Question

How do you find f'(x) using the limit definition given `3x^2-5x+2`?

Answer 1

`f'(x)=6x-5`

Explanation:

The limit definition of a derivative states that

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

Substituting `f(x)=3x^2-5x+2` into `f'(x)`,

`f'(x)=lim_(hrarr0)((3(x+h)^2-5(x+h)+2)-(3x^2-5x+2))/h`

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

`f'(x)=lim_(hrarr0)(3x^2+6xh+3h^2-5x-5h+2-3x^2+5x-2)/h`

`f'(x)=lim_(hrarr0)(color(red)cancelcolor(black)(3x^2)+6xh+3h^2color(blue)cancelcolor(black)(-5x)-5hcolor(darkorange)cancelcolor(black)(+2)color(red)cancelcolor(black)(-3x^2)color(blue)cancelcolor(black)(+5x)color(darkorange)cancelcolor(black)(-2))/h`

`f'(x)=lim_(hrarr0)(6xh+3h^2-5h)/h`

Factor out `h` from the numerator.

`f'(x)=lim_(hrarr0)(h(6x+3h-5))/h`

`f'(x)=lim_(hrarr0)6x+3h-5`

Plugging in `h=0`,

`f'(x)=6x+3(0)-5`

`color(green)(|bar(ul(color(white)(a/a)color(black)(f'(x)=6x-5)color(white)(a/a)|)))`