given $f(x)=3 - 2x^{2}$, find $f(x)$ using the limit definition of the derivative. $f(x)= square$ question…

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative is $f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=3 - 2x^{2}$, then $f(x + h)=3-2(x + h)^{2}=3-2(x^{2}+2xh+h^{2})=3-2x^{2}-4xh - 2h^{2}$.

Step2: Substitute into the formula

$\frac{f(x + h)-f(x)}{h}=\frac{(3-2x^{2}-4xh - 2h^{2})-(3 - 2x^{2})}{h}=\frac{3-2x^{2}-4xh - 2h^{2}-3 + 2x^{2}}{h}=\frac{-4xh-2h^{2}}{h}$.

Step3: Simplify the expression

$\frac{-4xh-2h^{2}}{h}=\frac{h(-4x - 2h)}{h}=-4x-2h$ for $h\neq0$.

Step4: Find the limit as $h\rightarrow0$

$f^{\prime}(x)=\lim_{h\rightarrow0}(-4x - 2h)$. As $h\rightarrow0$, we have $f^{\prime}(x)=-4x$.

Answer:

$-4x$