Question

how do i start doing this problem....
use limit defintion to find the derivative at teh indicated point
f(x) = - 2x^2-x-3 at (1,-6)

Answer 1

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Answer 2

\[\lim_{h->0}\frac{[-2(1+h)^2-(1+h)-3]-[-2(1)^2-1-3]}{h}\]

Answer 3

it should maybe be -5

Answer 4

\[\lim_{h->0}\frac{-2(1+2h+h^2)-1-h-3+2+1+3}{h}\]
\[\lim_{h->0}\frac{-2-4h-2h^2-1-h-3+2+1+3}{h}\]
\[\lim_{h->0}\frac{(-2+2)-4h-2h^2+(-1+1)-h+(-3+3)}{h}\]
\[\lim_{h->0}\frac{-4h-2h^2-h}{h}\]
\[\lim_{h->0}-4-2h-1\]
\[-4 -2(0)-1 = -5\]