Question

I tried to find the critical points of f(x)=2cot(x); [-pi,pi] and I thought they'd be at x=0, -pi, and pi, but the answer sheet says no critical points exist. Can someone explain to me why they don't exist?

Answer 1

cot x is undefined at x = 0, so it cant be at x = 0; look at the graph of 2 cot(x) to see whats happening

Answer 2

x = 0 is a vertical asymptote for cot x

Answer 3

But I thought critical points could be at where x=0 or in undefined. Is that not trues?

Answer 4

*true

Answer 5

Thats true, but then find the derivative and you'll see that it cant occur at x = 0. It should be ruled out right away because cot x is nonexistent at x = 0.

Sometimes the derivative at x = 0 is nonexistent but at least continuous there, like y = absolute value of x; but in cot x there's a complete discontinuity there

Answer 6

To see it from a different angle, 2 cot x = 2cos x/sin x...now this expression is undefined when x = 0, pi, 2pi, etc...

Answer 7

At these points, there's a vertical asymptote, and total discontuity

Answer 8

Okay, I think I get it now. So the function has to be continuous at that point for there to be a critical point?

Answer 9

Yes.

Answer 10

So no critical points where the original function has an asymptote?

Answer 11

Put 2 cot x on your graphing calculator..a picture is worth a thousand words.

Answer 12

true

Answer 13

I did, I was just trying to do it without it because I can't use a calculator on my test.

Answer 14

graph the function on [-pi/2, pi/2]

Answer 15

I understand that, but use it here to see what we are talking about so that you are convinced about the concept of vertical asymptotes and differentiation.

Recall: If a function is differentiuable, it is continuous. Do you remember that???????????

Answer 16

So that true statement implies that If a function is NOT continuos at a certain point, it is NOT differentiable at that point!!

Answer 17

Got it?