MHB Finding domain of a trigonometric function

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The function f(x) = (1+x) / e^(cos(x)) is defined for all x. The numerator, 1+x, is defined for all real numbers. The denominator, e^(cos(x)), is always positive since e^(cos(x)) is never zero. Therefore, there are no restrictions on the domain. The domain of the function is all real numbers.
tmt1
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I need to find the domain of this function:

$$f(x) = \frac{1+x}{ e^{cos(x)}}$$

I set $${ e^{cos(x)}}> 0$$

But I'm not sure what to do after this.
 
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tmt said:
I need to find the domain of this function:

$$f(x) = \frac{1+x}{ e^{cos(x)}}$$

I set $${ e^{cos(x)}}> 0$$

But I'm not sure what to do after this.

The top is obviously defined for all x.

The bottom, as you have established, is always positive. As both cos(x) and e^x are defined for all x, so is their composition.

So the top and bottom are defined for all x. The only place where the quotient of the two functions might not be defined is where the denominator is 0. But you already established that this doesn't happen anywhere.

So what is the domain of your function?
 
Prove It said:
The top is obviously defined for all x.

The bottom, as you have established, is always positive. As both cos(x) and e^x are defined for all x, so is their composition.

So the top and bottom are defined for all x. The only place where the quotient of the two functions might not be defined is where the denominator is 0. But you already established that this doesn't happen anywhere.

So what is the domain of your function?

All real numbers?
 
tmt said:
All real numbers?

Correct :)
 
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