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

Aug 4, 2017

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 $x = 0$.

Well, you can multiply by ${e}^{- 1 / x}$ on the top and bottom to get:

$f \left(x\right) = x \left(\frac{1 - {e}^{- 1 / x}}{1 + {e}^{- 1 / x}}\right)$

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

$f \left(0\right) = 0 \cdot \left(\frac{1 - {e}^{- 1 / 0}}{1 + {e}^{- 1 / 0}}\right)$

$= 0 \cdot 1 = 0$

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:

$f ' \left(x\right) = x \cdot \left(\frac{\left(1 + {e}^{- 1 / x}\right) \cdot {e}^{- 1 / x} \cdot \frac{1}{x} ^ 2 - \left(1 - {e}^{- 1 / x}\right) \cdot \left(- {e}^{- 1 / x} \cdot \frac{1}{x} ^ 2\right)}{1 + {e}^{- 1 / x}} ^ 2\right) + \frac{1 - {e}^{- 1 / x}}{1 + {e}^{- 1 / x}}$

Simplify by factoring out ${e}^{- 1 / x} \cdot \frac{1}{x} ^ 2$ and cancelling out the $x$ that is outside of the first grouped terms with the $\frac{1}{x} ^ 2$.

$= \frac{- 2 {e}^{- 1 / x}}{x {\left(1 + {e}^{- 1 / x}\right)}^{2}} + \frac{1 - {e}^{- 1 / x}}{1 + {e}^{- 1 / x}}$

Multiply the second fraction by $\frac{x \left(1 + {e}^{- 1 / x}\right)}{x \left(1 + {e}^{- 1 / x}\right)}$.

$= \frac{- 2 {e}^{- 1 / x} + x \left(1 - {e}^{- 2 / x}\right)}{x {\left(1 + {e}^{- 1 / x}\right)}^{2}}$

Let's choose $\left[- 0.1 , 0.1\right]$ as our "neighborhood". Then, we get:

$f ' \left(- 0.1\right) = \frac{- 2 {e}^{- 1 / - 0.1} + - 0.1 \left(1 - {e}^{- 2 / - 0.1}\right)}{- 0.1 {\left(1 + {e}^{- 1 / - 0.1}\right)}^{2}} \approx - 1$

$f ' \left(0.1\right) = \frac{- 2 {e}^{- 1 / 0.1} + 0.1 \left(1 - {e}^{- 2 / 0.1}\right)}{0.1 {\left(1 + {e}^{- 1 / 0.1}\right)}^{2}} \approx + 1$

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 $x = 0$.