# Is x/(1+e^(1/x)) differentiable at x=0

## Homework Statement

is x/(1+e1/x) differentiable at x =0

## The Attempt at a Solution

i calculated the right and left hand derivative and got them as 0 and 1 respectively , so it should not be differentiable, is it right my book says that it is differentiable ?

SammyS
Staff Emeritus
Homework Helper
Gold Member

Show what went into your results.

WolframAlpha says that you're correct!

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Show what went into your results.

WolframAlpha says that you're correct!

my book says that i am wrong , but i think iam convinced that i am right , thanks!!

HallsofIvy
Homework Helper

The function you give is not differentiable at x= 0 because it is not even defined at x= 0. What is the function value at x= 0?

Assuming that you also have f(0)= 0, then the difference quotient is
$$\frac{f(h)- f(0)}{h}= \frac{h}{1+ e^{1/h}}\frac{1}{h}= \frac{1}{1+ e^{1/h}}$$
which goes to 1 as h goes to either $\infty$ or $-\infty$.

You are wrong and your book is right.

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Char. Limit
Gold Member

The function you give is not differentiable at x= 0 because it is not even defined at x= 0. What is the function value at x= 0?

Assuming that you also have f(0)= 0, then the difference quotient is
$$\frac{f(h)- f(0)}{h}= \frac{h}{1+ e^{1/h}}\frac{1}{h}= \frac{1}{1+ e^{1/h}}$$
which goes to 0 as x goes to either $\infty$ or $-\infty$.

You are wrong and your book is right.

How can the book be right when the book says that it is differentiable, and you clearly say that it isn't?

How can the book be right when the book says that it is differentiable, and you clearly say that it isn't?

Because it is differentiable. I don't think he means what you think he means by non existent at 0 at least I hope so or I am confused.

Easiest way to solve this argument is to graph it.

At x=0, y = "0" though just plugging in the numbers, that should tell you something division by what is what?

This is differentiable because of the asymptote I think not despite it. To put what he said in words rather than latex, maybe that will help?

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The function you give is not differentiable at x= 0 because it is not even defined at x= 0. What is the function value at x= 0?

Assuming that you also have f(0)= 0, then the difference quotient is
$$\frac{f(h)- f(0)}{h}= \frac{h}{1+ e^{1/h}}\frac{1}{h}= \frac{1}{1+ e^{1/h}}$$
which goes to 0 as x goes to either $\infty$ or $-\infty$.

You are wrong and your book is right.

i guess by x you mean 1/h , but when 1/h goes to -infinity , e1/h goes to 0 , this means the last expression 1/(1+e1/h) becomes 1
and yes f(0)=0

HallsofIvy
Homework Helper

Because it is differentiable. I don't think he means what you think he means by non existent at 0 at least I hope so or I am confused.
If you are referring to my post, yes, I did mean that- in the first post, we were told that the function was $f(x)= x/(1+ e^{1/x})$ which is NOT defined at x= 0. If we assume that f(0)= 0, then the derivative does exist.

Easiest way to solve this argument is to graph it.

At x=0, y = "0" though just plugging in the numbers, that should tell you something division by what is what?
Plugging numbers into what? You cannot set x= 0 and get y. If you mean that values of x close to 0 give values of y close to 0, then to conclude that the value of f(0) is 0, you would have to know that the function is continuous at x= 0 and we were also not told that.

This is differentiable because of the asymptote I think not despite it. To put what he said in words rather than latex, maybe that will help?
There is no "asymptote". Assuming that f(0)= 0, this is just a basic differentiable function.

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$$f(x)= x/(1+ e^{1/x})$$ and

$$g(x) = \left\{ \begin{array}{lr} x/(1+ e^{1/x}),~ x \neq 0\\ 0,~ x = 0 \end{array} \right.$$

are different functions. If you calculate the corresponding limit of the first function as x approaches 0 and then define the value of said function as the limit, you'll have made the function continuous and then and only then does it make sense to calculate the derivative via the limit definition, as HallsofIvy has pointed out.

However, as the original function was given to us, I have to say it is not differentiable as it's not even defined for x = 0. It is indeed natural to "fill in" the removable discontinuity, but the author of the problem has to give us the precise definition of a function if he expects us to give precise answers.

SammyS
Staff Emeritus
Homework Helper
Gold Member

The function you give is not differentiable at x= 0 because it is not even defined at x= 0. What is the function value at x= 0?

Assuming that you also have f(0)= 0, then the difference quotient is
$$\frac{f(h)- f(0)}{h}= \frac{h}{1+ e^{1/h}}\frac{1}{h}= \frac{1}{1+ e^{1/h}}$$
which goes to 1 as h goes to either $\infty$ or $-\infty$.

You are wrong and your book is right.
@ HallsofIvy, (assuming that the removable discontinuity has been removed.)

Why did you take the limits as h → +∞ & h → ‒∞, rather than h → 0+ & h → 0 ?

If you are referring to my post, yes, I did mean that- in the first post, we were told that the function was $f(x)= x/(1+ e^{1/x})$ which is NOT defined at x= 0. If we assume that f(0)= 0, then the derivative does exist.

Plugging numbers into what? You cannot set x= 0 and get y. If you mean that values of x close to 0 give values of y close to 0, then to conclude that the value of f(0) is 0, you would have to know that the function is continuous at x= 0 and we were also not told that.

There is no "asymptote". Assuming that f(0)= 0, this is just a basic differentiable function.

My bad obviously I typed the wrong function into a grapher. Or the grapher was ****.

Either way ignore me. :D

obviously put a bracket in the wrong place and it all went pear shaped. :D

Not literally obv. :)

@ HallsofIvy, (assuming that the removable discontinuity has been removed.)

Why did you take the limits as h → +∞ & h → ‒∞, rather than h → 0+ & h → 0 ?

Reviewing this message, I have to say, my thoughts exactly. If I'm not mistaken, on one side, the derivative is 1, and approaching from the other side, it's 0. Hence, even when the discontinuity has been dealt with, the function is not differentiable.