How to find f'(0) ?

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Answer 1 · 2017-03-18T15:05:47

$f'(0)=-6$

As $f(x)=(x^2+3)/(2x-1)$

$f'(x)=(2x xx(2x-1)-2xx(x^2+3))/(2x-1)^2$

= $(4x^2-2x-2x^2-6)/(2x-1)^2$

= $(2x^2-2x-6)/(2x-1)^2$

= $(2(x^2-x-3))/(2x-1)^2$

and $f'(0)=(2(0^2-0-3))/(2xx0-1)^2=-6/1=-6$

Answer 2 · 2017-03-18T15:11:17

$f'(0) = -6$

$f(x) = (x^2+3)/(2x-1)$

$(df)/dx = ((2x-1) d/dx(x^2+3) - (x^2+3) d/dx(2x-1))/(2x-1)^2$

$(df)/dx = (2x (2x-1) - 2(x^2+3))/(2x-1)^2$

$(df)/dx = (4x^2 -2x - 2x^2-6)/(2x-1)^2$

$(df)/dx = (2x^2 -2x -6)/(2x-1)^2$

and for $x=0$

$[(df)/dx]_(x=0) = -6$

Answer 3 · 2017-03-18T15:21:56

$f'(0)=-6.$

Recall that, $f'(a)=lim_(x to a) {(f(x)-f(a))/(x-a)}.$

$:. f'(0)=lim_(xto0){(f(x)-f(0))/(x-0)}=lim_(xto0){(f(x)-f(0))/x}.$

$f(x)=(x^2+3)/(2x-1) rArr f(0)=(0+3)/(0-1)=-3.$

$:. f'(0)=lim_(xto0)[1/x{(x^2+3)/(2x-1)-(-3)}],$

$=lim_(xto0)1/x[{x^2+3+3(2x-1)}/(2x-1)],$

$=lim_(xto0)1/x{(x^2+6x)/(2x-1)},$

$=lim_(xto0){x(x+6)}/{x(2x-1)},$

$=lim_(xto0){(x+6)/(2x-1)}.$

$rArr f'(0)=(0+6)/(0-1)=-6.$

Enjoy Maths.!