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How do you find the critical numbers of #f(x) = (4 x - 6)e^{-6 x} #?

1 Answer

May 5, 2017

The only critical number is #5/3#

Explanation:

A critical number for #f# is a number, #c#, in the domain of #f# with #f'(c) = 0# or f'(c)# does not exist.

For #f(x) = (4x-6)e^(-6x)# the domain is #(-oo,oo)#.

#f'(x) = 4e^(-6x) + (4x-6)(-6e^(-6x))#

# = 4e^(-6x)-24xe^(-6x)+36e^(-6x)#

# = -24xe^(-6x)+40e^(-6x)#

# = -8e^(-6x)(3x-5)#

#f'(x)# is defined for all real #x# and #f'(x) = 0# at #x=5/3#

The only critical number is #5/3#

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