given the function $f(x)=-2x^{2}$. find the difference quotient $\frac{f(x + h)-f(x)}{h}$. find $f(x)$ by…

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)=-2x^{2}$. So $f(x + h)=-2(x + h)^{2}=-2(x^{2}+2xh+h^{2})=-2x^{2}-4xh - 2h^{2}$.

Step2: Calculate the difference - quotient

[ \begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{(-2x^{2}-4xh - 2h^{2})-(-2x^{2})}{h}\ &=\frac{-2x^{2}-4xh - 2h^{2}+2x^{2}}{h}\ &=\frac{-4xh - 2h^{2}}{h}\ &=-4x - 2h \end{align*} ]

Step3: Find the derivative $f^{\prime}(x)$

[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}\ &=\lim_{h\rightarrow0}(-4x - 2h) \end{align*} ] As $h\rightarrow0$, we substitute $h = 0$ into $-4x - 2h$. So $f^{\prime}(x)=-4x$.

Answer:

Difference quotient: $-4x - 2h$ $f^{\prime}(x)$: $-4x$