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#.
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#.
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Answer 2

To verify that ( f(x) ) is differentiable at ( x = 0 ), we need to check if the derivative exists at that point.

Given ( f(x) = x\frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1} ):

  1. First, we check the continuity of ( f(x) ) at ( x = 0 ).
  2. Then, we find the derivative of ( f(x) ) with respect to ( x ).
  3. Finally, we evaluate the limit of the derivative as ( x ) approaches 0.

Let's proceed with the calculations:

  1. ( f(x) ) is a rational function, and both the numerator and denominator approach 0 as ( x ) approaches 0. Therefore, ( f(x) ) is continuous at ( x = 0 ).

  2. To find the derivative of ( f(x) ), we use the quotient rule: [ f'(x) = \frac{(e^{\frac{1}{x}} + 1)(1) - (e^{\frac{1}{x}} - 1)(1)}{(e^{\frac{1}{x}} + 1)^2} ] [ = \frac{2}{(e^{\frac{1}{x}} + 1)^2} ]

  3. To evaluate the limit of ( f'(x) ) as ( x ) approaches 0, we substitute ( x = 0 ) into ( f'(x) ): [ \lim_{x \to 0} f'(x) = \frac{2}{(e^{\frac{1}{0}} + 1)^2} = \frac{2}{(e^{\infty} + 1)^2} = \frac{2}{(\infty + 1)^2} = \frac{2}{\infty} = 0 ]

Since the limit of ( f'(x) ) as ( x ) approaches 0 exists and is finite, ( f(x) ) is differentiable at ( x = 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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