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$$f(x) = 2x - \frac{1}{4} x^2$$

How could I know calculate with limits the derivative of this function when $x_0 = -4$? I started already like this:

$$f'(-4) = \lim_{x\to-4} (2x-\frac{1}{4}x^2-4)/???$$

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Do you mean $2x-\frac{1}{4}x^2$? It reads like $2x-\frac{1}{4x^2}$ –  Thomas Andrews Oct 19 '11 at 20:25
    
you forgot to divide by $x-a$... And then all you need to do is...divide... –  N. S. Oct 19 '11 at 20:27
    
Not quite, you also have $f(-4)= -4$, so when you subtract $f(-4)$, you need to add $4$. –  Thomas Andrews Oct 19 '11 at 20:30
    
@Thomas: Are you sure? $2(-4) - (1/4)(-4)^2 = -8 - (1/4)(16) = -8-4=-12$. –  Arturo Magidin Oct 19 '11 at 20:36
    
Whoops, you're right @ArturoMagidin, but it still makes his $-4$ wrong. –  Thomas Andrews Oct 19 '11 at 20:38

1 Answer 1

up vote 4 down vote accepted

By definition: $$f'(x_0) = \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}.$$ Here, $$f(x) = 2x - \frac{1}{4}x^2,$$ and so $$f(-4) = 2(-4) - \frac{1}{4}(-4)^2 = -8 - \frac{1}{4}(16) = -8-4 =-12.$$ Plugging everything into the general formula, $$f'(-4) = \lim_{x\to -4}\frac{f(x)-f(-4)}{x-(-4)} = \lim_{x\to -4}\frac{(2x-\frac{1}{4}x^2) - (-12)}{x+4} = \lim_{x\to -4}\frac{2x - \frac{1}{4}x^2 + 12}{x+4}.$$ Can you take it from here?

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