91Ó°ÊÓ

The given table is

Consider a function, $f\left(x\right)=\frac{3x^2-7x+2}{x-2}$

Substitute $x=1.9$in the function to calculate the value.

$$\begin{gathered} f(1.9)=\frac{3{(1.9)}^2-7(1.9)+2}{1.9-2} \\ =\frac{10.83-13.3+2}{-0.1} \\ =\frac{-0.47}{-0.1} \\ =4.7 \end{gathered}$$

Make the table for different values of x for the function.

As the value of x approaches 2, the denominator of the function approaches 0. Thus, the function is not defined as the x approaches 2.

Hence, the function that satisfies the points on the table is$f\left(x\right)=\frac{3x^2-7x+2}{x-2}$

We will write the solution set $(2.75,3)∪(3,3.25)\mathrm{in}\mathrm{the}\mathrm{form}\mathit{}(\mathrm{c}-\mathrm{δ},\mathrm{c})∪(\mathrm{c},\mathrm{c}+\mathrm{δ})$to obtain the values of $c\mathrm{and}\mathrm{δ}$,

role="math" localid="1648055197331" $$\begin{gathered} c=3, \\ \mathrm{δ}=0.25 \end{gathered}$$

Substitute the values of role="math" localid="1648055174467" $c\mathrm{and}\mathrm{δ}\mathrm{in}\mathrm{the}\mathrm{inequality}0<|\mathrm{x}-\mathrm{c}|<\mathrm{δ}$

$$\begin{gathered} 0<|x-3|<0.25 \\ 0<|x-3|<\frac{1}{4} \\ 0<|4x-12|<1 \end{gathered}$$

We will write the solution set $(-0.01,0)∪(0,0.01)$in the form of $(\mathrm{c}-\mathrm{δ},\mathrm{c})∪(\mathrm{c},\mathrm{c}+\mathrm{δ})$to obtain the values of $c\mathrm{and}\mathrm{δ}$,

$$\begin{gathered} c=0, \\ \mathrm{δ}=0.01 \end{gathered}$$

Substitute the values of $c\mathrm{and}\mathrm{δ}\mathrm{in}\mathrm{the}\mathrm{inequality}0<|x-c|<\mathrm{δ}$

$$\begin{gathered} 0<|x-0|<0.01 \\ 0<x<\frac{1}{100} \\ 0<\left|100x\right|<1 \end{gathered}$$