How do I find all points of discontinuity of the function $f(x)= |x+2|+|x+3|$?

Question

Basically it is a modulus $\operatorname{fun}^c$. To begin with we'll redefine the function, i.e. $f(x)=|x+2|+|x-3|$.

Taking $$ u = |x+2| = \begin{cases} -(x+2), & \text{if $x<0$,} \\ x+2, & \text{if $x=0$,} \\ x+2, & \text{if $x>0$,} \end{cases} $$ and $$ v = |x-3| = \begin{cases} -(x-3), & \text{if $x<0$,} \\ x-3, & \text{if $x=0$,} \\ x-3, & \text{if $x>0$.} \end{cases} $$ we have $f(x)=u+v$.

LHL for u = -(0+2) = -2 RHL for u = 0+2 = 2 Therefore, RHL not equal to LHL. It is not cont at x=0

LHL for v = -(0-3) = 3 RHL for v = 0-3 = 3 Therefore, RHL not equal to LHL. It is not cont at x=0

Therefore, it cont at all R-{0}. Point of discontinuity is 1.

Where I am gng wrong pls anyone help. Note: THIS NOT A HOME WORK QUESTION. I have used some abbreviation. Like, cont-continuous.

Answer 1

Do you mean differentiable? The function $g(x)=|x+2|$ is continuous everywhere because it's just the absolute function shifted 2 to the left, and that is continuous. Also $h(x)=|x+3|$ is continuous because it's just the absolute value function shifted 3 to the left. And $f(x)=|x+2|+|x+3|$ is the sum of continuous functions, so it's also continuous.

Answer 2

You didn't redefine the modulus function correctly. It should be $$u=\begin{cases} x+2 &, x\ge -2\\ -(x+2) &, x<-2\end{cases}$$ and $$v=\begin{cases} x-3 &, x\ge 3\\ -(x-3) &, x< 3\end{cases}$$