f'(-4) if f(x) = x^2-x

So I'm calculating

f´(-4) = lim(h goes against 0) (f(-4+h)-f(-4))/h

f´(-4) = lim(h goes against 0) ((-4+h)^2+4+4^2-4)/h

f´(-4) = lim(h goes against 0)(4^2-4^2-8h+h^2)/h

f´(-4) = lim(h goes against 0)h(-8+h)/h

f´(-4) = lim(h goes against 0) -8+h

f´(-4) = -8

I know the answer should be -9 since the diverate of f(x) = x^2-x should be 2x - 1 (-4*2-1) but I can't make it work :(

Guest May 29, 2017

#1**+1 **

By definition, we have

[ (x + h)^2 - (x + h) - ( x^2 - x ) ] / h =

[ x^2 + 2xh + h^2 - x - h - x^2 + x ] / h =

[ 2xh + h^2 - h] / h =

[ (h)( 2x + h - 1 )] / h = divide by h on top/bottom

[ 2x + h - 1 ] let h → 0 and we have

[ 2x - 1 ] = f ' (x)

So

f '(-4) = 2(-4) - 1 = - 8 - 1 = - 9

CPhill May 29, 2017