What is the derivative of f(x)=-x^3+4x^2 at (-1,5) using the limit definition?

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To find the derivative of the function f(x) = -x^3 + 4x^2 at the point (-1, 5) using the limit definition, the correct approach involves evaluating f(-1) first, which is 5. The limit definition for the derivative is f'(-1) = lim (x -> -1) [(f(x) - f(-1)) / (x + 1)]. After substituting and simplifying, the limit leads to the expression -3x^2 + 8x. Evaluating this at x = -1 gives a final derivative value of -11. The discussion highlights the importance of correctly applying the limit definition and algebraic manipulation to arrive at the solution.
  • #61
sutupidmath said:
Well, with shortcut here i meant to use the other alternative of the deff. of the derivative,

\lim_{x\rightarrow -1}\frac{f(x)-f(-1)}{x+1} it takes you much faster to the answer.

I'd prefer not to resort to synthetic division.
 
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  • #62
yourdadonapogostick said:
I'd prefer not to resort to synthetic division.

thats what i did at first...
and i got the -x^3+4x^2-5 to go into (x+1)(-x^2 +5x -5)/(x+1) and so the (x+1) canceled...
but when i told everyone that i got -x^2 +5x -5 for my answer they said that there was no x+1 in the numerator and i had to go through the other equation...
and i just realized if i would have plugged in the -1 for all the x's i would have gotten -11 baack in like post number 15 XD
 
  • #63
sutupidmath said:
Well, you will learn this stuff, don't worry about that. I didn't know you were still in high school...lol...

is this for like mostly college people or something?
 
  • #64
Precal_Chris said:
is this for like mostly college people or something?


Well, one usually learns this stuff in calc 1. So, i wrongly guessed...lol..
 
  • #65
Precal_Chris said:
is this for like mostly college people or something?

I learned calculus in high school and I never went to college. I did join the military as a Naval Nuclear Mechanic, though.
 
  • #66
oh yeah she said this last six weeks we are going to be doing a lot of calc stuff
 

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