Thread: Sketching y = x arccos(x)

The question:
Sketch the following curve, showing its main features:


I'm not sure how to attempt this. I know that the inverse of cosine is restricted to [-1, 1] in the domain, and must be restricted on the range to allow it to have an inverse, i.e. I know the graph passes through the origin too. I'm stumped in regards to sketching it. I don't think I have all the information. Any assistance would be great!

Originally Posted by Glitch

The question:
Sketch the following curve, showing its main features:


I'm not sure how to attempt this. I know that the inverse of cosine is restricted to [-1, 1] in the domain, and must be restricted on the range to allow it to have an inverse, i.e. I know the graph passes through the origin too. I'm stumped in regards to sketching it. I don't think I have all the information. Any assistance would be great!

The domain is [-1, 1]. Yes, (0, 0) is on the curve. I suggest you plot points eg. endpoints (the points where x = 1, -1), 1/2, -1/2 etc. and then try to fit a curve. Are you expected to find turning points? If so, calculus is required.

Thanks. The question didn't mention anything about finding turning points. I did that anyway, and it only has one in the interval.

Originally Posted by Glitch

Thanks. The question didn't mention anything about finding turning points. I did that anyway, and it only has one in the interval.

plot y = x Arccos[x] - Wolfram|Alpha

It's an ugly thing, isn't it? :P

Originally Posted by Glitch

It's an ugly thing, isn't it? :P

Everything is relative ....