yes I do (sorry had to lengthen my post).

6. Dec 24, 2004

### UrbanXrisis

it's is zero when x=0, x=pi, x=2pi

so it hits three times!

thanks! Also, why doesnt the e^x make a difference?

7. Dec 24, 2004

### dextercioby

It never annulates.Not even for complex arguments.

Daniel.

EDIT:Cause it never annulates,it does not affect the zero-s of the function.Plot the graph of 'f'.U'll see quite an interesting behavior.It has no limit for x->+infty.At minus infty it goes to zero.

Last edited: Dec 24, 2004
8. Dec 24, 2004

### UrbanXrisis

so e^x only increases the sinX amplitude, never now far it streatchs, so it doesnt effect how many times sinX crosses the x-axis

9. Dec 25, 2004

### HallsofIvy

"Annulates"??? I assume you mean "is never equal to 0" but the only definition I can find of "annulate" is "ring shaped".

UrbanXrises: It's not so much that it is 'always increasing'. In order to solve AB= 0, you solve A= 0 and B= 0. Since ex is never 0, The only solutions of exsin(x)= 0 are where sin(x)= 0.

Last edited by a moderator: Dec 25, 2004