What is the instantaneous rate of change of #f(x)=(x^2-3x)e^(x-1) # at #x=0 #?

1 Answer
Jim G.
Aug 1, 2016

#-3/e#

Explanation:

The #color(blue)"instantaneous rate of change"# is the value of f'(0)

differentiate f(x) using the #color(blue)"product rule"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(f(x)=g(x)h(x)rArrf'(x)=g(x)h'(x)+h(x)g'(x))color(white)(a/a)|)))#

here #g(x)=(x^2-3x)rArrg'(x)=2x-3#

and #h(x)=e^(x-1)rArrh'(x)=e^(x-1).d/dx(x-1)=e^(x-1)#
#color(blue)"----------------------------------------------------------------------"#

#rArrf'(x)=(x^2-3x)e^(x-1)+e^(x-1).(2x-3)#

#=e^(x-1)(x^2-3x+2x-3)=e^(x-1)(x^2-x-3)#
#color(blue)"-------------------------------------------------------------------"#

#rArrf'(0)=e^(0-1)(0-0-3)=-3e^-1=-3/e#

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