Is Ln X1 Continuous at X0

2. Is $|\ln|x||$ differentiable?

Is $|\ln|x||$ differentiable?

Solution 1

At points where $f(x)\ne0$ there is no problem in differentiating $|f|$, because the function is positive/negative in a neighborhood of the point.

So all you need is to check where $f(x)=0$. In this case at $1$ and $-1$. However, since the function is clearly even, we can just look at $1$: $$ \lim_{x\to1^+}\frac{|\log|x|\,|-|\log|1|\,|}{x-1}= \lim_{x\to1^+}\frac{\log x}{x-1}=1 $$ whereas $$ \lim_{x\to1^-}\frac{|\log|x|\,|-|\log|1|\,|}{x-1}= \lim_{x\to1^-}\frac{-\log x}{x-1}=-1 $$ So the function is not differentiable at $1$ and $-1$, but it is everywhere else (provided it is defined to begin with).

Yes, your attempt is good.

Solution 2

Assuming $f(x) : \mathbb{R}\setminus \{0\} \to \mathbb{R}$.

We can rewrite $f(x) = \left|\ln\left|x\right|\right|$ as:

$$ f(x) = \begin{cases} \ln(x), & x \geq 1 \\ -\ln(x), & 1 \geq x > 0 \\ -\ln(-x), & 0 > x \geq -1 \\ \ln(-x), & -1 \geq x \end{cases} $$

We see that on each of the intervals $(-\infty, -1), (-1, 0), (0, 1), (1, \infty)$, $f(x)$ is both continuous and differentiable, due to the continuity and differentiability of $\ln(x)$ on $(0, 1)$.

We thus have 2 possible $x$ values where $f$ is not differentiable, $-1$ and $1$. Note that we don't have to consider $x=0$ as $f$ is not defined there.

Lets look at $x = 1$.

For $x \geq 1$ we have $f'_+(x) = \frac{1}{x}$, and for $1 \geq x > 0$ we have $f'_{\vphantom+-} (x) = -\frac{1}{x}$. Thus:

$$\lim_{x \to 1+} f'_+(x) = 1 \neq \lim_{x \to 1-} f'_{\vphantom+-}(x) = -1$$

Solution 3

The facts:

1. $|x|$ is not differentiable at $x = 0$
2. The derivative of $\ln |x|$ vanishes no-where on $\mathbb{R}$

implies that $ |\ln |x| |$ is not differentiable at $x$ for which $\ln |x| = 0$ i.e. $x= \pm 1$

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Comments

• Is $|\ln|x||$ differentiable for all $x$ is defined and continuous?

  I can see that on the graph that it is not differentiable at $-1$ and $1$, but how can I prove it?

  So I look at $\lim_{h\to 0+} \frac{|\ln(1-h)|-|\ln(1)|}{h}=\lim_{h\to 0+} \frac{|\ln(1-h)|}{h}$ because it is positive $(0^{+})$ we can say $\lim_{h\to 0+} \frac{\ln(1-h)}{h}$ applying L'Hôpital $\lim_{h\to 0+} \frac{\frac{-1}{1-h}}{1}=-1$?

  • Prove in general that if $f(x_0)=0$ and $f'(x_0)\ne 0$, then $x\mapsto|f(x)|$ is not differentiable at $x_0$. The limit-based definition of "differentiable" can be used directly for this.

  • I have try to solve int

  • How do you define your function for $x=0$ ??

  • @Nizar it is not defined, but I do not need to look at points that are not defined

  • Okay, so please state the domain of your function.

• is there a way to turn a function that is not differentiable to be differentiable? Or the way you solve it, I can use everywhere?

• And this is supposed to solve what exactly? It looks to me like blind and foolish application of the derivative without seeing any details and consequences.

• I gave him the derivative, now the OP can take the effort to see the consequences

• @gbox No, you can not use it when the outer function has zero derivative at zero.

• @yo' This method says nothing about $|x|^2$ or any other powers of $|x|$

• @yo' strictly increasing functions

• I should not have used the term well behaved. Sorry about that.

• @yo' oops. strictly increasing and non-zero at x=0. I understood why this method is wrong. But I am trying to find the class of functions for which it is valid

• can we say that graph exhibits cusp at $x=\pm 1$

• @EkaveeraKumarSharma Not really a cusp. It's a jump (of 1st order discontinuity) for the derivative, but this does not change the fact that the derivative does not exist for $x=\pm1$.

• I think the limit calculation I did is wrong, I get the same result for $1^{+} $ and $ 1^{-}$

• @yo' yes, you are right. $f(x)$ must have non-vanishing derivative.

• @EkaveeraKumarSharma In Italian it's called "punto angoloso" (angle point, the direct English translation).

• @gbox You have surely an error, because the numerator is positive by definition, while the denominator is negative for the limit from the left and positive for the limit from the right.

Recents

dunnclook1975.blogspot.com

Source: https://9to5science.com/is-ln-x-differentiable

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