Question

a function (f(x)) is said to have a removable discontinuity at (x = a) if: 1. (f) is either not defined or not continuous at (x = a). 2. (f(a)) could either be defined or re - defined so that the new function is continuous at (x = a). let (f(x)=\begin{cases}\frac{9}{x}+\frac{-8x + 27}{x(x - 3)},& \text{if }x
eq0,3\\2,&\text{if }x = 0end{cases}). show that (f(x)) has a removable discontinuity at (x = 0) and determine what value for (f(0)) would make (f(x)) continuous at (x = 0). must redefine (f(0)=square). hint: try combining the fractions and simplifying. the discontinuity at (x = 3) is not a removable discontinuity, just in case you were wondering.

Explanation:

Step1: Simplify the rational - part of the function

For \(x
eq0,3\), we have \(f(x)=\frac{9}{x}+\frac{- 8x + 27}{x(x - 3)}\). First, find a common denominator \(x(x - 3)\). Then \(\frac{9}{x}\cdot\frac{x - 3}{x - 3}+\frac{-8x + 27}{x(x - 3)}=\frac{9(x - 3)+(-8x + 27)}{x(x - 3)}=\frac{9x-27-8x + 27}{x(x - 3)}=\frac{x}{x(x - 3)}=\frac{1}{x - 3}\).

Step2: Find the limit as \(x\to0\)

We want to find \(\lim_{x\to0}f(x)\). Since \(f(x)=\frac{1}{x - 3}\) for \(x
eq0,3\), then \(\lim_{x\to0}f(x)=\lim_{x\to0}\frac{1}{x - 3}=-\frac{1}{3}\).

Step3: Redefine \(f(0)\) for continuity

For the function \(y = f(x)\) to be continuous at \(x = 0\), we need \(f(0)=\lim_{x\to0}f(x)\). So we should re - define \(f(0)=-\frac{1}{3}\).

Answer:

\(-\frac{1}{3}\)