Modules in need!

Lets break up this function into several pieces.

Case 1 When $x \geq 2$ , $f(x)= x-2 + x+3 \implies 2x+1$ No maxima-minima,

Case II When $-3 < x \leq 2$ $f(x)= -x+2+x+3 \implies 5$

Case III When $x \leq -3$

$f(x) = -x+2 - x-3 = -2x-1$

No maxima, minima,

So $f(x)_{min}=5$

Using AM-GM inequality , we have

$\begin{aligned} |x-2| + |x+3| & \ge 2 \sqrt{|x-2| |x+3|} & \small \color{#3D99F6} \text{Equality occurs when } |x-2|= |x+3| \text{ or } x = -\frac 12 \\ & \ge 2 \sqrt{\frac 52 \cdot \frac 52} = \boxed{5} \end{aligned}$

We have

$\begin{aligned} f(x) = |x-2| + |x+3| & := \begin{cases} (2-x) + (-3-x), & x < -3 \\ (2-x) + (x+3), & -3 \le x < 2 \\ (x-2) + (x+3), & x \ge 2 \end{cases} \\ \\ & := \begin{cases} -1-2x, & x < -3 \\ 5, & -3 \le x < 2 \\ 2x+1, & x \ge 2 \end{cases} \end{aligned}$

Observe that

$\begin{aligned} x<-3 & \implies & f(x) > 5 \\ -3 \le x < 2 & \implies & f(x) = 5 \\ x \ge 2 & \implies & f(x) \ge 5 \end{aligned}$

We conclude that $\min (f(x)) = \boxed{5}$ .