i have to find left and right derivative can somebody help me?

Question

I have 1/|x+4| I don't even know how to start i'm sure is not that hard but can somedoby explain with steps please

Thank you

Answer 1

We know the function is well defined everywhere except at $x=-4$.

So consider $p=-4+\delta$, with $\delta >0$.

$f(p)=\frac 1{|p+4|}=\frac 1{|-4+\delta+4|}=\frac 1{\delta}$ since $\delta >0$

$f(p+h)=\frac 1{|p+h+4|}=\frac 1{|-4+\delta+h+4|}=\frac 1{\delta+h}$

Let $f'(p)=\lim _{h \rightarrow 0} \frac {f(p+h)-f(p)}{h}$

So $f'(p)=\lim _{h \rightarrow 0} \frac {\frac 1{\delta+h}-\frac 1{\delta}}{h}=\lim _{h \rightarrow 0} \frac {\delta-(\delta+h)}{h\delta(\delta+h)}=\lim _{h \rightarrow 0} \frac {-h}{h\delta(\delta+h)}=\lim _{h \rightarrow 0} \frac {-1}{\delta(\delta+h)}=\frac {-1}{\delta^2}$

As $\delta \rightarrow 0$ from the right, we have $f'(p) \rightarrow -\infty$

Can do something similar from the left.