Question

In Problem, determine whether f is continuous at c.

$f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-4}{x+2},x\ne -2\\ 4,x=-2\end{array}\right.,c=-2$

Short answer

The given polynomial function$f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-4}{x+2},x\ne -2\\ 4,x=-2\end{array}\right.,c=-2$
is not continuous

Step-by-step solution

Step.1 Given information 

Polynomial function$f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-4}{x+2},x\ne -2\\ 4,x=-2\end{array}\right.,c=-2$

c=-2

We know that

A function is called continuous function when at x= a,

$\underset{x\to a}{lim}f\left(x\right)=f\left(a\right)$

and defined for every number of domain

and$$\begin{gathered} \underset{x\to c}{f(x})=A \\ andf(c)=A \end{gathered}$$

Step2. Calculations of checking of continuity 

Assume that x=-2 is in the domain of the given function, so we can proceed to find the continuity, for continuity x is the approach to -2,

so according to question

$$\begin{gathered} f\left(x\right)=\frac{x^2-4}{x+2} \\ f\left(x\right)=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)} \\ f\left(x\right)=x-2 \\ \underset{x\to -2}{f\left(x\right)}=-2-2=-4 \end{gathered}$$

but in question f(-2)= 4

so,$\underset{x\to c}{f\left(x\right)\ne }f\left(c\right)$, the function is not continuous