What Causes Undefined Derivatives in Trigonometric Functions?

[Nick_OCD]

Hello All,

My problem is to find the relative extrema of abs(sin 2x); 0< x < 2pi.

I got the derivative right, I think: 2 cos 2x if x > 0, and -2 cos 2x if x < 0.

However, the solution says that f ' (x) does not exist at x = pi/2, x = pi, and x = (3pi)/2.

Glancing at my trigonometry charts, I see that sin and cos are never undefined,

so why does 2 cos 2x or -2 cos 2x not exist at pi/2, pi, or 3pi/2?

Thanks,

Nick

 

[LCKurtz]

Hello All,

My problem is to find the relative extrema of abs(sin 2x); 0< x < 2pi.

I got the derivative right, I think: 2 cos 2x if x > 0, and -2 cos 2x if x < 0.

But your interval only includes x > 0. And it isn't the sign of x that is important, it is the sign of sin(x) because of the absolute value signs.

However, the solution says that f ' (x) does not exist at x = pi/2, x = pi, and x = (3pi)/2.

Glancing at my trigonometry charts, I see that sin and cos are never undefined,

so why does 2 cos 2x or -2 cos 2x not exist at pi/2, pi, or 3pi/2?

Thanks,

Nick

Draw the graph of |sin(2x)| and you will answer your own questions.

 

[Nick_OCD]

Thank you. I had a brain freeze on that one.