Question

How to find f'(0) ?

Answer 1

`f'(0)=-6`

Explanation:

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

and using quotient rule,

`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

`f'(0) = -6`

Explanation:

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

Using the quotient rule:

`(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

`f'(0)=-6.`

Explanation:

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.!