Determine if Continuous f(x)=1/(x^2-25)

$f (x) = \frac{1}{x^{2} - 25}$

Step 1

Set the denominator in $\frac{1}{x^{2} - 25}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 25 = 0$

Step 2

Solve for $x$ .

Add $25$ to both sides of the equation.

$x^{2} = 25$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 5$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 5$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 5$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 5, - 5$

Step 3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 5) \cup (- 5, 5) \cup (5, \infty)$

Set-Builder Notation:

${x | x \neq 5, - 5}$

Step 4

Since the domain is not all real numbers, $f (x) = \frac{1}{x^{2} - 25}$ is not continuous over all real numbers.

Not continuous

Step 5