Is #f(x)=(x-2)e^x # increasing or decreasing at #x=-2 #?

1 Answer
Oct 12, 2016

decreasing.

Explanation:

You need to remember these rules:

  • If #f'(x)>0# for every #x# on some interval I, then #f(x)# is increasing on
    the interval.
  • If #f((x)<0# for every #x# on some interval I, then #f(x)# is
    decreasing on the interval.
  • If #f((x)=0# for every #x# on some interval I, then #f(x)# is constant on the interval

So #f(x)=(x-2)e^x#; to differentiate we need to use the product rule:
#d/dx(uv)=u(dv)/dx+(du)/dxv#

Applying this rule gives us:

# f'(x)=(x-2)(e^x)+(1)(e^x) #
# :. f'(x)=xe^x-2e^x+e^x #
# :. f'(x)=xe^x-e^x#
# :. f'(x)=(x-1)e^x #

When # x=-2=>f'(-2)=(-2-1)e^-2 #
When # :. f'(-2)=-3e^-2 #

Remember that # A^x > 0 AA x in RR => e^-2 > 0#

Hence, # f'(-2)=-3e^-2 < 0 # so we can conclude that when #x=-2# then #f(x)# is decreasing .

A graph always helps to visualise and confirm these solutions:
graph{(x-2)e^x [-10.114, 7.666, -4.41, 4.48]}