Question 996088
Rule: if y = |x|, then dy/dx = |x|/x


Apply the derivative
g(x) = x*|x+5|
g ' (x) = |x+5| + x*(|x+5|)/(x+5) ... product rule (also use the rule above)
g ' (x) = |x+5| * [1 + x/(x+5)]
g ' (x) = |x+5| * [1*(x+5)/(x+5) + x/(x+5)]
g ' (x) = |x+5| * [(x+5)/(x+5) + x/(x+5)]
g ' (x) = |x+5| * [(x+5+x)/(x+5)]
g ' (x) = |x+5|*(2x+5)/(x+5)



Solve g ' (x) = 0 for x to find the critical values
g ' (x) = |x+5|*(2x+5)/(x+5)
0 = |x+5|*(2x+5)/(x+5)
|x+5| = 0 or (2x+5)/(x+5) = 0
|x+5| = 0 or 2x+5 = 0


If |x+5| = 0, then x+5 = 0 and x = -5
If 2x+5 = 0, then x = -5/2 = -2.5


The critical numbers are <font color="red">-5 and -2.5</font>