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

1 Answer
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(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#.