$$f(x) = 2x - \frac{1}{4} x^2$$
How could I know calculate with limits the derivative of this function when $x_0 = -4$? I started already like this:
$$f'(-4) = \lim_{x\to-4} (2x-\frac{1}{4}x^2-4)/???$$
$$f(x) = 2x - \frac{1}{4} x^2$$
How could I know calculate with limits the derivative of this function when $x_0 = -4$? I started already like this:
$$f'(-4) = \lim_{x\to-4} (2x-\frac{1}{4}x^2-4)/???$$
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?