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I want to determine if the following function is differentiable:

$$y=\ln\left(\left|x^{2}-4x+3\right|\right)$$ Logically, I should find the one-sided derivatives at points that are roots of the polynomial in the modulus, but the function is undefined at these points ($y = \ln(0)$).

So what to do here? Find derivatives at points in which $y = 0$?

Vivaan Daga
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user492754
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    As real function from real variable it's defined only when modulus$>0$. And on this set is composition of differentiable functions. – zkutch Feb 12 '23 at 14:49

1 Answers1

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First we need notice where the function is well-defined as a real valued function, that is, we need know the natural domain of the map $f:x\mapsto \ln|x^2-4x+3|$. Since the map $x\mapsto \ln x$ is well-defined for $x>0$ and for the map $x\mapsto |x|$ we have for all $x$ that $|x|\geqslant 0$, we need only to see in $x^2-4x+3=0$ but then, $$x^2-4x+3=0\iff (x-3)(x-1)=0\iff x=3\vee x=1.$$ Thus, the domain of $f$ is given by $${\rm dom}_{f}={\bf R}\setminus \{1,3\}$$ Then, \begin{align*} f: {\bf R}\setminus \{1,3\}&\longrightarrow {\bf R}\\ x&\longmapsto \ln|x^2-4x+3| \end{align*} Now, over that domain of $f$ noticed that we have the composition of derivables function. Indeed, if $g: x\mapsto x^2-4x+3$, then

  • If $g(x)>0$, then $(f\circ g)(x)$ is derivable, since we have composition of derivables functions.
  • If $g(x)<0$, then $h=-g$ by definition of absolute value and then give $(f\circ h)(x)$ is derivable, since we have the composition of derivables functions.
  • Remark that $g(x)=0$ is not in the domain of $f$.

Therefore, $f$ is derivable over ${\bf R}\setminus\{1,3\}$. Moreover, the derivative is given by $$f'(x)=\frac{1}{|x^2-4x+3|}(|x^2-4x+3|)'=\frac{2(-2+x)}{x^2-4x+3}$$ Recall the map $x\mapsto |x|$ is not derivable at $x=0$, then $x\mapsto |x^2-4x+3|$ is not derivable at $x=1$ or $x=3$ but these values are outside the domain of $f$.

A. P.
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