given the function $f(x)=-3x^{2}$. find the difference quotient $\frac{f(x + h)-f(x)}{h}$. find $f(x)$ by…

given the function $f(x)=-3x^{2}$. 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 $f(1)$. graph the function $f(x)=-3x^{2}$ and draw the tangent lines to the graph at points whose $x$ - coordinates are -2, 0, and 1.
Answer
Explanation:
Step1: Find the difference - quotient
First, find (f(x + h)): [ \begin{align*} f(x + h)&=-3(x + h)^2\ &=-3(x^{2}+2xh + h^{2})\ &=-3x^{2}-6xh-3h^{2} \end{align*} ] Then, calculate (\frac{f(x + h)-f(x)}{h}): [ \begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{-3x^{2}-6xh - 3h^{2}-(-3x^{2})}{h}\ &=\frac{-3x^{2}-6xh-3h^{2}+3x^{2}}{h}\ &=\frac{-6xh-3h^{2}}{h}\ &=-6x - 3h \end{align*} ]
Step2: Find the derivative (f^{\prime}(x))
[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}\ &=\lim_{h\rightarrow0}(-6x - 3h)\ &=-6x \end{align*} ]
Step3: Find (f^{\prime}(-2))
Substitute (x=-2) into (f^{\prime}(x)): [ f^{\prime}(-2)=-6\times(-2)=12 ]
Step4: Find (f^{\prime}(0))
Substitute (x = 0) into (f^{\prime}(x)): [ f^{\prime}(0)=-6\times0 = 0 ]
Step5: Find (f^{\prime}(1))
Substitute (x = 1) into (f^{\prime}(x)): [ f^{\prime}(1)=-6\times1=-6 ]
Answer:
- Difference - quotient: (-6x - 3h)
- (f^{\prime}(x)): (-6x)
- (f^{\prime}(-2)): (12)
- (f^{\prime}(0)): (0)
- (f^{\prime}(1)): (-6)