Finding the Zeros of f(x)=e^xsin(x) on [0,2π]

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SUMMARY

The function f(x) = e^x sin(x) has three zeros on the interval [0, 2π]. The zeros occur at x = 0, x = π, and x = 2π. The exponential component e^x does not affect the number of zeros since it is never equal to zero, thus the zeros are solely determined by the sine function. The derivative f'(x) = e^x(sin(x) + cos(x)) was incorrectly used to find the zeros, which led to confusion regarding the function's behavior.

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UrbanXrisis
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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?
 
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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?
 
you mean when 0=e^xsin(x)?
 
UrbanXrisis said:
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?

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

Daniel.
 
UrbanXrisis said:
you mean when 0=e^xsin(x)?


Yes, yes I do (sorry had to lengthen my post).
 
it's is zero when x=0, x=pi, x=2pi

so it hits three times!

thanks! Also, why doesn't the e^x make a difference?
 
UrbanXrisis said:
it's is zero when x=0, x=pi, x=2pi
so it hits three times!
thanks! Also, why doesn't the e^x make a difference?

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:
so e^x only increases the sinX amplitude, never now far it streatchs, so it doesn't effect how many times sinX crosses the x-axis
 
"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.
 
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