Question

How to find the points of of non-differentiability of a given function without graphs?

Given a function :
`f(x) = max{2 sinx, 1-cosx}, x in (0,pi)`
Find the points of non-differentiability.

I tried thinking analytically, and wrongly concluded that there are no points of non-differentiablilty as the function is defined at all point.

I would like to know:

1. What is the correct approach to such questions?
2. How to solve it intuitively, if possible?

Regards & Thanks
Aditya

Answer 1

Please see below.

Explanation:

1. I assume that there are multiple correct approaches to such questions. (I'm not sure what you mean by a "correct approach", but I would call any approach that yields a correct answer to the question a correct approach.)

2. I don't know what you would count as an intuitive versus and non- (or un-) intuitive solution.

Here is how I would find an answer to the question:

Note that `y=2sinx` and `z = 1-cosx` are differentiable everywhere.
The only place(s) that `f` might fail to be differentiable are the intersection(s) of `y` and `z`. (Where the maximum changes from one to the other.)

Find the `x` values if the intersection.

Solve `2sinx=1-cosx` to get
`x=0` or `x = cos^-1(-3/5)`

Check to see it the transition from one being maximum to the other is a "smooth" transition. (See if the derivatives are equal.)

`dy/dx = 2cosx` and `dz/dx=sinx`

At `x=0`, we get `{:dy/dx]_0=2` and `{:dz/dx]_0=0`, so

`f` is not differentiable at `x = 0`.

At `x = cos^-1(-3/5)`, we get `{:dy/dx]_(cos^-1(-3/5)) = -6/5` and `{:dz/dx]_(cos^-1(-3/5))=4/5`, so

`f` is not differentiable at `x=cos^-1(-3/5)`.