Question

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

Step1: Recall limit - definition of derivative

The limit - definition of the derivative is $f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=6 - 4x^{2}$, then $f(x + h)=6-4(x + h)^{2}$.

Step2: Expand $f(x + h)$

Expand $6-4(x + h)^{2}=6-4(x^{2}+2xh+h^{2})=6-4x^{2}-8xh - 4h^{2}$.

Step3: Substitute $f(x + h)$ and $f(x)$ into the derivative formula

$\frac{f(x + h)-f(x)}{h}=\frac{(6-4x^{2}-8xh - 4h^{2})-(6 - 4x^{2})}{h}=\frac{6-4x^{2}-8xh - 4h^{2}-6 + 4x^{2}}{h}=\frac{-8xh-4h^{2}}{h}$.

Step4: Simplify the expression

$\frac{-8xh-4h^{2}}{h}=\frac{h(-8x - 4h)}{h}=-8x-4h$ for $h
eq0$.

Step5: Find the limit as $h

ightarrow0$
$f^{\prime}(x)=\lim_{h
ightarrow0}(-8x - 4h)$. As $h
ightarrow0$, we have $f^{\prime}(x)=-8x$.

Answer:

$-8x$