Q. 2 - Calculus

Question

Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading. 

(a) A function f with values given in the following table but whose limit as x → 2 is not equal to 5:

(b) An inequality involving absolute values whose solution set is $(2.75,3)\cup (3,3.25)$

(c) An inequality involving absolute values whose solution set is $(-0.01,0)\cup (0,0.01)$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{Part}\left(\mathrm{a}\right) f\left(x\right)=\frac{3x^2-7x+2}{x-2} \\ \mathrm{Part}\left(\mathrm{b}\right) 0<|4x-12|<1 \\ \mathrm{Part}\left(\mathrm{c}\right) 0<\left|100x\right|<1 \end{gathered}$$

Part(a) Step 1. Given Information

The given table is 

Part(a) Step 2. Explanation

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

Part(b) Step 1. Explanation

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

$$\begin{gathered} c=3, \\ \delta =0.25 \end{gathered}$$

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

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

Part(c) Step 1. Explanation

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

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

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

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