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

given the function $f(x)=-2x^{2}$. find the difference quotient $\frac{f(x + h)-f(x)}{h}$. find $f(x)$ by determining $lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$. use the previous answer to find the values indicated below. the values should match the slopes of the tangent lines you will draw on the graph at the bottom. find $f(-2)$. find $f(0)$. find $f(1)$.

given the function $f(x)=-2x^{2}$. find the difference quotient $\frac{f(x + h)-f(x)}{h}$. find $f(x)$ by determining $lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$. use the previous answer to find the values indicated below. the values should match the slopes of the tangent lines you will draw on the graph at the bottom. find $f(-2)$. find $f(0)$. find $f(1)$.

Answer

Explanation:

Step1: Find the difference quotient

First, find $f(x + h)$: [ \begin{align*} f(x+h)&=-2(x + h)^2\ &=-2(x^{2}+2xh+h^{2})\ &=-2x^{2}-4xh - 2h^{2} \end{align*} ] Then, calculate $\frac{f(x + h)-f(x)}{h}$: [ \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*} ]

Step2: Find the derivative $f'(x)$

[ \begin{align*} f'(x)&=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}\ &=\lim_{h\rightarrow0}(-4x - 2h)\ &=-4x \end{align*} ]

Step3: Find $f'(-2)$

Substitute $x=-2$ into $f'(x)$: [ f'(-2)=-4\times(-2)=8 ]

Step4: Find $f'(0)$

Substitute $x = 0$ into $f'(x)$: [ f'(0)=-4\times0 = 0 ]

Step5: Find $f'(1)$

Substitute $x = 1$ into $f'(x)$: [ f'(1)=-4\times1=-4 ]

Answer:

Difference quotient: $-4x - 2h$ $f'(x)$: $-4x$ $f'(-2)$: $8$ $f'(0)$: $0$ $f'(1)$: $-4$