Question

Verify that `f(x)` is differentiable at `x = 0`? `f(x) = x((e^(1//x) - 1)/(e^(1//x) + 1))`

Answer 1

The function is NOT differentiable at `x = 0`... When you say verify, you would mean that we already ASSUME the truth, not that we are checking whether it is true or not.

graph{x((e^(1/x) - 1)/(e^(1/x) + 1)) [-1.065, 1.0684, -0.533, 0.5335]}

There are in fact different slopes from either side of `x = 0`.

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Well, you can multiply by `e^(-1//x)` on the top and bottom to get:

`f(x) = x((1 - e^(-1//x))/(1 + e^(-1//x)))`

Then, plugging in `x = 0` is one of two tests to show that the function is differentiable: it must, for one, be continuous at `x = 0`.

`f(0) = 0 * ((1 - e^(-1//0))/(1 + e^(-1//0)))`

`= 0 * 1 = 0`

Since we have successfully found a determinate form and evaluated it, this function is continuous at `x = 0`.

Now, we could very well have a function that has a corner or cusp at `x = 0`. To check that, we could see if the slope in the NEIGHBORHOOD of `x = 0` is large, or nonsmall, just in case the function is wacky at `x = 0`.

The derivative is:

`f'(x) = x cdot (((1 + e^(-1//x)) cdot e^(-1//x) cdot 1/x^2 - (1 - e^(-1//x))cdot(-e^(-1//x) cdot 1/x^2))/(1 + e^(-1//x))^2) + (1 - e^(-1//x))/(1 + e^(-1//x))`

Simplify by factoring out `e^(-1//x) cdot 1/x^2` and cancelling out the `x` that is outside of the first grouped terms with the `1/x^2`.

`= (-2e^(-1//x))/(x(1 + e^(-1//x))^2) + (1 - e^(-1//x))/(1 + e^(-1//x))`

Multiply the second fraction by `(x(1 + e^(-1//x)))/(x(1 + e^(-1//x)))`.

`= (-2e^(-1//x) + x(1 - e^(-2//x)))/(x(1 + e^(-1//x))^2)`

Let's choose `[-0.1, 0.1]` as our "neighborhood". Then, we get:

`f'(-0.1) = (-2e^(-1//-0.1) + -0.1(1 - e^(-2//-0.1)))/(-0.1(1 + e^(-1//-0.1))^2) ~~ -1`

`f'(0.1) = (-2e^(-1//0.1) + 0.1(1 - e^(-2//0.1)))/(0.1(1 + e^(-1//0.1))^2) ~~ +1`

Since the slopes very near `x = 0` are equal in magnitude and opposite in sign, but not close to zero, there is either a cusp or a corner at `x = 0`, or at least, the slope differs from either side.

Therefore, this function is NOT differentiable at `x = 0`.