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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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    @Thomas: Oh, absolutely.2011-10-19

1 Answers 1

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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?