Question #21f32

1 Answer
Dharma R.
Sep 17, 2015

#f(x+h)#= #(2(x+h)-1)/(x+h+3)#
#f(x)# = #(2x-1)/(x+3#
#f(x+h)-f(x)#= #(2(x+h)-1)/(x+h+3)#-#(2x-1)/(x+3#
numerator is
#2x^2+2(h+3)x+6h-x-3-(2x^2+2hx+6x-x-h-3)#
=#7h#
demonimator
#(x+h+3)*(x+3)#
#f'(x)#=#Lt_h->0##(f(x+h)-f(x))/h#
so #Lt_h->_0# #(7h)/((x+h+3)h(x+3)#
cancelling #h# and applying #Lt#gives
#7/(x+3)^2#
#=>f'(x)=7/(x+3)^2#
#f'(3)#=#7/36#

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