Question

Find the numbers at which f is continuous. At which numbers is f discontinuous?

$f\left(x\right)=\frac{2x+5}{x^2-5}$

Short answer

f is continuous at all real numbers except $x=2,x=-2$.

Step-by-step solution

Step 1. Given information 

We have been given a function $f\left(x\right)=\frac{2x+5}{x^2-5}$.

We have to find the numbers at which f is continuous and at which numbers f is discontinuous.

Step 2. Discontinuous function property

A function f is discontinuous at the number where:

• f is undefined
• the limit does not exist, i.e., its one-sided limits are not equal
• the limit exists, i.e., its one-sided limits equal, but is not equal to the function value

Step 3. Analyze the function

Analyzing the given function,

$f\left(x\right)=\frac{2x+5}{x^2-5}$

We know that it is a rational function in the form of $R\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}$whose domain is

$\left\{x\left|Q\right(x)\ne 0\right\}$

Step 4. Solve for the denominator

Solve for the $Q\left(x\right)$,

$f\left(x\right)=\frac{2x+5}{x^2-4}=\frac{2x+5}{(x+2)(x-2)}$

Thus,

$Q\left(x\right)=(x+2)(x-2)$

which implies that,

$x\ne 2;x\ne -2$

Hence domain of f is$\left\{x|x\ne 2,x\ne -2\right\}$

Step 5. Determine the continuity of the function

According to the properties of continuities, if a function is rational, then the function's graph must be continuous at every number in the domain and has a hole or vertical asymptote where the function is undefined.

This implies that at x = -2 and x = 2 f is undefined.