Question

question 6
given $f(x)=4-4x^2$, find $f(x)$ using the limit definition of the derivative.
$f(x)=$
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Explanation:

Step1: Recall limit definition of derivative

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Step2: Substitute $f(x+h)$ and $f(x)$

$$f'(x) = \lim_{h \to 0} \frac{[4 - 4(x+h)^2] - (4 - 4x^2)}{h}$$

Step3: Expand $(x+h)^2$

$$f'(x) = \lim_{h \to 0} \frac{4 - 4(x^2 + 2xh + h^2) - 4 + 4x^2}{h}$$

Step4: Simplify numerator terms

$$f'(x) = \lim_{h \to 0} \frac{4 - 4x^2 - 8xh - 4h^2 - 4 + 4x^2}{h} = \lim_{h \to 0} \frac{-8xh - 4h^2}{h}$$

Step5: Factor and cancel $h$

$$f'(x) = \lim_{h \to 0} \frac{h(-8x - 4h)}{h} = \lim_{h \to 0} (-8x - 4h)$$

Step6: Evaluate limit as $h\to0$

$$f'(x) = -8x - 4(0)$$

Answer:

$-8x$