What are the critical points of #g(x)=e^(x^2)(x-10)#?

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Answer 1 · 2017-08-21T01:26:21

#x= (2+-sqrt23)/2#

Domain of #g# is #RR#

#g'(x) = 2xe^(x^2)(x-10) + e^(x^2)#

# = e^(x^2)(2x^2-10x+1)#

#g'(x)# is never undefined and #g'(x) = 0# at

#2x^2-10x+1=0#

#x = (-(-10)+-sqrt((-10)^2-4(2)(1)))/(2(2))#

# = (10+-sqrt92)/4 = (10+-2sqrt23)/4 = (2+-sqrt23)/2#