f(x) = \left\{ {\matrix{
{3\left( {{{1 - \left| x \right|} \over 2}} \right)} & {if\,\left| x \right| \le 2} \cr
0 & {if\,\left| x \right| > 2} \cr
} } \right.
g(x)=f(x+2)−f(x−2)
f(x) = \left\{ {\matrix{
{0,} & {x < - 2} \cr
{{3 \over 2}(1 + x),} & { - 2 \le x < 0} \cr
{{3 \over 2}(1 - x),} & {0 \le x < 2} \cr
{0,} & {x > 2} \cr
} } \right.
f(x + 2) = \left\{ {\matrix{
{0,} & {x < - 4} \cr
{{3 \over 2}( 3 + x),} & { - 4 \le x < - 2} \cr
{{3 \over 2}( - 1 - x),} & { - 2 \le x < 0} \cr
{0,} & {x > 4} \cr
} } \right.
f(x - 2) = \left\{ {\matrix{
{0,} & {x < 0} \cr
{{3 \over 2}(x - 1),} & {0 \le x < 2} \cr
{{3 \over 2}( - 1 - x),} & {2 \le x < 4} \cr
{0,} & {x > 4} \cr
} } \right.
g(x)=f(x+2)+f(x−2)
= \left\{ {\matrix{
{{{3x} \over 2} + 6,} & { - 4 \le x \le 2} \cr
{ - {{3x} \over 2},} & { - 2 < x < 2} \cr
{{{3x} \over 2} - 6,} & {2 \le x \le 4} \cr
{0,} & {\left| x \right| > 4} \cr
} } \right.
So, n = 0 and m = 4
∴ m + n = 4
