Chapter 1, Problem 1CRE

To determine

The expression $\left(4,10\right)$ as a set $\left\{x:\left|x–a\right|<c\right\}$ for suitable value of $a$ and $c$

Answer to Problem 1CRE

Solution:

To express Interval $\left(4,10\right)$ as a set $\left\{x:\left|x–a\right|<c\right\}$, the value of $a\text{ = 7}$ and $c\text{ = 3}$

Explanation of Solution

The absolute value, also defined as modulus |x|, of any real number x, is referred to as the non-negative value of x, without considering the sign.

Given:

A set as $\left\{x:\left|x–a\right|<c\right\}$

Also, an interval $\left(4,10\right)$

Formula used:

For the given inequality $\text{|}x|<a$, the solution is given as $–a<x<a$.

Calculation:

Given inequality is $\left|x–a\right|<c$

Thus, the solution for above inequality is given as

$\text{- }c<x–a<c$

$\text{- }c\text{ + }a<x<c\text{ + }a$

Also, the interval is $\left(4,10\right)$, so we get $\text{4 }<x<\text{ 10}$

On substituting the values, we get

$c\text{ + }a\text{ = 10}$

$a\text{ - }c\text{ = 4}$

Solving the above two equations, we get

$a\text{ = 7}$ And $c\text{ = 3}$

Conclusion:

To express $\left(4,10\right)$ as a set $\left\{x:\left|x–a\right|<c\right\}$, the value of $a\text{ = 7}$ and $c\text{ = 3}$