Answer
a) The average rate of change of y=$m_{sec}$=$\frac{-3}{4}$
b) Instantaneous rate of change of y at specific value of $x_{0} $=$m_{tan}$=$-2$
c) Instantaneous rate of change of y at specific value of $x_{0} $=$m_{tan}$=$-\frac{2}{x_{0}^3}$
d) points (1,1) and (2, \frac{1}{4})
Draw graph using these two points
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^{2}}$
let y=f(x)=$\frac{1}{x^2}$
since we have $ x_{0}$=1 and $ x_{1}$=2 then we get
f($ x_{0}$)=f(2)=$\frac{1}{2^{2}}$= $\frac{1}{4}$
and f($ x_{1}$)=f(1)=$\frac{1}{1}$=1
Now,
The average rate of change of y=$m_{sec}$=$\frac{f( x_{1})-f(x_{0})}{x_{1}-x_{0}}$
putting values,
The average rate of change of y=$m_{sec}$=$\frac{\frac{1}{4}-1}{2-1}$
=$\frac{\frac{-3}{4}}{1}$
a) The average rate of change of y=$m_{sec}$=$\frac{-3}{4}$
b)
As we know that the specific value of $x_{0}$ =1, then
$f( x_{1})=\frac{1}{x_{1}^2}$ and $f(x_{0})=1$
Instantaneous rate of change of y at specific value of $x_{0} $=$m_{tan}$
=$\lim\limits_{x_{1} \to 1}$$\frac{f( x_{1})-f(x_{0})}{x_{1}-x_{0}}$
=$\lim\limits_{x_{1} \to 1}$$\frac{(\frac{1}{x_{1}^2}-1)}{x_{1}-1}$
=$\lim\limits_{x_{1} \to 1}$$ \frac{1-x_{1}^2}{x_{1}^2({x_{1}-1})}$
=$\lim\limits_{x_{1} \to 1}$$ -(\frac{-1+x_{1}^2}{x_{1}^2({x_{1}-1})})$
=$\lim\limits_{x_{1} \to 1}$$ -(\frac{x_{1}^2-1}{x_{1}^2({x_{1}-1})})$
=$\lim\limits_{x_{1} \to 1}$$ -(\frac{(x_{1}-1)({x_{1}+1})}{x_{1}^2({x_{1}-1})})$
=$\lim\limits_{x_{1} \to 1}$$ -(\frac{{x_{1}+1}}{x_{1}^2})$
Applying limit
=$-(\frac{(1)+1}{(1)^2})$
=$-(\frac{2}{1})$
=$-2$
Instantaneous rate of change of y at specific value of $x_{0} $=$m_{tan}$=$-2$
c) The arbitrary value of $x_{0}$ =$x_{0}$ then
$f( x_{1})=\frac{1}{x_{1}^2}$ and $f(x_{0})=\frac{1}{{x_{0}^2}}$
Instantaneous rate of change of y at arbitrary value of $x_{0} $=$m_{tan}$
=$\lim\limits_{x_{1} \to x_{0}}$$\frac{f( x_{1})-f(x_{0})}{x_{1}-x_{0}}$
=$\lim\limits_{x_{1} \to x_{0}}$$\frac{( \frac{1}{x_{1}^2}-\frac{1}{ x_{0}^2})}{x_{1}- x_{0}}$
=$\lim\limits_{x_{1} \to x_{0}}$$\frac{( \frac{ x_{0}^2-x_{1}^2}{ x_{1}^2x_{0}^2})}{x_{1}- x_{0}}$
=$\lim\limits_{x_{1} \to x_{0}}$$\frac{( \frac{ -x_{1}^2+x_{0}^2}{ x_{1}^2x_{0}^2})}{x_{1}- x_{0}}$
=$\lim\limits_{x_{1} \to x_{0}}$$\frac{-( \frac{ x_{1}^2-x_{0}^2}{ x_{1}^2x_{0}^2})}{x_{1}- x_{0}}$
=$\lim\limits_{x_{1} \to x_{0}}$$ -(\frac{(x_{1}-x_{0})({x_{1}+x_{0}})}{x_{1}^2x_{0}^2({x_{1}-x_{0}})})$
=$\lim\limits_{x_{1} \to x_{0}}$$-\frac{1(x_{1}+x_{0})}{ x_{1}^2x_{0}^2(1)}$
=$\lim\limits_{x_{1} \to x_{0}}$$-\frac{(x_{1}+x_{0})}{ x_{1}^2x_{0}^2}$
Applying limit
=$-\frac{((x_{0})+x_{0})}{( x_{0}^2) x_{0}^2}$
Instantaneous rate of change of y at specific value of $x_{0} $=$m_{tan}$=$-\frac{2x_{0}}{x_{0}^4}$
Instantaneous rate of change of y at specific value of $x_{0} $=$m_{tan}$=$-\frac{2}{x_{0}^3}$
take value of ${x_{0}}$ as 1,2,3,4,5,......
d)
put value of ${x_{0}}$= 1 and ${x_{1}}$=2 in y=$\frac{1}{x^2}$
at $x_{0}$=1
f($ x_{0}$)=f(2)=$\frac{1}{1}=1$
(2, 1)
at $x_{1}$=2
f($ x_{1}$)=f(3)=$(\frac{1}{2})^2$
(2, $\frac{1}{4}$)
graph : draw graph using the given points (1,1) and (2,$\frac{1}{4}$)
