What is the instantaneous rate of change of f(x)=(x^2-3x)e^(x-1) at x=0 ?
1 Answer
Aug 1, 2016
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
