Question 901758
f(x)=(2x+1)e^(3x+1)
Critical Value: where 1st derivative = 0
f' = e^(3x+1)(6x + 5)critical vale = -5/6,  extrema is (-5/6,f(-5/6) = (5/6, -.1488) (minimum value)
Inflection Pt when f "(x) = 0 
f" = 3(6x + 5)e^(3x+1)+ 6e^(3x+1)
= e^(3x+1)(18x + 21) point of inflection is (-21/18, f(-21/18)) 0r(-7/6, -.1094)
So recommend using a graphing Calculator to Verify