a. use the definition mtan = lim(x→a) (f(x) - f(a))/(x - a) to find the slope of the line tangent to the…

Answer:

a. $-\frac{1}{3}$ b. $y = -\frac{1}{3}x - 2$ c. To plot the graph of $f(x)=\frac{3}{x}$ and the tangent line $y = -\frac{1}{3}x - 2$, for $f(x)=\frac{3}{x}$, we can find some points such as when $x = - 6,y=-\frac{1}{2}$; when $x=-1,y = - 3$; when $x = 1,y=3$; when $x = 6,y=\frac{1}{2}$. For the tangent - line $y=-\frac{1}{3}x - 2$, we know the $y$-intercept is $-2$ and the slope is $-\frac{1}{3}$. We can use graphing software (like Desmos) or a graphing calculator to plot both the curve $y = \frac{3}{x}$ and the line $y=-\frac{1}{3}x - 2$.

Explanation:

Step1: Find the slope of the tangent line

Given $f(x)=\frac{3}{x}$ and $a=-3,f(a)=f(-3)=-1$. [ \begin{align*} m_{tan}&=\lim_{x\rightarrow - 3}\frac{f(x)-f(-3)}{x + 3}\ &=\lim_{x\rightarrow - 3}\frac{\frac{3}{x}+1}{x + 3}\ &=\lim_{x\rightarrow - 3}\frac{\frac{3 + x}{x}}{x + 3}\ &=\lim_{x\rightarrow - 3}\frac{1}{x}\ &=-\frac{1}{3} \end{align*} ]

Step2: Find the equation of the tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$. Here $x_1=-3,y_1=-1,m = -\frac{1}{3}$. [ \begin{align*} y+1&=-\frac{1}{3}(x + 3)\ y+1&=-\frac{1}{3}x-1\ y&=-\frac{1}{3}x - 2 \end{align*} ]

Step3: Plotting instructions

As described above, find points on the function $f(x)=\frac{3}{x}$ and use the slope - intercept form of the tangent line $y=-\frac{1}{3}x - 2$ to plot both on a graph.