if f(x)=e^xcos(1/2x), then the number of zeros of f on the closed interval [o,2pi]is? 0,1,2,3 or 4?

Question

calculusfunctions · Answer

$$f(x)=e ^{x}\cos \frac{ x }{ 2 }$$
$$e ^{x} $$ never equals zero. Now
$$\cos \frac{ x }{ 2 }=0$$ on the interval [0, 2π] Realize that the given interval is for x so the interval for x/2 would be [0, π]$$\frac{ x }{ 2 }=\frac{ \pi }{ 2 }$$
∴ x = π

Thus there is only one zero in the interval.

anonymous · Answer

its not cos(x/2)...It is cos(1/2x)

calculusfunctions · Answer

Is x in the numerator or the denominator?

anonymous · Answer

denominator

anonymous · Answer

1/2 and x are separate.

calculusfunctions · Answer

How do you know, since it's not your question? How do you know what he meant?

anonymous · Answer

So I guess x/2 is accurate

anonymous · Answer

What about 3pi/2? Why doesn't that work? I thought there was two 0 but I was wrong on my test

calculusfunctions · Answer

$$\frac{ 1 }{ 2 }x =\frac{ x }{ 2 }$$ It is exactly the same thing. Thank you!

anonymous · Answer

Since the interval includes all the way up to 2pi

calculusfunctions · Answer

Did you read my explanation? Because I explained about the intervals.
If the given interval for x was 0 ≤ x ≤ 2π, then to get the interval for x/2, you must divide the inequality by 2 to obtain $$0\le \frac{ x }{ 2 }\le \pi $$ Do you understand? You always have to modify the given interval to satisfy the angle.