How to graph rational trigonometric function y = cosx/x+(pi/2) when x is negative

[calc67x]

I am having loads of trouble graphing this function.
The function is
cosx/x+(pi/2)

Now the first steps are easy:
if x=0 then f(x) or y =2/pi
if x= pi/2 then f(x) or y =0

However, what I don't understand is when x becomes negative
For example, I found that f(x)=1 for -pi/2, why is this so?
I would think that since cos(-pi/2) =0 that it would be 0 or undefined.
Can someone explain to me?
Also, what is the rule for obtaining the amplitude of this function?
Thanks so much for any help!!

 

[Dr.Peterson]

I am having loads of trouble graphing this function.
The function is
cosx/x+(pi/2)

Now the first steps are easy:
if x=0 then f(x) or y =2/pi
if x= pi/2 then f(x) or y =0

However, what I don't understand is when x becomes negative
For example, I found that f(x)=1 for -pi/2, why is this so?
I would think that since cos(-pi/2) =0 that it would be 0 or undefined.
Can someone explain to me?
Also, what is the rule for obtaining the amplitude of this function?
Thanks so much for any help!!

Judging by the values you found, you omitted some essential parentheses. You apparently meant

f(x) = cos(x)/(x+pi/2)

so that the denominator is x + pi/2, rather than just x as you wrote it.

Given that, you are correct that

f(0) = cos(0)/(0+pi/2) = 1/(pi/2) = 2/pi
f(pi/2) = cos(pi/2)/(pi/2+pi/2) = 0/pi = 0

But how did you "find" f(-pi/2) = 1? Are you saying, perhaps, that this is what you see on a graph, but you can't obtain it yourself?

Clearly,

f(-pi/2) = cos(-pi/2)/(-pi/2+pi/2) = 0/0, which is undefined.

It may be that the limit as x approaches -pi/2 is 1 (I haven't checked that, not knowing if that is what you saw), so that a graphing calculator would lead you to think the value is 1. But it isn't.

This is not a general issue with negative values of x; it is specific to this one value, -pi/2.

As for amplitude, that concept doesn't apply here.