Find x for Differentiability of |x2-4x+3|

[Yankel]

Hello all,

I wish the find the values of x for which the following function is differentiable:

\[\left | x^{2}-4x+3 \right |\]

I got the point that the function is continuous apart from x=1,3.

I need to find if it is differentiable at x=1,3, using the limit definition of the derivative. I am not sure how. Can you assist please ?

Thank you in advance.

 

[Prove It]

Hello all,

I wish the find the values of x for which the following function is differentiable:

\[\left | x^{2}-4x+3 \right |\]

I got the point that the function is continuous apart from x=1,3.

I need to find if it is differentiable at x=1,3, using the limit definition of the derivative. I am not sure how. Can you assist please ?

Thank you in advance.

The function is continuous everywhere...

The only points where the function isn't differentiable is where the derivatives are not continuous.

I would start by writing the function in its hybrid function form, and then taking the derivatives, and testing the limit of the derivatives.

 

[Greg]

Hello all,

I wish the find the values of x for which the following function is differentiable:

\[\left | x^{2}-4x+3 \right |\]

I got the point that the function is continuous apart from x=1,3.

I need to find if it is differentiable at x=1,3, using the limit definition of the derivative. I am not sure how. Can you assist please ?

Thank you in advance.

Let's start with

$$\lim_{h\to0}\frac{|x+h|-|x|}{h}$$

Conjugate:

$$\lim_{h\to0}\frac{x^2+2hx+h^2-x^2}{h(|x+h|+|x|)}$$

Simplify and take the limit:

$$\lim_{h\to0}\frac{2hx+h^2}{h(|x+h|+|x|)}=\frac{2x}{2|x|}=\frac{x}{|x|}$$

If you're allowed to apply the chain rule then

$$|f(x)|'=\frac{f(x)}{|f(x)|}f'(x)$$

else you've got some tedious algebra if you choose the method I outlined above.

 

[Fernando Revilla]

As $f(x)=(x-1)(x-3)$ and according with the sign of $f(x)$ we can write $$f(x)=\begin{cases}{x^2-4x+3}&\text{if}& x\le1\\-\left(x^2-4x+3\right) & \text{if}& 1<x<3\\x^2-4x+3 & \text{if}& x\ge 3\end{cases}$$ For every $x\in \mathbb{R}\setminus\{1,3\}$ there is neighborhood $V$ of $x$ such that $f:V\to \mathbb{R}$ is a polynomical function, so $f$ is differentiable at $x.$ For $x=1$: $$f_-^{\prime}(1)=\lim_{h\to 0^-}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0^-}\frac{[(1+h)^2-4(1+h)+3]-0}{h}$$ $$=\lim_{h\to 0^-}\frac{h^2-2h}{h}=\lim_{h\to 0^-}(h-2)=-2.$$ $$f_+^{\prime}(1)=\lim_{h\to 0^+}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0^+}\frac{-[(1+h)^2-4(1+h)+3]-0}{h}$$ $$=\lim_{h\to 0^+}\frac{-h^2+2h}{h}=\lim_{h\to 0^+}(-h+2)=2.$$ We have $f_+^{\prime}(1)\ne f_-^{\prime}(1)$, so $f$ is not differentiable at $x=1.$ Same arguments for $x=3$.

 

[Yankel]

Thank you all.

Fernando, your explanation is very clear. I just have one small question. Why when you take 1 from the left, h goes to 0- and when you take 1 from the right h goes to 0+ ?

 

[Fernando Revilla]

Thank you all. Fernando, your explanation is very clear. I just have one small question. Why when you take 1 from the left, h goes to 0- and when you take 1 from the right h goes to 0+ ?

The symbol $h\to 0^+$ means $h\to 0$ when $h>0.$ When whe find $f_+^{\prime}(1)$, whe consider the function only to the right of $1,$ that is $1+h>1.$

Simmetrically, the symbol $h\to 0^-$ means $h\to 0$ when $h<0.$ When whe find $f_-^{\prime}(1)$, whe consider the function only to the left of $1,$ that is $1+h<1.$

 

[HallsofIvy]

The function is continuous everywhere...

The only points where the function isn't differentiable is where the derivatives are not continuous.

This is not true. The derivative of a function is not necessarily continuous so it is not true that "the function isn't differentiable is where the derivatives are not continuous". For example, the function f(x)= xsin(1/x) for x not 0, f(0)= 0
is differentiable for all x but its derivative, f'(x)= sin(1/x)- (1/x^2)cos(1/x) if x is not 0, f'(0)= 0, is not continuous at x= 0
.

I would start by writing the function in its hybrid function form, and then taking the derivatives, and testing the limit of the derivatives.

Still, good advice.