Question

How do you find f'(x) using the limit definition given `f(x) = 3/(x-2)`?

Answer 1

`f'(x)=-3/(x-2)^2`

Explanation:

The limit definition of a derivative states that

`f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h`

Substituting `f(x)=4+9x-x^2` into `f'(x)`,

`f'(x)=lim_(hrarr0)(3/(x+h-2)-3/(x-2))/h`

From this point on, you want to expand and simplify.

`f'(x)=lim_(hrarr0)(3(x-2)-3(x+h-2))/((x+h-2)(x-2))*1/h`

`f'(x)=lim_(hrarr0)(3x-6-3x-3h+6)/((x+h-2)(x-2))*1/h`

`f'(x)=lim_(hrarr0)(color(red)cancelcolor(black)(3x)color(blue)cancelcolor(black)(-6)color(red)cancelcolor(black)(-3x)-3hcolor(blue)cancelcolor(black)(+6))/((x+h-2)(x-2))*1/h`

`f'(x)=lim_(hrarr0)(-3h)/((x+h-2)(x-2))*1/h`

`f'(x)=lim_(hrarr0)(-3color(darkorange)cancelcolor(black)h)/((x+h-2)(x-2))*1/color(darkorange)cancelcolor(black)h`

`f'(x)=lim_(hrarr0)-3/((x+h-2)(x-2))`

Plugging in `h=0`,

`f'(x)=-3/((x+0-2)(x-2))`

`f'(x)=color(green)(|bar(ul(color(white)(a/a)color(black)(-3/(x-2)^2)color(white)(a/a)|)))`