What are the extrema of #f(x) = e^x(x^2+2x+1)#? Calculus Graphing with the First Derivative Identifying Turning Points (Local Extrema) for a Function 1 Answer Bdub Apr 5, 2016 x=-3 or x = -1 Explanation: #f=e^x, g=x^2+2x+1# #f'=e^x, g'=2x+2# #f'(x)=fg'+gf'=e^x (2x+2)+e^x (x^2+2x+1)=0# #e^x (2x+2+x^2+2x+1)=0# #e^x (x^2+4x+3)=0# #e^x(x+3)(x+1)=0# #e^x = 0 or x+3=0 or x+1 =0# not possible, #x=-3 or x = -1# #f(-3)=e^-3(9-6+1)=0.199#->max #f(-1)=e^-1(1-2+1) = 0#->min Answer link Related questions How do you find the x coordinates of the turning points of the function? How do you find the turning points of a cubic function? How many turning points can a cubic function have? How do you find the coordinates of the local extrema of the function? How do you find the local extrema of a function? How many local extrema can a cubic function have? How do I find the maximum and minimum values of the function #f(x) = x - 2 sin (x)# on the... If #f(x)=(x^2+36)/(2x), 1 <=x<=12#, at what point is f(x) at a minimum? How do you find the maximum of #f(x) = 2sin(x^2)#? How do you find a local minimum of a graph using the first derivative? See all questions in Identifying Turning Points (Local Extrema) for a Function Impact of this question 1520 views around the world You can reuse this answer Creative Commons License