So with this problem I am completely lost on where to even start this problem to get to the answer that my professor wants me to prove. The questions is shown in the attached picture.
Let f(x) = abs(x), a) Show that f'(x) = 2*theta(x) - 1
I honestly would just like some help on where to start/ point me in the right direction with both part a in problems G and H. Ive looked everwhere online and cant find anything that would be useful. Ive even tried using the notes from class and there are no similar problems that were even gone over. Thanks so much!
By definition we have $$|x|=\begin{cases} x \space\space\space\space\space\space\text{if $x\geq 0$}\\ -x \space\space\space\text{if $x< 0$}\end{cases}$$
Thus $$f'(x)=\begin{cases} 1 \space\space\space\space\space\space\text{if $x> 0$}\\ -1 \space\space\space\text{if $x< 0$}\end{cases}$$ $$=2\theta(x)-1$$ where $$\theta(x)=\begin{cases} 1 \space\space\space\text{if $x> 0$}\\ 0 \space\space\space\text{if $x< 0$}\end{cases}$$
Next note that the derivative of the Heaviside's unit step function is the Dirac delta function.
You want to show that $f'(x)=2\theta(x)-1$, from the definition above we have $$f'(x)=\begin{cases} 1 \space\space\space\space\space\space\text{if $x> 0$}\\ -1 \space\space\space\text{if $x< 0$}\end{cases}$$
Do you agree with this?
Since we have defined $$\theta(x)=\begin{cases} 1 \space\space\space\text{if $x> 0$}\\ 0 \space\space\space\text{if $x< 0$}\end{cases}$$
then what can we do to $\theta(x)$ so that it gives $f'(x)$?
Well:
$$2\theta(x)=\begin{cases} 2 \space\space\space\text{if $x> 0$}\\ 0 \space\space\space\text{if $x< 0$}\end{cases}$$ so $$2\theta(x)-1=\begin{cases} 1 \space\space\space\text{if $x> 0$}\\ -1 \space\space\text{if $x< 0$}\end{cases}$$
which is $f'(x)$.