Question

a function (f(x)) is said to have a removable discontinuity at (x = a) if both of the following conditions hold: 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). show that (f(x)=\begin{cases}x^{2}+10x + 31&\text{if }xlt - 5\\-3&\text{if }x=-5\\-x^{2}-10x - 19&\text{if }xgt - 5end{cases}) has a removable discontinuity at (x=-5) by (a) verifying (1) in the definition above, and then (b) verifying (2) in the definition above by determining a value of (f(-5)) that would make (f) continuous at (x = - 5). (f(-5)=square) would make (f) continuous at (x=-5). now draw a graph of (f(x)). its just a couple of parabolas!

Explanation:

Step1: Recall the definition of continuity

A function $y = f(x)$ is continuous at $x = a$ if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. We need to find the left - hand limit and the right - hand limit of $f(x)$ as $x
ightarrow - 5$.

Step2: Calculate the left - hand limit

For $x\lt - 5$, $f(x)=x^{2}+10x + 31$. Then $\lim_{x
ightarrow - 5^{-}}f(x)=\lim_{x
ightarrow - 5}(x^{2}+10x + 31)$. Substitute $x=-5$ into $x^{2}+10x + 31$: $(-5)^{2}+10\times(-5)+31=25-50 + 31=6$.

Step3: Calculate the right - hand limit

For $x\gt - 5$, $f(x)=-x^{2}-10x - 19$. Then $\lim_{x
ightarrow - 5^{+}}f(x)=\lim_{x
ightarrow - 5}(-x^{2}-10x - 19)$. Substitute $x = - 5$ into $-x^{2}-10x - 19$: $-(-5)^{2}-10\times(-5)-19=-25 + 50-19=6$.

Step4: Determine the value of $f(-5)$ for continuity

Since $\lim_{x
ightarrow - 5^{-}}f(x)=\lim_{x
ightarrow - 5^{+}}f(x)=6$, to make $f(x)$ continuous at $x=-5$, we should define $f(-5)=6$.

Answer:

6