Answer
a) The average rate of change of y=$m_{sec}$=$\frac{-1}{6}$
b) Instantaneous rate of change of y at specific value of $x_{0} $=$m_{tan}$=$-\frac{1}{4}x_{y}$
Work Step by Step
a) find the average rate of change of y w.r.t x over intervals$ x_{0}$ and $ x_{1}$.
Given:
y=$\frac{1}{x}$
let y=f(x)=$\frac{1}{x}$
since we have $ x_{0}$=2 and $ x_{1}$=3 then we get
f($ x_{0}$)=f(2)=$\frac{1}{2}$ and f($ x_{1}$)=f(3)=$\frac{1}{3}$
Now,
The average rate of change of y=$m_{sec}$=$\frac{f( x_{1})-f(x_{0})}{x_{1}-x_{0}}$
putting values,
The average rate of change of y=$m_{sec}$=$\frac{\frac{1}{2}-\frac{1}{3}}{3-2}$
=$\frac{\frac{-1}{6}}{1}$
The average rate of change of y=$m_{sec}$=$\frac{-1}{6}$
b)
As we know that the specific value of $x_{0}$ =1, then
$f( x_{1})=\frac{1}{x_{1}}$ and $f(x_{0})=\frac{1}{2}$
Instantaneous rate of change of y at specific value of $x_{0} $=$m_{tan}$
=$\lim\limits_{x_{1} \to 2}$$\frac{f( x_{1})-f(x_{0})}{x_{1}-x_{0}}$
=$\lim\limits_{x_{1} \to 2}$$\frac{( \frac{1}{x_{1}})-\frac{1}{2})}{x_{1}-2}$
=$\lim\limits_{x_{1} \to 2}$$\frac{( \frac{2-x_{1}}{2x_{1}})}{x_{1}-2}$
=$\lim\limits_{x_{1} \to 2}$$-\frac{( \frac{-2+x_{1}}{2x_{1}})}{x_{1}-2}$
=$\lim\limits_{x_{1} \to 2}$$-\frac{( \frac{1}{2x_{1}})}{1}$
=$\lim\limits_{x_{1} \to 2}$$-\frac{1}{2x_{1}}$
Applying limit
=$-\frac{1}{2(2)}$
Instantaneous rate of change of y at specific value of $x_{0} $=$m_{tan}$=$-\frac{1}{4}$
c) The arbitrary value of $x_{0}$ =$x_{0}$ then
$f( x_{1})=\frac{1}{x_{1}}$ and $f(x_{0})=\frac{1}{{x_{0}}}$
Instantaneous rate of change of y at arbitrary value of $x_{0} $=$m_{tan}$
=$\lim\limits_{x_{1} \to x_{0}}$$\frac{f( x_{1})-f(x_{0})}{x_{1}-x_{0}}$
=$\lim\limits_{x_{1} \to x_{0}}$$\frac{( \frac{1}{x_{1}})-\frac{1}{ x_{0}})}{x_{1}- x_{0}}$
=$\lim\limits_{x_{1} \to x_{0}}$$\frac{( \frac{ x_{0}-x_{1}}{ x_{0}x_{1}})}{x_{1}- x_{0}}$
=$\lim\limits_{x_{1} \to x_{0}}$$-\frac{( \frac{- x_{0}+x_{1}}{ x_{0}x_{1}})}{x_{1}- x_{0}}$
=$\lim\limits_{x_{1} \to x_{0}}$$-\frac{( \frac{1}{ x_{0}x_{1}})}{1}$
=$\lim\limits_{x_{1} \to x_{0}}$$-\frac{1}{ x_{0}x_{1}}$
Applying limit
=$-\frac{1}{ x_{0}( x_{0})}$
Instantaneous rate of change of y at specific value of $x_{0} $=$m_{tan}$=$-\frac{1}{x_{0}^{2}}$
take value of ${x_{0}}$ as 1,2,3,4,5,......
d)
put value of ${x_{0}}$= 2 and ${x_{1}}$=3 in y=$\frac{1}{x}$
at $x_{0}$=2
f($ x_{0}$)=f(2)=$\frac{1}{2}$
(2, $\frac{1}{2}$ )
f($ x_{1}$)=f(3)=$\frac{1}{3}$
at $ x_{1}$ =3
(3, $\frac{1}{3}$)
graph :
