Question 991172
f'(x) = limit as h --> 0 of ( f(x+h) - f(x) ) / h
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we are given the function,
f(x) = (3x-1) / (x+2)
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f'(x) = limit as h --> 0 of (( 3(x+h)-1 / (x+h+2) ) - ( (3x-1) / (x+2) )) / h =
( 3x+3h-1 ) / (x+h+2) ) - ( (3x-1) / (x+2) ) / h =
( (x+2)*(3x+3h-1) - ( (x+h+2)*(3x-1) ) / (x+h+2)*(xh+2h) =
( (3x^2+3hx+5x+6h-2) - (3x^2+3hx+5x-h-2) ) / (x^2h+xh^2+4xh+2h^2+4h) =
( 7h ) / (x^2h+xh^2+4xh+2h^2+4h) =
7 / (x^2 +xh +4x +2h + 4) =
as h -->0 xh and 2h terms go to 0, then we have
7 / (x^2 +4x + 4) =
7 / (x+2)^2